Free Math Worksheets

- Which derivative must be zero at an interior local extremum? - Check whether each critical point lies in the given interval. - How can the second derivative help classify each critical point?

1. Differentiate: \(f'(x) = 3x^2 - 3\). 2. Find the critical points by solving \(3x^2 - 3 = 0\). This gives \(x = -1\) and \(x = 1\), and both lie in the given interval. 3. Use \(f''(x) = 6x\). Since \(f''(-1) = -6 < 0\), there is a local maximum at \(x = -1\). Since \(f''(1) = 6 > 0\), there is a local minimum at \(x = 1\). 4. Therefore, the function has \(2\) local extrema on the interval.

\(2\)

- Find the first and second derivatives. - Critical numbers occur where the first derivative is zero or undefined. - What does the sign of the second derivative tell you about a critical point? - Check the signs carefully when solving the quadratic equation.

1. Differentiate: \(g'(x) = -3x^2 + 9x - 6\) and \(g''(x) = -6x + 9\). 2. Solve \(g'(x) = 0\): \(-3x^2 + 9x - 6 = -3(x - 1)(x - 2) = 0\). Thus, the critical numbers are \(x = 1\) and \(x = 2\). 3. Since \(g''(1) = 3 > 0\), \(g\) has a local minimum at \(x = 1\). 4. Since \(g''(2) = -3 < 0\), \(g\) has a local maximum at \(x = 2\). 5. Evaluate the function at the critical numbers: \(g(1) = -0.5\) and \(g(2) = 0\). Therefore, the local minimum is \((1, -0.5)\), and the local maximum is \((2, 0)\).

Local minimum: \((1, -0.5)\) Local maximum: \((2, 0)\)

- What equation identifies the critical numbers? - What does the sign of the second derivative indicate? - How do you obtain the full coordinates after finding each critical number? - Differentiate carefully before applying the test.

1. Differentiate: \(f'(x) = x^2 - 4x + 3\) and \(f''(x) = 2x - 4\). 2. Solve \(f'(x) = 0\): \(x^2 - 4x + 3 = (x - 1)(x - 3) = 0\). Thus, the critical numbers are \(x = 1\) and \(x = 3\). 3. Since \(f''(1) = -2 < 0\), \(f\) has a local maximum at \(x = 1\). Since \(f''(3) = 2 > 0\), \(f\) has a local minimum at \(x = 3\). 4. Evaluate the function: \(f(1) = \frac{16}{3}\) and \(f(3) = 4\). 5. The local maximum is \((1, \frac{16}{3})\), and the local minimum is \((3, 4)\).

Local maximum: \((1, \frac{16}{3})\) Local minimum: \((3, 4)\)

- First find where the tangent line is horizontal. - How does the second derivative distinguish a local maximum from a local minimum? - Substitute each critical number into the original function to find the corresponding \(y\)-coordinate. - Check the factorization of the quadratic derivative.

1. Differentiate: \(f'(x) = 3x^2 - 3x - 6\) and \(f''(x) = 6x - 3\). 2. Solve \(f'(x) = 0\): \(3(x^2 - x - 2) = 3(x + 1)(x - 2) = 0\). Thus, \(x = -1\) and \(x = 2\). 3. Since \(f''(-1) = -9 < 0\), \(f\) has a local maximum at \(x = -1\). Since \(f''(2) = 9 > 0\), \(f\) has a local minimum at \(x = 2\). 4. Evaluate the function: \(f(-1) = 5.5\) and \(f(2) = -8\). 5. The local maximum is \((-1, 5.5)\), and the local minimum is \((2, -8)\).

Critical numbers: \(x = -1\) and \(x = 2\) Local maximum: \((-1, 5.5)\) Local minimum: \((2, -8)\)

- What condition must the first derivative satisfy at an interior local extremum? - How can the second derivative classify each critical point? - Remember to calculate the \(y\)-coordinates. - Be careful when substituting negative values into the original function.

1. Differentiate: \(f'(x) = 3x^2 - 6x - 9\) and \(f''(x) = 6x - 6\). 2. Solve \(f'(x) = 0\): \(3(x^2 - 2x - 3) = 3(x + 1)(x - 3) = 0\). Thus, \(x = -1\) and \(x = 3\). 3. Since \(f''(-1) = -12 < 0\), \(f\) has a local maximum at \(x = -1\). Since \(f''(3) = 12 > 0\), \(f\) has a local minimum at \(x = 3\). 4. Evaluate the function: \(f(-1) = 15\) and \(f(3) = -17\). 5. The local maximum is \((-1, 15)\), and the local minimum is \((3, -17)\).

Local maximum: \((-1, 15)\) Local minimum: \((3, -17)\)

- Apply the chain rule to the exponential term. - A local extremum can occur where the first derivative is zero. - Use the second derivative to determine whether the point is a maximum or minimum. - Evaluate the original function at the critical number.

1. Differentiate using the chain rule: \(f'(x)=e^{0.5x}-1\). 2. Set the first derivative equal to zero: \(e^{0.5x}=1\), so \(0.5x=0\) and \(x=0\). 3. The second derivative is \(f''(x)=0.5e^{0.5x}\). Since \(f''(0)=0.5>0\), the point is a local minimum. 4. Evaluate the function: \(f(0)=2\).

The function has a local minimum at \((0,2)\).

- What condition identifies a horizontal tangent? - How can the second derivative classify a critical point? - Substitute each critical number into the original function to obtain the full coordinates.

1. Differentiate: \(f'(x) = 3x^2 - 9x + 6\) and \(f''(x) = 6x - 9\). 2. Solve \(f'(x) = 0\): \(3(x^2 - 3x + 2) = 3(x - 1)(x - 2) = 0\). Thus, \(x = 1\) and \(x = 2\). 3. Since \(f''(1) = -3 < 0\), \(f\) has a local maximum at \(x = 1\). Since \(f''(2) = 3 > 0\), \(f\) has a local minimum at \(x = 2\). 4. Evaluate the function: \(f(1) = 4.5\) and \(f(2) = 4\). 5. Therefore, the local maximum is \((1, 4.5)\), and the local minimum is \((2, 4)\).

Local maximum: \((1, 4.5)\) Local minimum: \((2, 4)\)

- Rewrite the fraction using a negative exponent. - Check that the first derivative is zero at the given value. - Use the second derivative test. - What does a positive second derivative mean geometrically?

1. Rewrite \(f(x) = x + 4x^{-2}\), so \(f'(x) = 1 - \frac{8}{x^3}\). 2. Evaluate the first derivative: \(f'(2) = 1 - \frac{8}{8} = 0\). 3. Differentiate again: \(f''(x) = \frac{24}{x^4}\). 4. Since \(f''(2) = \frac{3}{2} > 0\), the second derivative test confirms a local minimum at \(x = 2\). 5. The point is \((2, f(2)) = (2, 3)\).

Because \(f'(2) = 0\) and \(f''(2) = \frac{3}{2} > 0\), the graph has a local minimum at \((2, 3)\).

- Set the first derivative equal to zero. - Factor out the common factor. - Use the sign of the second derivative at each critical number. - Evaluate the original function for the coordinates.

1. Differentiate: \(f'(x) = \frac{1}{2}x^3 - 2x\) and \(f''(x) = \frac{3}{2}x^2 - 2\). 2. Solve \(f'(x) = 0\): \(\frac{1}{2}x(x^2 - 4) = 0\). The critical numbers are \(x = -2\), \(x = 0\), and \(x = 2\). 3. Since \(f''(0) = -2 < 0\), \((0, 3)\) is a local maximum. 4. Since \(f''(-2) = f''(2) = 4 > 0\), \((-2, 1)\) and \((2, 1)\) are local minima.

Local maximum: \((0, 3)\) Local minima: \((-2, 1)\) and \((2, 1)\)

- Set the first derivative equal to zero. - Factor the resulting quadratic. - Use the second derivative to classify each critical point. - Substitute into the original function for the coordinates.

1. Differentiate: \(f'(x) = x^2 - x - 2 = (x + 1)(x - 2)\), and \(f''(x) = 2x - 1\). 2. The critical numbers are \(x = -1\) and \(x = 2\). 3. Since \(f''(-1) = -3 < 0\), \(f\) has a local maximum at \(x = -1\). Evaluate \(f(-1) = \frac{31}{6}\). 4. Since \(f''(2) = 3 > 0\), \(f\) has a local minimum at \(x = 2\). Evaluate \(f(2) = \frac{2}{3}\).

Local maximum: \((-1, \frac{31}{6})\) Local minimum: \((2, \frac{2}{3})\)

- Set the first derivative equal to zero. - Use the second derivative to classify the critical points. - Evaluate the original function for the coordinates. - Compare the results with the turning points on the graph.

1. Differentiate: \(f'(x) = -3x^2 + 3\) and \(f''(x) = -6x\). 2. Solve \(f'(x) = 0\): \(-3x^2 + 3 = 0\), so \(x = -1\) and \(x = 1\). 3. Since \(f''(-1) = 6 > 0\), \(f\) has a local minimum at \(x = -1\). Evaluate \(f(-1) = -1\), giving \((-1, -1)\). 4. Since \(f''(1) = -6 < 0\), \(f\) has a local maximum at \(x = 1\). Evaluate \(f(1) = 3\), giving \((1, 3)\).

Local minimum: \((-1, -1)\) Local maximum: \((1, 3)\)

- Find the zeros of \(f'\). - At each critical value, read the sign of \(f''\). - If \(f''(x_0)>0\), the graph is concave up and has a local minimum. - If \(f''(x_0)<0\), the graph is concave down and has a local maximum.

1. The zeros of \(f'\) are \(x=-2\), \(x=0\), and \(x=2\). 2. At \(x=-2\), \(f''(-2)=2>0\), so \(f\) has a local minimum. 3. At \(x=0\), \(f''(0)=-1<0\), so \(f\) has a local maximum. 4. At \(x=2\), \(f''(2)=2>0\), so \(f\) has a local minimum.

Local minima at \(x=-2\) and \(x=2\); local maximum at \(x=0\)

- First find all horizontal tangents. - Use the second derivative to distinguish maxima from minima. - The even symmetry of the function provides a useful check.

1. Differentiate: \(f'(x) = -4x^3 + 4x\) and \(f''(x) = -12x^2 + 4\). 2. Solve \(f'(x) = 0\): \(-4x(x - 1)(x + 1) = 0\), so the critical numbers are \(x = -1\), \(x = 0\), and \(x = 1\). 3. Since \(f''(0) = 4 > 0\), \((0, 1)\) is a local minimum. 4. Since \(f''(-1) = f''(1) = -8 < 0\), \((-1, 2)\) and \((1, 2)\) are local maxima.

Local maxima: \((-1, 2)\) and \((1, 2)\) Local minimum: \((0, 1)\)

- Find the zeros of the first derivative. - Read the sign of the second derivative at each critical value. - A negative second derivative gives a local maximum; a positive second derivative gives a local minimum.

1. The zeros of \(f'\) are \(x=1\) and \(x=3\). 2. At \(x=1\), \(f''(1)=-2<0\), so the second derivative test gives a local maximum. 3. At \(x=3\), \(f''(3)=2>0\), so the second derivative test gives a local minimum.

Local maximum at \(x=1\); local minimum at \(x=3\)

- The critical values are already given. - Read the sign of \(f''\) at each critical value. - Positive means local minimum; negative means local maximum.

1. At \(x=-4\), \(f''(-4)=12>0\), so \(f\) has a local minimum. 2. At \(x=0\), \(f''(0)=-4<0\), so \(f\) has a local maximum. 3. At \(x=4\), \(f''(4)=12>0\), so \(f\) has a local minimum.

Local minima at \(x=-4\) and \(x=4\); local maximum at \(x=0\)

- A horizontal tangent means \(f'(x)=0\). - Read the sign of \(f''\) at each given value. - Positive second derivative means local minimum; negative second derivative means local maximum.

1. At \(x=-4\), the graph shows \(f''(-4)<0\), so \(f\) has a local maximum. 2. At \(x=-1\), the graph shows \(f''(-1)>0\), so \(f\) has a local minimum. 3. At \(x=1\), the graph shows \(f''(1)<0\), so \(f\) has a local maximum. 4. At \(x=4\), the graph shows \(f''(4)>0\), so \(f\) has a local minimum.

Local maxima at \(x=-4\) and \(x=1\); local minima at \(x=-1\) and \(x=4\)

- The first derivative gives the ramp's slope. - A horizontal tangent has slope \(0\). - Use higher derivatives to distinguish an extremum from a stationary inflection point. - Check the sign of the first derivative on both sides.

1. The derivatives are \(f'(x)=(x-2)^2\), \(f''(x)=2x-4\), and \(f'''(x)=2\). 2. A horizontal tangent requires \(f'(x)=0\), so \(x=2\). 3. At \(x=2\), \(f'(2)=0\), \(f''(2)=0\), and \(f'''(2)=2\ne0\). Therefore, the point is a stationary inflection point, not a local extremum. 4. Since \(f'(x)=(x-2)^2\ge0\) on the interval, the ramp rises before and after \(x=2\); it is only momentarily horizontal there.

a) \(x=2\,\text{m}\) b) It is a stationary inflection point. The ramp remains increasing on both sides and is momentarily horizontal at \(x=2\).

- Look for points with horizontal tangents. - Inspect the graph's concavity near \(x=0\).

1. Horizontal tangents occur at the visible local extrema. From the graph, they occur at approximately \(x=-2.2\), \(x=0\), and \(x=2.2\). 2. Near \(x=0\), the graph is concave down. Therefore, \(f''(0)<0\).

a) \(x\approx-2.2\), \(x=0\), and \(x\approx2.2\) b) \(f''(0)<0\)

- A smooth local extremum has a horizontal tangent. - Inspect the graph's concavity at \(P\) to determine the sign of the second derivative.

1. The marked point \(P\) at \(x=-2\) is a local maximum, so the tangent is horizontal. Therefore, \(f'(-2)=0\). 2. The graph is concave down at \(x=-2\), so \(f''(-2)<0\).

\(f'(-2)=0\) and \(f''(-2)<0\)

- Horizontal tangents occur where the first derivative is \(0\). - Differentiate the given derivative to obtain the second derivative. - A positive second derivative at a critical point indicates a local minimum; a negative second derivative indicates a local maximum.

1. Set the derivative equal to \(0\): \(-x^2+4x-3=-(x-1)(x-3)=0\). Thus the critical values are \(x=1\) and \(x=3\). 2. Differentiate again: \(f''(x)=-2x+4\). 3. Since \(f''(1)=2>0\), \(f\) has a local minimum at \(x=1\). 4. Since \(f''(3)=-2<0\), \(f\) has a local maximum at \(x=3\).

The graph has horizontal tangents at \(x=1\) and \(x=3\). There is a local minimum at \(x=1\) and a local maximum at \(x=3\).

- Find where the first derivative equals zero. - Use the sign of the second derivative to classify each critical point. - Compare the y-coordinates of the extrema with the x-axis. - Use the degree and leading coefficient to determine both ends of the graph.

1. Differentiate: \(f'(x)=x^3-3x^2+2x=x(x-1)(x-2)\). Thus, the critical values are \(x=0\), \(x=1\), and \(x=2\). The second derivative is \(f''(x)=3x^2-6x+2\). Since \(f''(0)=2>0\), \((0, -4)\) is a local minimum. Since \(f''(1)=-1<0\), \((1, -\frac{15}{4})\) is a local maximum. Since \(f''(2)=2>0\), \((2, -4)\) is a local minimum. 2. Because \(f\) has even degree and a positive leading coefficient, \(f(x)\to\infty\) as \(x\to\pm\infty\). All three local-extremum y-coordinates are negative, including the local maximum \(-\frac{15}{4}\). Therefore, the graph stays below the x-axis between the two local minima, crosses once to the left of \(x=0\), and crosses once to the right of \(x=2\). Hence, \(f\) has exactly two real zeros.

1. Local minima: \((0, -4)\) and \((2, -4)\). Local maximum: \((1, -\frac{15}{4})\). 2. The function has exactly two real zeros.

- Find the first and second derivatives. - Carefully evaluate the function at each critical value. - What must happen between an extremum below the x-axis and an extremum above it? - Determine the graph's end behavior from its degree and leading coefficient.

1. Differentiate: \(f'(x)=2x^3-4x=2x(x^2-2)\). The critical values are \(x=0\) and \(x=\pm\sqrt{2}\). 2. The second derivative is \(f''(x)=6x^2-4\). Since \(f''(0)=-4<0\), \((0, 1.5)\) is a local maximum. Since \(f''(\pm\sqrt{2})=8>0\), \((-\sqrt{2}, -0.5)\) and \((\sqrt{2}, -0.5)\) are local minima. 3. Because \(f\) has even degree and a positive leading coefficient, \(f(x)\to\infty\) as \(x\to\pm\infty\). Each local minimum lies below the x-axis, while the local maximum between them lies above the x-axis. The graph therefore crosses the x-axis once on each of the four monotonic intervals determined by the critical values. Thus, \(f\) has exactly four real zeros.

The local maximum is \((0, 1.5)\). The local minima are \((-\sqrt{2}, -0.5)\) and \((\sqrt{2}, -0.5)\). The function has exactly four real zeros.

- Start with the general form of a polynomial that has the required symmetry. - Translate the local-extremum conditions into conditions on the first and second derivatives. - You may choose simple coefficient values as long as all conditions are satisfied. - How many zeros must the first derivative have to produce the stated number of extrema?

1. For part a, symmetry about the y-axis requires an even function, so use \(f(x)=ax^4+bx^2+c\). Choose \(f(x)=x^4-2x^2\). Then \(f'(x)=4x(x^2-1)\), so the critical values are \(x=-1\), \(x=0\), and \(x=1\). Because \(f''(0)=-4<0\), there is a local maximum at \(x=0\); because \(f''(\pm1)=8>0\), there are local minima at \(x=\pm1\). 2. For part b, symmetry about the origin requires an odd function, so use \(f(x)=ax^3+cx\). The condition \(f'(-2)=0\) gives \(12a+c=0\). Choosing \(a=1\) gives \(c=-12\), so \(f(x)=x^3-12x\). Since \(f''(-2)=-12<0\), the critical point at \(x=-2\) is a local maximum.

a) One possible function is \(f(x)=x^4-2x^2\). b) One possible function is \(f(x)=x^3-12x\).

- A stationary inflection point is an inflection point with a horizontal tangent. - Begin with a translated version of the basic cubic function. - Symmetry about the y-axis requires only even powers of \(x\). - Make the first derivative have exactly the required number of distinct real zeros.

1. For part a, a translated cubic power function has a stationary inflection point at its horizontal shift. Choose \(g(x)=(x-2)^3\). Then \(g'(x)=3(x-2)^2\ge0\), with equality only at \(x=2\), so \(g\) is strictly increasing. Also, \(g''(2)=0\), and the concavity changes at \(x=2\). 2. For part b, symmetry about the y-axis requires an even polynomial, so use \(g(x)=ax^4+bx^2+c\). Passing through \((0, 1)\) gives \(c=1\). Choose \(a=1\) and \(b=-4\), giving \(g(x)=x^4-4x^2+1\). Its derivative is \(g'(x)=4x(x^2-2)\), which has three distinct zeros, \(x=0\) and \(x=\pm\sqrt{2}\). These produce one local maximum and two local minima.

a) One possible function is \(g(x)=(x-2)^3\). b) One possible function is \(g(x)=x^4-4x^2+1\).

- What does the sign of the second derivative tell you about concavity? - What two conditions are used in the second derivative test for a local maximum? - Find the first and second derivatives before substituting a value. - Solve an inequality to determine where the second derivative is positive.

1. Differentiate: \(f'(x) = x^2 - 4x + 3\) and \(f''(x) = 2x - 4\). 2. For \(x > 2\), \(2x - 4 > 0\). Therefore, \(f''(x)>0\), so the graph is concave up on \((2, \infty)\). 3. At \(x=1\), \(f'(1)=1-4+3=0\), so \(x=1\) is a critical point. Also, \(f''(1)=2-4=-2<0\). By the second derivative test, \(f\) has a local maximum at \(x=1\).

a) Since \(f''(x)=2x-4>0\) for \(x>2\), the graph is concave up on \((2, \infty)\). b) Since \(f'(1)=0\) and \(f''(1)=-2<0\), \(f\) has a local maximum at \(x=1\).

- What must the first derivative equal at an interior local extremum? - Compute both derivatives at \(x=0\). - Nonzero concavity alone does not create a peak or valley. - State all hypotheses of the second derivative test.

1. Differentiate: \(f'(x)=2x+6\) and \(f''(x)=2\). 2. At \(x_0=0\), \(f''(0)=2\neq0\), so the classmate's stated condition holds. 3. However, \(f'(0)=6\neq0\). The graph does not have a horizontal tangent at \(x=0\), so it cannot have a local extremum there. 4. The second derivative test requires both \(f'(x_0)=0\) and \(f''(x_0)\neq0\).

The claim is false. Although \(f''(0)=2\neq0\), \(f'(0)=6\neq0\), so \(x=0\) is not a critical point and no local extremum occurs there.

- Recall what must be true about the slope at a differentiable interior local maximum or minimum. - Determine whether \(e^x\) can ever equal \(0\). - A sufficient condition must guarantee the stated conclusion whenever the condition holds.

1. If a differentiable function has a local extremum at an interior point \(x_0\), then \(f'(x_0)=0\). 2. For \(f(x)=e^x\), \(f'(x)=e^x\) and \(f''(x)=e^x\). Because \(e^x>0\) for every real \(x\), \(f''(x)\neq 0\) everywhere. Also, \(f'(x)=e^x\) is always positive, so \(f\) is strictly increasing and has no local extrema. 3. The condition \(f''(x_0)\neq 0\) is not sufficient by itself. The example satisfies that condition everywhere but has no local extremum. The critical-point condition \(f'(x_0)=0\) is also needed before the second derivative can classify the point.

1. \(f'(x_0)=0\). 2. \(f''(x)=e^x>0\) for all \(x\), while \(f'(x)=e^x>0\), so \(f\) is strictly increasing and has no local extrema. 3. No. The condition \(f''(x_0)\neq 0\) alone is not sufficient.

- Use the powers in each polynomial to check symmetry. - Factor each function to count its distinct real zeros. - Use the first and second derivatives to identify extrema and stationary inflection points. - A binomial identity may simplify one of the functions.

1. For \(f(x)=2x^2-x^4\), only even powers occur, so the graph is symmetric about the y-axis (A). Factoring gives \(f(x)=x^2(2-x^2)\), so its distinct zeros are \(x=0\) and \(x=\pm\sqrt{2}\) (E). Since \(f'(x)=4x(1-x^2)\) has three sign-changing zeros, the function has three local extrema, not two. It has no stationary inflection point. 2. For \(g(x)=x^3-3x\), only odd powers occur, so the graph is symmetric about the origin (B). The zeros are \(x=0\) and \(x=\pm\sqrt{3}\) (E). Since \(g'(x)=3(x^2-1)\) changes sign at \(x=-1\) and \(x=1\), the function has exactly two local extrema (C). 3. Since \(h(x)=(x-1)^3\), it has only one distinct zero and neither listed symmetry. Its derivative is \(h'(x)=3(x-1)^2\), which is zero at \(x=1\) but does not change sign. The function is increasing through a horizontal tangent, and its concavity changes there, so the graph has exactly one stationary inflection point (D).

(1): A, E (2): B, C, E (3): D

- Differentiate each function and count the real zeros of the derivative. - A stationary inflection point has a horizontal tangent and a concavity change. - Determine whether each function is increasing or decreasing through a critical point. - Rewrite the quadratic in vertex form to identify its axis of symmetry.

1. For \(f(x)=x^4+1\), \(f'(x)=4x^3\) has exactly one real zero, \(x=0\), so A applies. The function has a local minimum there, so B and C do not apply. Its symmetry axis is \(x=0\), not \(x=1\), so D does not apply. 2. Since \(g(x)=(x+1)^3\), its derivative is \(g'(x)=3(x+1)^2\), which has exactly one real zero, so A applies. The derivative does not change sign, and the graph has a horizontal tangent with a concavity change at \(x=-1\), so B applies. The function is strictly increasing and has no local extremum, so C applies. D does not apply. 3. For \(h(x)=x^2-2x+5=(x-1)^2+4\), the derivative \(h'(x)=2x-2\) has exactly one real zero, so A applies. The vertex is a local minimum, so B and C do not apply. The vertex form shows that the graph is symmetric about \(x=1\), so D applies.

(1): A (2): A, B, C (3): A, D

- Locate the local maximum and local minima. - The number of solutions equals the number of intersections with the horizontal line \(y=a\). - Use the end behavior of the quartic. - Use the extrema and end behavior to compare all horizontal-line cases.

1. The derivative is \(f'(x)=x^3-4x=x(x-2)(x+2)\), so the critical values are \(x=-2\), \(x=0\), and \(x=2\). 2. The corresponding function values are \(f(-2)=-1\), \(f(0)=3\), and \(f(2)=-1\). The outer critical points are local minima, and \(x=0\) is a local maximum. Also, \(f(x)\to\infty\) as \(x\to\pm\infty\). 3. The number of solutions of \(f(x)=a\) is the number of intersections between the graph and the horizontal line \(y=a\). There are four solutions for \(-1<a<3\), three solutions for \(a=3\), two solutions for \(a=-1\) or \(a>3\), no solutions for \(a<-1\), and no value of \(a\) gives exactly one solution.

Four solutions: \(-1<a<3\) Three solutions: \(a=3\) Two solutions: \(a=-1\) or \(a>3\) One solution: no values of \(a\) No solutions: \(a<-1\)

- What equation must the first derivative satisfy at a local extremum? - Substitute the given critical number into your expression in terms of \(k\). - What sign of the second derivative identifies a local maximum? - Evaluate the original function at the critical number.

1. Differentiate: \(f_k'(x) = 2kx - 6\). 2. Set the derivative equal to zero: \(2kx - 6 = 0\). Since \(k \ne 0\), the critical number is \(x = \frac{3}{k}\). 3. Require the critical number to equal \(-1\): \(\frac{3}{k} = -1\), so \(k = -3\). 4. The second derivative is \(f_k''(x) = 2k\). For \(k = -3\), \(f_{-3}''(-1) = -6 < 0\), so the critical point is a local maximum. 5. Evaluate the function: \(f_{-3}(-1) = -3(-1)^2 - 6(-1) - 1 = 2\). The local maximum is \((-1, 2)\).

a) \(f_k'(x) = 2kx - 6\); critical number: \(x = \frac{3}{k}\) b) \(k = -3\) c) Local maximum at \((-1, 2)\)

- A known extremum gives information about both the function value and the derivative. - Use \(f(1) = 2\) and \(f'(1) = 0\) to form two equations. - Use the sign of the second derivative to classify the extremum.

1. Differentiate: \(f'(x) = 2ax + b\). 2. Because \(x = 1\) is a critical number, \(f'(1) = 0\). Thus, \(2a + b = 0\). 3. Because the point \((1, 2)\) lies on the graph, \(f(1) = 2\). Thus, \(a + b + 3 = 2\), or \(a + b = -1\). 4. Solve the system. From \(2a + b = 0\), \(b = -2a\). Substituting into \(a + b = -1\) gives \(-a = -1\), so \(a = 1\) and \(b = -2\). 5. The second derivative is \(f''(x) = 2a = 2 > 0\). By the second derivative test, the extremum is a local minimum.

a) \(f'(x) = 2ax + b\) b) \(a = 1\), \(b = -2\) c) The point \((1, 2)\) is a local minimum.

- Use the point condition and the second-derivative condition to form two equations. - Check every zero of the second derivative for a concavity change. - Find all zeros of the first derivative and test for sign changes. - A repeated zero of the first derivative may not produce an extremum.

1. The point condition gives \(a+b=-6\). Also, \(f''(x)=12ax^2+6bx\), so the inflection condition \(f''(2)=0\) gives \(48a+12b=0\). Solving the system gives \(a=2\) and \(b=-8\). Thus, \(f(x)=2x^4-8x^3\). 2. The second derivative is \(f''(x)=24x(x-2)\), so the possible inflection values are \(x=0\) and \(x=2\). Since \(f'''(0)=-48\ne0\) and \(f'''(2)=48\ne0\), both are inflection values. The points are \((0, 0)\) and \((2, -32)\). 3. The first derivative is \(f'(x)=8x^2(x-3)\). The critical values are \(x=0\) and \(x=3\). At \(x=0\), the derivative does not change sign, so there is no local extremum; the point is a stationary inflection point. Since \(f''(3)=72>0\), \((3, -54)\) is a local minimum.

a) \(a=2\), \(b=-8\) b) \((0, 0)\) and \((2, -32)\) c) Local minimum: \((3, -54)\). The point \((0, 0)\) is a stationary inflection point, not an extremum.

- Apply the necessary condition for an inflection value twice. - Solve the resulting system of two equations. - Factor the first derivative and use the discriminant to count real critical values. - Use the second derivative to classify the critical point.

1. The second derivative is \(g''(x)=12x^2+6ax+2b\). The conditions \(g''(1)=0\) and \(g''(2)=0\) give \(12+6a+2b=0\) and \(48+12a+2b=0\). Subtracting yields \(a=-6\), and substitution gives \(b=12\). 2. Thus, \(g''(x)=12(x-1)(x-2)\), which changes sign at both \(x=1\) and \(x=2\). Therefore, both required values are inflection values. Also, \(g(x)=x^4-6x^3+12x^2\), and \(g'(x)=2x(2x^2-9x+12)\). The quadratic factor has discriminant \(81-96=-15<0\), so the only real critical value is \(x=0\). 3. Since \(g''(0)=24>0\), the point \((0, 0)\) is a local minimum. It is the only local extremum.

a) \(a=-6\), \(b=12\) b) The only local extremum is the local minimum \((0, 0)\).

- What must the first derivative equal at a local extremum? - Substitute \(x = 2\) into the derivative and solve for \(k\). - Verify that the critical point is an extremum by using the second derivative.

1. Differentiate: \(f_k'(x) = 2x^3 - 2kx\). 2. A necessary condition for a local extremum at \(x = 2\) is \(f_k'(2) = 0\). Substitute: \(2(2)^3 - 2k(2) = 16 - 4k\). 3. Solve \(16 - 4k = 0\): \(k = 4\). 4. Verify the result with the second derivative. For \(k = 4\), \(f_4''(x) = 6x^2 - 8\), so \(f_4''(2) = 16 > 0\). Therefore, \(x = 2\) is a local minimum.

\(k = 4\)

- Shift an even power so its minimum occurs at \(x=5\). - Compare a quadratic with a higher even power. - When the second derivative test is inconclusive, use the first derivative test. - A local minimum has negative slopes before it and positive slopes after it.

1. For part a), choose \(f(x)=(x-5)^2\). Then \(f'(x)=2(x-5)\) and \(f''(x)=2\), so \(f'(5)=0\) and \(f''(5)>0\). Therefore, \(x=5\) is a local minimum. 2. For part b), choose \(f(x)=(x-5)^4\). Then \(f'(x)=4(x-5)^3\) and \(f''(x)=12(x-5)^2\), so both derivatives equal \(0\) at \(x=5\). 3. For \(x<5\), \(f'(x)<0\); for \(x>5\), \(f'(x)>0\). By the first derivative test, \(f\) has a local minimum at \(x=5\).

a) One example is \(f(x)=(x-5)^2\). b) One example is \(f(x)=(x-5)^4\). Its derivative changes from negative to positive at \(x=5\), so it has a local minimum there.

- Find the first two derivatives of each function. - A positive second derivative at a critical point indicates a local minimum. - A zero second derivative makes the test inconclusive, not false. - Check the sign of the first derivative for extrema and the sign of the second derivative for a change in concavity.

1. The derivatives are \(f'(x)=2(x-1)\), \(f''(x)=2\); \(g'(x)=3(x-1)^2\), \(g''(x)=6(x-1)\); and \(h'(x)=4(x-1)^3\), \(h''(x)=12(x-1)^2\). 2. Since \(f'(1)=0\) and \(f''(1)=2>0\), \(f\) has a local minimum at \(x=1\). 3. For \(g\), the second derivative test is inconclusive. Because \(g'(x)>0\) on both sides of \(1\), there is no extremum. Also, \(g''(x)=6(x-1)\) changes sign at \(x=1\), so the graph changes concavity there. Thus \(x=1\) is a stationary inflection point. 4. For \(h\), the second derivative test is inconclusive. Since \(h'\) changes from negative to positive at \(1\), \(h\) has a local minimum there.

a) The second derivative test proves that \(f\) has a local minimum at \(x=1\). b) \(g\) has a stationary inflection point at \(x=1\), and \(h\) has a local minimum at \(x=1\).

- Evaluate the function and its first two derivatives at \(x = 0\). - Factor out the greatest common power of \(x\) to find the zeros. - Find the second critical number and express its function value in terms of \(t\). - Set the local minimum value equal to \(-4\).

1. Since \(g_t(0) = 0\), every graph passes through the origin. 2. Differentiate: \(g_t'(x) = 3x^2 - 6tx\) and \(g_t''(x) = 6x - 6t\). At \(x = 0\), \(g_t'(0) = 0\) and \(g_t''(0) = -6t < 0\), so the origin is a local maximum. 3. Factor: \(g_t(x) = x^2(x - 3t)\). Thus, \(x = 0\) is a zero of multiplicity \(2\), and \(x = 3t\) is a simple zero. 4. Solve \(g_t'(x) = 3x(x - 2t) = 0\). The other critical number is \(x = 2t\). Since \(g_t''(2t) = 6t > 0\), it is a local minimum. 5. Its value is \(g_t(2t) = 8t^3 - 12t^3 = -4t^3\). Set \(-4t^3 = -4\). Because \(t > 0\), \(t = 1\).

a) Every graph has a local maximum at \((0, 0)\). b) \(x = 0\), with multiplicity \(2\), and \(x = 3t\), with multiplicity \(1\) c) \(t = 1\)

- Factor the function to find its zeros. - Solve the first-derivative equation in terms of \(k\). - Use \(k>0\) when classifying the critical points. - Use the second and third derivatives to locate the inflection point.

1. Factor the function: \(f_k(x)=x(x^2-3k^2)\). The x-intercepts are \((0, 0)\) and \((\pm\sqrt{3}k, 0)\). The y-intercept is also \((0, 0)\). 2. The derivative is \(f_k'(x)=3(x^2-k^2)\), so the critical values are \(x=\pm k\). Since \(f_k''(x)=6x\), the point \((-k, 2k^3)\) is a local maximum and \((k, -2k^3)\) is a local minimum. 3. The second derivative is zero at \(x=0\), and \(f_k'''(x)=6\ne0\). Therefore, the inflection point is \((0, 0)\).

Intercepts: \((0, 0)\) and \((\pm\sqrt{3}k, 0)\) Local maximum: \((-k, 2k^3)\) Local minimum: \((k, -2k^3)\) Inflection point: \((0, 0)\)

- Factor the quartic to find its zeros. - Use the sign of the second derivative to classify the critical points. - Substitute parameterized x-values carefully into the original function. - Use the graph's even symmetry to check paired points.

1. Factor: \(g_a(x)=\frac{1}{4}x^2(x^2-4a^2)\). The x-intercepts are \((0, 0)\) and \((\pm2a, 0)\). The y-intercept is \((0, 0)\). 2. The derivative is \(g_a'(x)=x(x^2-2a^2)\), so the critical values are \(x=0\) and \(x=\pm\sqrt{2}a\). Since \(g_a''(0)=-2a^2<0\), \((0, 0)\) is a local maximum. Since \(g_a''(\pm\sqrt{2}a)=4a^2>0\), the local minima are \((\pm\sqrt{2}a, -a^4)\). 3. Solving \(g_a''(x)=3x^2-2a^2=0\) gives \(x=\pm\sqrt{\frac{2}{3}}a\). The third derivative is nonzero at both values, so the inflection points are \((\pm\sqrt{\frac{2}{3}}a, -\frac{5}{9}a^4)\).

Intercepts: \((0, 0)\) and \((\pm2a, 0)\) Local maximum: \((0, 0)\) Local minima: \((\pm\sqrt{2}a, -a^4)\) Inflection points: \((\pm\sqrt{\frac{2}{3}}a, -\frac{5}{9}a^4)\)

- Distinguish necessary and sufficient conditions. - Test claims with simple powers such as \(x^2\), \(x^3\), and \(x^4\). - Recall the sign convention for concavity. - State all conditions of the second derivative test.

1. Statement a) is false. For \(f(x)=x^3\), \(f'(0)=0\), but \(x=0\) is a stationary inflection point, not a local extremum. 2. Statement b) is true. Because \(f''\) is continuous, a change in concavity requires \(f''\) to change sign and therefore equal \(0\) at the inflection point. 3. Statement c) is false. A negative second derivative indicates concave down, not concave up. 4. Statement d) is true by the second derivative test. A positive second derivative gives a local minimum; a negative second derivative gives a local maximum.

a) False b) True c) False d) True

- Factor the function before finding its zeros. - The sign of \(k\) affects the second derivative at each critical number. - Use the first two derivatives to find and classify the local extrema. - Substitute each x-coordinate into the original function to find the point coordinates.

1. Factor the function: \(f_k(x)=x^2(x-3k)\). The zeros are \(x=0\), with multiplicity \(2\), and \(x=3k\). 2. Differentiate: \(f_k'(x)=3x(x-2k)\), \(f_k''(x)=6(x-k)\), and \(f_k'''(x)=6\). 3. The critical numbers are \(x=0\) and \(x=2k\). 4. When \(k>0\), \(f_k''(0)=-6k<0\), so \((0, 0)\) is a local maximum, while \(f_k''(2k)=6k>0\), so \((2k, -4k^3)\) is a local minimum. 5. When \(k<0\), the classifications reverse: \((0, 0)\) is a local minimum and \((2k, -4k^3)\) is a local maximum. 6. The equation \(f_k''(x)=0\) gives \(x=k\). Since \(f_k'''(k)=6\neq0\), the inflection point is \((k, -2k^3)\).

Zeros: \(x=0\) with multiplicity \(2\), and \(x=3k\). For \(k>0\): local maximum \((0, 0)\); local minimum \((2k, -4k^3)\). For \(k<0\): local minimum \((0, 0)\); local maximum \((2k, -4k^3)\). Inflection point: \((k, -2k^3)\).

- Factor before finding the zeros. - Substitute the given point to solve for the parameter. - Use the first derivative for critical points and the second derivative for classification. - Track how the sign of the parameter affects the second derivative.

1. Factor \(g_a(x)=ax^2(x^2-4)=ax^2(x-2)(x+2)\). Since \(a\ne0\), the x-intercepts occur at \(x=-2\), \(x=0\), and \(x=2\), independent of \(a\). 2. Substituting \(Q(1, 6)\) gives \(6=a(1-4)=-3a\), so \(a=-2\). 3. The first derivative is \(g_a'(x)=4ax(x^2-2)\), so the critical points are \(x=0\) and \(x=\pm\sqrt{2}\), independent of \(a\). 4. The second derivative is \(g_a''(x)=a(12x^2-8)\). Thus \(g_a''(0)=-8a\), while \(g_a''(\pm\sqrt{2})=16a\). If \(a>0\), \(x=0\) is a local maximum and \(x=\pm\sqrt{2}\) are local minima. If \(a<0\), the classifications reverse.

a) X-intercepts: \((-2, 0)\), \((0, 0)\), and \((2, 0)\) b) \(a=-2\) c) Critical-point x-coordinates: \(0\) and \(\pm\sqrt{2}\). For \(a>0\), \(x=0\) is a local maximum and \(x=\pm\sqrt{2}\) are local minima; for \(a<0\), the classifications reverse.

- Find critical numbers by setting the first derivative equal to zero. - Use unit-circle cosine values. - Apply the second derivative test. - Keep only interior points of the stated interval.

1. Differentiate: \(f'(x)=\cos x+\frac{1}{2}\). 2. Set the derivative equal to zero: \(\cos x=-\frac{1}{2}\). In \((0, 2\pi)\), the critical numbers are \(x=\frac{2\pi}{3}\) and \(x=\frac{4\pi}{3}\). 3. The second derivative is \(f''(x)=-\sin x\). At \(x=\frac{2\pi}{3}\), \(f''\left(\frac{2\pi}{3}\right)=-\frac{\sqrt{3}}{2}<0\), so there is a local maximum. At \(x=\frac{4\pi}{3}\), \(f''\left(\frac{4\pi}{3}\right)=\frac{\sqrt{3}}{2}>0\), so there is a local minimum.

Local maximum at \(x=\frac{2\pi}{3}\); local minimum at \(x=\frac{4\pi}{3}\)

- Find critical numbers from the first derivative. - Use exact unit-circle values. - Apply the second derivative test. - Keep only interior points of the interval.

1. Differentiate twice: \(g'(x)=\sqrt{2}\cos x-1\) and \(g''(x)=-\sqrt{2}\sin x\). 2. Set the first derivative equal to zero: \(\cos x=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\). In \((0, 2\pi)\), the critical numbers are \(x=\frac{\pi}{4}\) and \(x=\frac{7\pi}{4}\). 3. At \(x=\frac{\pi}{4}\), \(g''\left(\frac{\pi}{4}\right)=-1<0\), so there is a local maximum. At \(x=\frac{7\pi}{4}\), \(g''\left(\frac{7\pi}{4}\right)=1>0\), so there is a local minimum.

Local maximum at \(x=\frac{\pi}{4}\); local minimum at \(x=\frac{7\pi}{4}\)

- Factor out the common factor \(e^x\) before differentiating. - Use the fact that \(e^x\) is never zero. - Use the second derivative to classify each critical point. - Check that each x-coordinate lies in the stated interval.

1. Combine like terms: \(f(x)=(x^2+2x-1)e^x\). 2. Differentiate using the product rule: \(f'(x)=(2x+2)e^x+(x^2+2x-1)e^x=(x^2+4x+1)e^x\). 3. Since \(e^x>0\), critical points satisfy \(x^2+4x+1=0\). Thus \(x=-2\pm\sqrt{3}\), and both values lie in \([-5,0]\). 4. The second derivative is \(f''(x)=(x^2+6x+5)e^x\). At \(x=-2-\sqrt{3}\), \(f''(x)=-2\sqrt{3}e^{-2-\sqrt{3}}<0\), so there is a local maximum. At \(x=-2+\sqrt{3}\), \(f''(x)=2\sqrt{3}e^{-2+\sqrt{3}}>0\), so there is a local minimum. 5. Evaluate \(f\): \(f(-2-\sqrt{3})=(2+2\sqrt{3})e^{-2-\sqrt{3}}\approx 0.131\), and \(f(-2+\sqrt{3})=(2-2\sqrt{3})e^{-2+\sqrt{3}}\approx -1.120\).

Local maximum: \(\left(-2-\sqrt{3},(2+2\sqrt{3})e^{-2-\sqrt{3}}\right)\approx(-3.732,0.131)\). Local minimum: \(\left(-2+\sqrt{3},(2-2\sqrt{3})e^{-2+\sqrt{3}}\right)\approx(-0.268,-1.120)\).

- Substitute the entire inner function into the outer function. - Use the chain rule to differentiate the exponential function. - The exponential factor is never zero. - A point on the horizontal line \(y=4\) has y-coordinate \(4\).

1. Composing the functions gives \(h_k(x)=f(g_k(x))=e^{-x^2+k}\). 2. By the chain rule, \(h_k'(x)=-2xe^{-x^2+k}\). Since the exponential factor is positive, the only critical point is \(x=0\). 3. The second derivative is \(h_k''(x)=e^{-x^2+k}(4x^2-2)\). Because \(h_k''(0)=-2e^k<0\), the point is a local maximum. Also, \(-x^2+k\le k\) for every real \(x\), so it is the absolute maximum. Its y-coordinate is \(h_k(0)=e^k\), so the maximum point is \((0, e^k)\). 4. For this point to lie on \(y=4\), solve \(e^k=4\). Therefore, \(k=\ln(4)\approx1.386\).

a) \(h_k(x)=e^{-x^2+k}\) b) \((0, e^k)\) c) \(k=\ln(4)\approx1.386\)

- Recall the general solutions of \(\sin(u)=0\). - Use the chain rule to differentiate \(\sin(x^2)\). - Determine which possible values of \(x^2\) lie between \(0\) and \(4\). - Use the second derivative to classify the critical points.

1. For the zeros, solve \(\sin(x^2)=0\). Thus \(x^2=k\pi\) for an integer \(k\). Since \(0\le x^2\le 4\), the possible values are \(x^2=0\) and \(x^2=\pi\). Therefore, the zeros are \(x=0\) and \(x=\pm\sqrt{\pi}\). 2. Differentiate using the chain rule: \(f'(x)=2x\cos(x^2)\). Critical numbers satisfy \(x=0\) or \(\cos(x^2)=0\). In the given interval, this gives \(x=0\) and \(x=\pm\sqrt{\frac{\pi}{2}}\). 3. The second derivative is \(f''(x)=2\cos(x^2)-4x^2\sin(x^2)\). Since \(f''(0)=2>0\), \((0,0)\) is a local minimum. At \(x=\pm\sqrt{\frac{\pi}{2}}\), \(f''(x)=-2\pi<0\), so both points are local maxima. Their function value is \(1\).

a) \(x=0\) and \(x=\pm\sqrt{\pi}\approx\pm1.772\). b) Local minimum: \((0,0)\). Local maxima: \(\left(-\sqrt{\frac{\pi}{2}},1\right)\) and \(\left(\sqrt{\frac{\pi}{2}},1\right)\), with x-coordinates approximately \(\pm1.253\).

- Substitute the given point into the first function. - A local maximum at a specified x-value requires a zero first derivative there. - Use the product and chain rules for \(g_k\). - Check the sign of the second derivative.

1. Substitute \((2, 10)\) into \(f_a\): \(10=8a-10+4\). Thus, \(8a=16\), so \(a=2\). 2. Differentiate \(g_k\): \(g_k'(x)=e^{0.5x}(0.5k-0.5x-1)\). The condition \(g_k'(4)=0\) gives \(e^2(0.5k-3)=0\), so \(k=6\). 3. The second derivative is \(g_k''(x)=e^{0.5x}(0.25k-0.25x-1)\). For \(k=6\), \(g_6''(4)=-0.5e^2<0\), confirming a local maximum at \(x=4\).

a) \(a=2\) b) \(k=6\)

- Find critical points from the first derivative. - Use the sign of the second derivative and consider both signs of \(k\). - Find the inflection point and eliminate the parameter.

1. The derivatives are \(f_k'(x)=3x^2-6kx=3x(x-2k)\) and \(f_k''(x)=6x-6k\). 2. The critical points occur at \(x=0\) and \(x=2k\). Their y-coordinates are \(2\) and \(-4k^3+2\), respectively. 3. Since \(f_k''(0)=-6k\) and \(f_k''(2k)=6k\), for \(k>0\), \((0, 2)\) is a local maximum and \((2k, -4k^3+2)\) is a local minimum. For \(k<0\), the classifications reverse. 4. Since \(f_k''(x)=6(x-k)\) changes sign at \(x=k\), the inflection point has x-coordinate \(k\). Its y-coordinate is \(-2k^3+2\). Eliminating \(k\) gives the locus \(y=-2x^3+2\).

a) For \(k>0\): local maximum \((0, 2)\), local minimum \((2k, -4k^3+2)\). For \(k<0\): the classifications reverse. b) \(y=-2x^3+2\)

- Factor the derivative to find the critical points. - Classify the points separately for positive and negative \(a\). - Eliminate \(a\) from the inflection-point coordinates.

1. The derivatives are \(f_a'(x)=3x^2-12ax+9a^2=3(x-a)(x-3a)\) and \(f_a''(x)=6x-12a\). 2. The critical points occur at \(x=a\) and \(x=3a\). Their coordinates are \((a, 4a^3)\) and \((3a, 0)\). 3. Since \(f_a''(a)=-6a\) and \(f_a''(3a)=6a\), for \(a>0\), \((a, 4a^3)\) is a local maximum and \((3a, 0)\) is a local minimum. For \(a<0\), the classifications reverse. 4. Since \(f_a''(x)=6(x-2a)\) changes sign at \(x=2a\), the inflection point is \((2a, 2a^3)\). Because \(a=\frac{x}{2}\), its y-coordinate is \(2\left(\frac{x}{2}\right)^3=\frac{1}{4}x^3\). Thus every inflection point lies on \(g\).

a) For \(a>0\): local maximum \((a, 4a^3)\), local minimum \((3a, 0)\). For \(a<0\): the classifications reverse. b) Inflection point \((2a, 2a^3)\), and its locus is \(y=\frac{1}{4}x^3\).

- Treat \(k\) as a constant when differentiating. - Set the first derivative equal to \(0\). - Use the natural logarithm to solve an exponential equation. - After finding the critical number, use the monotonicity of the natural logarithm to determine how it changes with \(k\).

1. Differentiate: \(f_k'(x)=ke^x-1\). 2. Set the derivative equal to \(0\): \(ke^x=1\), so the critical number is \(x=\ln\left(\frac{1}{k}\right)=-\ln(k)\). 3. The second derivative is \(f_k''(x)=ke^x>0\), so this critical point is a local minimum. 4. As \(k\) increases, \(\ln(k)\) increases. Therefore, \(-\ln(k)\) decreases, and the minimum moves to the left along the x-axis.

1. \(f_k'(x)=ke^x-1\) 2. The x-coordinate is \(-\ln(k)\). 3. As \(k\) increases, the x-coordinate decreases, so the minimum moves left.

- Use the product rule together with the chain rule. - The exponential factor cannot equal zero. - Use the second derivative to classify the critical point.

1. Differentiate using the product and chain rules: \(f'(x)=4e^{-0.2x}+(4x+2)(-0.2)e^{-0.2x}=(3.6-0.8x)e^{-0.2x}\). 2. Since the exponential factor is positive, the critical-point equation is \(3.6-0.8x=0\), which gives \(x=4.5\). 3. Differentiate again: \(f''(x)=(0.16x-1.52)e^{-0.2x}\). Then \(f''(4.5)=-0.8e^{-0.9}<0\), so the point is a local maximum. 4. Its y-coordinate is \(f(4.5)=20e^{-0.9}\).

The function has a local maximum at \(\left(4.5,20e^{-0.9}\right)\).

- Factor the critical-point equation using a common exponential factor. - Use the inverse relationship between \(e^x\) and \(\ln(x)\). - Use the second derivative to classify the critical point.

1. Differentiate using the chain rule: \(f'(x)=e^{2x}-4e^x\). 2. Set the first derivative equal to zero: \(e^{2x}-4e^x=e^x(e^x-4)=0\). Since \(e^x>0\), \(e^x=4\), so \(x=\ln(4)\). 3. The second derivative is \(f''(x)=2e^{2x}-4e^x\). At the critical number, \(f''(\ln(4))=2\cdot16-4\cdot4=16>0\), so the point is a local minimum. 4. Evaluate the function: \(f(\ln(4))=\frac{1}{2}\cdot16-4\cdot4=-8\).

The function has a local minimum at \((\ln(4),-8)\).

- Use the product rule. - Factor out the exponential term when solving \(f'(x)=0\). - Use the second derivative to classify each critical point. - Substitute the x-coordinates into the original function.

1. Differentiate using the product rule: \(f'(x)=(2x-5)e^x+(x^2-5x+7)e^x=(x^2-3x+2)e^x\). 2. Since \(e^x>0\), critical numbers satisfy \(x^2-3x+2=0\). Factoring gives \((x-1)(x-2)=0\), so \(x=1\) and \(x=2\). 3. The second derivative is \(f''(x)=(x^2-x-1)e^x\). Since \(f''(1)=-e<0\), \(x=1\) gives a local maximum. Since \(f''(2)=e^2>0\), \(x=2\) gives a local minimum. 4. Evaluate the function: \(f(1)=3e\), and \(f(2)=e^2\).

Local maximum: \((1,3e)\). Local minimum: \((2,e^2)\).

- Find the critical point by setting the first derivative equal to zero. - Use the second derivative to classify the critical point. - Substitute the critical x-value into the original function. - A point on \(y=2\) must have y-coordinate \(2\).

1. Differentiate: \(f_a'(x)=(0.5a-1-0.5x)e^{0.5x}\). Since the exponential factor is positive, the critical point satisfies \(0.5a-1-0.5x=0\), so \(x=a-2\). 2. The second derivative is \(f_a''(x)=(0.25a-1-0.25x)e^{0.5x}\). At \(x=a-2\), \(f_a''(a-2)=-0.5e^{0.5a-1}<0\), so the critical point is a local maximum. 3. Its y-coordinate is \(f_a(a-2)=2e^{0.5a-1}\). Therefore, the local maximum point is \(H=(a-2, 2e^{0.5a-1})\). 4. For this point to lie on \(y=2\), require \(2e^{0.5a-1}=2\). Thus, \(0.5a-1=0\), so \(a=2\).

a) \(H=(a-2, 2e^{0.5a-1})\) b) \(a=2\)

- Factor out the common exponential term. - Use the fact that \(e^x\) is always positive. - Apply the second derivative test to the critical point. - Compare the growth rates of \(e^x\) and \(e^{2x}\).

1. Factor out \(e^x\): \(f(x)=e^x(4-e^x)\). Since \(e^x>0\), the zero satisfies \(e^x=4\), so \(x=\ln(4)\). 2. Differentiate: \(f'(x)=4e^x-2e^{2x}=2e^x(2-e^x)\). The critical-point equation gives \(e^x=2\), so \(x=\ln(2)\). 3. The second derivative is \(f''(x)=4e^x-4e^{2x}\). Since \(f''(\ln(2))=8-16=-8<0\), the point is a local maximum. Its y-coordinate is \(f(\ln(2))=8-4=4\). 4. As \(x\to-\infty\), both exponential terms approach \(0\), so \(f(x)\to0\). As \(x\to\infty\), the term \(-e^{2x}\) dominates, so \(f(x)\to-\infty\).

a) \(f(x)=e^x(4-e^x)\); zero at \(x=\ln(4)\). b) Local maximum: \((\ln(2),4)\). c) \(\lim_{x\to-\infty}f(x)=0\), and \(\lim_{x\to\infty}f(x)=-\infty\).

- Set the first derivative equal to \(0\). - Use the second derivative to classify the critical point. - A point on the y-axis has \(x=0\). - The first derivative gives the tangent slope.

1. Differentiate: \(f_a'(x)=1-e^{a-x}\) and \(f_a''(x)=e^{a-x}\). 2. Set the first derivative equal to \(0\): \(e^{a-x}=1\), so \(x=a\). 3. Since \(f_a''(a)=1>0\), the point is a local minimum. Its coordinates are \((a, a+1)\). 4. For the minimum to lie on \(y=5\), solve \(a+1=5\), giving \(a=4\). 5. At the y-intercept, \(x=0\), so the slope is \(1-e^a\). Set \(1-e^a=1-e\), which gives \(a=1\).

a) Local minimum: \((a, a+1)\) b) \(a=4\) c) \(a=1\)

- The exponential factor in the derivative is never zero. - A point on the x-axis has y-coordinate \(0\). - Parallel lines have equal slopes.

1. Differentiate: \(g_k'(x)=ke^x-e^{2x}=e^x(k-e^x)\). 2. The only critical number satisfies \(e^x=k\), so \(x=\ln(k)\). 3. The second derivative is \(g_k''(x)=ke^x-2e^{2x}\). At \(x=\ln(k)\), it equals \(-k^2<0\), so the point is a local maximum. 4. Its y-coordinate is \(\frac{1}{2}k^2-2\). Thus, the local maximum is \(\left(\ln(k), \frac{1}{2}k^2-2\right)\). 5. For the point to lie on the x-axis, solve \(\frac{1}{2}k^2-2=0\). Since \(k>0\), \(k=2\). 6. At \(x=0\), the slope is \(k-1\). Parallelism to \(y=3x-5\) requires \(k-1=3\), so \(k=4\).

a) Local maximum: \(\left(\ln(k), \frac{1}{2}k^2-2\right)\) b) \(k=2\) c) \(k=4\)

- Test whether \(g_k(-x)=g_k(x)\). - Symmetry suggests a possible critical point at \(x=0\). - Use derivatives to verify the local minimum. - A point on the x-axis has y-coordinate \(0\).

1. Evaluate at \(-x\): \(g_k(-x)=k(e^{-x}+e^x)-k^2=g_k(x)\). Therefore, each function is even and each graph is symmetric about the y-axis. 2. Differentiate: \(g_k'(x)=k(e^x-e^{-x})\). Since \(k>0\), the only critical number is \(x=0\). 3. The second derivative is \(g_k''(x)=k(e^x+e^{-x})\), and \(g_k''(0)=2k>0\). Thus, the point at \(x=0\) is a local minimum. 4. Its y-coordinate is \(g_k(0)=2k-k^2\). Set this equal to \(0\): \(k(2-k)=0\). 5. Since \(k>0\), \(k=2\).

a) \(g_k(-x)=g_k(x)\), so each graph is symmetric about the y-axis. b) \(k=2\)

- Relate the number of local extrema to the number of distinct real zeros of the first derivative. - Use the discriminant of the quadratic derivative. - For \(k = 5\), find the critical numbers and apply the second derivative test. - Evaluate the original function at each critical number.

1. Differentiate: \(f_k'(x) = x^2 - kx + 4\). 2. The graph has two local extrema when the derivative has two distinct real zeros. Its discriminant is \(D = k^2 - 16\). 3. Solve \(k^2 - 16 > 0\): \(k < -4\) or \(k > 4\). 4. For \(k = 5\), \(f_5'(x) = x^2 - 5x + 4 = (x - 1)(x - 4)\), so the critical numbers are \(x = 1\) and \(x = 4\). 5. The second derivative is \(f_5''(x) = 2x - 5\). Since \(f_5''(1) = -3 < 0\), there is a local maximum at \(x = 1\). Its value is \(f_5(1) = \frac{1}{3} - \frac{5}{2} + 4 = \frac{11}{6}\). 6. Since \(f_5''(4) = 3 > 0\), there is a local minimum at \(x = 4\). Its value is \(f_5(4) = \frac{64}{3} - 40 + 16 = -\frac{8}{3}\).

a) \(k < -4\) or \(k > 4\) b) Local maximum at \(\left(1, \frac{11}{6}\right)\); local minimum at \(\left(4, -\frac{8}{3}\right)\)

- Set the first derivative equal to zero. - Use logarithms to solve an equation with \(x\) in an exponent. - Use the second derivative to classify the critical point. - Evaluate the original function to obtain the y-coordinate.

1. Differentiate: \(f'(x)=2e^{2x}-6\). 2. Set the first derivative equal to zero: \(e^{2x}=3\), so \(2x=\ln(3)\) and \(x=\frac{1}{2}\ln(3)\). 3. The second derivative is \(f''(x)=4e^{2x}>0\) for every real \(x\), so the critical point is a local minimum. 4. Its y-coordinate is \(f\left(\frac{1}{2}\ln(3)\right)=3-3\ln(3)\).

The function has a local minimum at \(\left(\frac{1}{2}\ln(3),3-3\ln(3)\right)\). The x-coordinate is approximately \(0.549\).

- Apply the quotient rule and the chain rule. - A fraction is zero when its numerator is zero and its denominator is nonzero. - The exponential factor is always positive. - Use the second derivative to classify each critical point.

1. Apply the quotient and chain rules: \(f'(x)=\frac{x\left(xe^{x^2/2}\right)-e^{x^2/2}}{x^2}=\frac{(x^2-1)e^{x^2/2}}{x^2}\). 2. Since \(e^{x^2/2}>0\) and \(x^2>0\) on the domain, critical numbers satisfy \(x^2-1=0\). Thus \(x=-1\) and \(x=1\). 3. The second derivative is \(f''(x)=\left(x-\frac{1}{x}+\frac{2}{x^3}\right)e^{x^2/2}\). Then \(f''(1)=2\sqrt{e}>0\), so \(x=1\) gives a local minimum. Also, \(f''(-1)=-2\sqrt{e}<0\), so \(x=-1\) gives a local maximum. 4. The corresponding function values are \(f(1)=\sqrt{e}\) and \(f(-1)=-\sqrt{e}\).

Local maximum: \((-1,-\sqrt{e})\). Local minimum: \((1,\sqrt{e})\).

- Find the critical numbers from the first derivative. - Use the second derivative to classify the critical points when \(k\ne0\). - Examine the sign of the first derivative separately when \(k=0\). - A point lies on the x-axis when its y-coordinate is zero.

1. Differentiate: \(f_k'(x)=x^2-k^2\) and \(f_k''(x)=2x\). When \(k=0\), the only critical number is \(x=0\). The derivative does not change sign there, so there is no local extremum. For \(k\ne0\), the critical numbers are \(x=k\) and \(x=-k\). Their function values are \(f_k(k)=-\frac23k^3+\frac23\) and \(f_k(-k)=\frac23k^3+\frac23\). If \(k>0\), then \(f_k''(k)>0\) and \(f_k''(-k)<0\), so there is a local minimum at \(\left(k, -\frac23k^3+\frac23\right)\) and a local maximum at \(\left(-k, \frac23k^3+\frac23\right)\). If \(k<0\), the classifications reverse: the point at \(x=k\) is a local maximum, and the point at \(x=-k\) is a local minimum. 2. Set each possible extremum value equal to zero: \(-\frac23k^3+\frac23=0\) gives \(k=1\), and \(\frac23k^3+\frac23=0\) gives \(k=-1\). In either case, the local minimum is \((1, 0)\).

1) For \(k=0\), there are no local extrema. For \(k>0\): local maximum \(\left(-k, \frac23k^3+\frac23\right)\) and local minimum \(\left(k, -\frac23k^3+\frac23\right)\). For \(k<0\): local maximum \(\left(k, -\frac23k^3+\frac23\right)\) and local minimum \(\left(-k, \frac23k^3+\frac23\right)\). 2) \(k=1\) or \(k=-1\); in both cases, the local minimum is \((1, 0)\).

- Use y-axis symmetry to determine which powers of \(x\) may appear. - Translate the given point and the local-extremum condition into equations. - Solve for the coefficients in terms of \(k\). - Use the second derivative to classify the extremum.

1. Symmetry about the y-axis gives the form \(f_k(x)=ax^4+bx^2+c\). Since the graph passes through the origin, \(c=0\). 2. The point \((1, k)\) and the critical-point condition give \(a+b=k\) and \(4a+2b=0\). From the second equation, \(b=-2a\). Substitution gives \(-a=k\), so \(a=-k\) and \(b=2k\). Therefore, \(f_k(x)=-kx^4+2kx^2\). 3. The second derivative is \(f_k''(x)=-12kx^2+4k\). At \(x=1\), \(f_k''(1)=-8k\). Thus, the point is a local maximum when \(k>0\) and a local minimum when \(k<0\).

a) \(f_k(x)=-kx^4+2kx^2\) b) Local maximum for \(k>0\); local minimum for \(k<0\).

- Begin with the domain of the logarithm. - Set the first derivative equal to zero. - Check when the resulting critical number lies in the domain. - Use the second derivative to classify the critical point.

1. Differentiate: \(f_a'(x)=\frac ax-1\) and \(f_a''(x)=-\frac{a}{x^2}\). 2. A critical number must satisfy \(\frac ax-1=0\), so \(x=a\). Because the domain requires \(x>0\), this critical number exists exactly when \(a>0\). 3. For \(a>0\), \(f_a''(a)=-\frac1a<0\), so the critical point is a local maximum. Its y-coordinate is \(f_a(a)=a\ln(a)-a=a(\ln(a)-1)\). For \(a\le0\), there is no critical number in the domain and therefore no local extremum.

A local extremum exists exactly when \(a>0\). It is a local maximum at \(\left(a, a(\ln(a)-1)\right)\).

- Differentiate using the chain rule. - Use the range of the exponential function to determine when the critical-point equation has a solution. - Take natural logarithms to solve for \(x\). - Use the second derivative to classify the critical point.

1. Differentiate: \(h_k'(x)=ke^{kx}-1\) and \(h_k''(x)=k^2e^{kx}\). 2. A critical number must satisfy \(ke^{kx}=1\), or \(e^{kx}=\frac1k\). Since an exponential is always positive, this equation has a real solution exactly when \(k>0\). 3. For \(k>0\), take natural logarithms: \(kx=-\ln(k)\), so \(x=-\frac{\ln(k)}{k}\). Also, \(h_k''(x)=k^2e^{kx}>0\) for every \(x\), so the critical point is a local minimum. 4. Its y-coordinate is \(h_k\left(-\frac{\ln(k)}{k}\right)=\frac1k+\frac{\ln(k)}{k}=\frac{1+\ln(k)}{k}\).

A local extremum exists exactly when \(k>0\). It is a local minimum at \(\left(-\frac{\ln(k)}{k}, \frac{1+\ln(k)}{k}\right)\).

- Which differentiation rule applies to a quotient? - Horizontal tangents occur where the derivative is zero. - Substitute the critical numbers into the original function. - Use the second derivative to classify the points.

1. Apply the quotient rule: \(f'(x) = \frac{(2x + 2)(x + 1) - (x^2 + 2x + 10)}{(x + 1)^2} = \frac{x^2 + 2x - 8}{(x + 1)^2}\). 2. Solve \(f'(x) = 0\). The numerator factors as \((x - 2)(x + 4)\), so the critical numbers are \(x = 2\) and \(x = -4\). 3. Evaluate the function: \(f(2) = 6\) and \(f(-4) = -6\). 4. Differentiate again: \(f''(x) = \frac{18}{(x + 1)^3}\). 5. Since \(f''(2) > 0\), \((2, 6)\) is a local minimum. Since \(f''(-4) < 0\), \((-4, -6)\) is a local maximum.

Local maximum: \((-4, -6)\) Local minimum: \((2, 6)\)

- Use the product rule and chain rule. - The exponential factor cannot equal zero. - Use the second derivative to classify the critical point. - Evaluate the original function to obtain the y-coordinate.

1. Differentiate using the product and chain rules: \(g'(x)=e^{-x}-(x+2)e^{-x}=(-x-1)e^{-x}\). 2. Since \(e^{-x}>0\), the critical-point equation is \(-x-1=0\), so \(x=-1\). 3. The second derivative is \(g''(x)=xe^{-x}\). Since \(g''(-1)=-e<0\), the critical point is a local maximum. 4. Evaluate the function: \(g(-1)=e\).

The graph has a local maximum at \((-1,e)\).

- Determine where \(\sqrt{x}\) is defined. - Use the zero-product property. - Differentiate and solve for an interior critical number. - A local maximum has a negative second derivative. - Rewrite \(a\sqrt{a}\) using exponents.

1. The square root requires \(x\geq0\), so the domain is \([0,\infty)\). The zeros are \(x=0\) and \(x=a\). 2. For \(x>0\), differentiate: \(h_a'(x)=\frac{a-3x}{2\sqrt{x}}\). 3. The only interior critical number is \(x=\frac{a}{3}\). The second derivative is \(h_a''(x)=\frac{-3x-a}{4x\sqrt{x}}<0\), so this is a local maximum. 4. Its y-coordinate is \(\left(a-\frac{a}{3}\right)\sqrt{\frac{a}{3}}=\frac{2a\sqrt{3a}}{9}\). Thus, the local maximum is \(\left(\frac{a}{3}, \frac{2a\sqrt{3a}}{9}\right)\). 5. Set \(\frac{2a\sqrt{3a}}{9}=2\). This is equivalent to \(a^{3/2}=3^{3/2}\), so \(a=3\).

a) Domain: \([0,\infty)\); zeros: \(x=0\) and \(x=a\) b) Local maximum: \(\left(\frac{a}{3}, \frac{2a\sqrt{3a}}{9}\right)\) c) \(a=3\)

- Apply the chain rule to the square-root function. - A fraction is zero when its numerator is zero and its denominator is nonzero. - At a critical number, terms containing the zero numerator may drop out of the second derivative. - Use the sign of the second derivative to classify each critical point.

1. By the chain rule, \(f'(x)=\frac{2x^3-2x}{\sqrt{x^4-2x^2+2}}=\frac{2x(x^2-1)}{\sqrt{x^4-2x^2+2}}\). 2. The denominator is always positive, so \(f'(x)=0\) when \(2x(x^2-1)=0\). The critical numbers are \(x=-1,0,1\). 3. At a critical number, the numerator of \(f'\) is \(0\), so differentiating the quotient shows \(f''(x)=\frac{6x^2-2}{\sqrt{x^4-2x^2+2}}\) at that number. 4. At \(x=0\), \(f''(0)=-\sqrt{2}<0\), so \((0,\sqrt{2})\) is a local maximum. 5. At \(x=1\) and \(x=-1\), \(f''(\pm1)=4>0\), so \((1,1)\) and \((-1,1)\) are local minima.

Local maximum: \((0,\sqrt{2})\) Local minima: \((-1,1)\) and \((1,1)\)

- Examine the function as \(x\) approaches the left endpoint of its domain. - A local extremum can occur where the first derivative is zero. - Use the second derivative to classify the critical point. - Analyze each coordinate of the minimum as a function of \(k\).

1. As \(x\to0^+\), \(\frac{x}{k}\to0\) and \(-\ln(x)\to\infty\). Therefore, \(f_k(x)\to\infty\), so \(x=0\) is a vertical asymptote. 2. Differentiate: \(f_k'(x)=\frac{1}{k}-\frac{1}{x}\). Setting the derivative equal to zero gives \(x=k\). 3. The second derivative is \(f_k''(x)=\frac{1}{x^2}\). Since \(f_k''(k)=\frac{1}{k^2}>0\), the critical point is a local minimum. 4. Its y-coordinate is \(f_k(k)=1-\ln(k)\). Thus, the local minimum is \((k, 1-\ln(k))\). 5. As \(k\) increases, the x-coordinate increases and \(1-\ln(k)\) decreases. The minimum moves to the right and downward.

a) \(x=0\) b) \((k, 1-\ln(k))\) c) It moves to the right and downward as \(k\) increases.

- The logarithm determines the domain. - Compare quadratic and logarithmic growth as \(x\to\infty\). - Set the first derivative equal to zero. - Use the second derivative to classify the critical point.

1. Since \(\ln(x)\) requires \(x>0\), the domain is \((0,\infty)\). 2. As \(x\to0^+\), \(-2\ln(x)\to\infty\), so \(f(x)\to\infty\). As \(x\to\infty\), the quadratic term dominates the logarithmic term, so \(f(x)\to\infty\). 3. The derivatives are \(f'(x)=x-\frac{2}{x}\) and \(f''(x)=1+\frac{2}{x^2}\). 4. Set \(f'(x)=0\): \(x^2=2\). The domain gives \(x=\sqrt{2}\). 5. Since \(f''(\sqrt{2})=2>0\), the point is a local minimum. Its y-coordinate is \(f(\sqrt{2})=1-\ln(2)\).

a) Domain: \((0,\infty)\). Also, \(\lim_{x\to0^+}f(x)=\infty\) and \(\lim_{x\to\infty}f(x)=\infty\). b) Local minimum: \((\sqrt{2},1-\ln(2))\).

- The logarithm determines the domain. - Compare logarithmic and power growth. - Apply the quotient rule carefully. - Use the second derivative to classify the critical point.

1. The domain is \((0,\infty)\). As \(x\to0^+\), \(\ln(x)\to-\infty\) while \(x^2\to0^+\), so \(g(x)\to-\infty\). As \(x\to\infty\), the power in the denominator grows faster than the logarithm, so \(g(x)\to0\). 2. The quotient rule gives \(g'(x)=\frac{1-2\ln(x)}{x^3}\). The second derivative is \(g''(x)=\frac{6\ln(x)-5}{x^4}\). 3. Set \(g'(x)=0\): \(\ln(x)=\frac{1}{2}\), so \(x=\sqrt{e}\). 4. Since \(g''(\sqrt{e})=-\frac{2}{e^2}<0\), the point is a local maximum. Its y-coordinate is \(g(\sqrt{e})=\frac{1}{2e}\).

a) Domain: \((0,\infty)\). Also, \(\lim_{x\to0^+}g(x)=-\infty\) and \(\lim_{x\to\infty}g(x)=0\). b) Local maximum: \(\left(\sqrt{e},\frac{1}{2e}\right)\).

- Find where the first derivative is zero. - Use the sign of the second derivative to identify a global maximum. - Compare the maximum y-value with \(0\).

1. Differentiate: \(f'(x)=2-e^{x+1}\). The critical-point equation gives \(e^{x+1}=2\), so \(x=\ln(2)-1\). 2. The second derivative is \(f''(x)=-e^{x+1}<0\) for every real \(x\), so the graph is concave down everywhere and the critical point is a global maximum. 3. The maximum value is \(f(\ln(2)-1)=2\ln(2)-4\), which is negative. 4. Because the greatest value of \(f\) is less than \(0\), every function value is negative. Therefore, the graph has no x-intercepts.

Global maximum: \(\left(\ln(2)-1,2\ln(2)-4\right)\). Since \(2\ln(2)-4<0\), the function has no real zeros.

- A horizontal tangent has slope zero. - The argument of a logarithm must be positive. - Determine what happens when the logarithm's argument approaches \(0^+\). - Use the second derivative to classify the critical point.

1. Differentiate using the chain rule: \(f_k'(x)=\frac{k-2x}{kx-x^2}\). 2. A horizontal tangent at \(x=2\) requires \(f_k'(2)=0\). Therefore, \(k-4=0\), so \(k=4\). 3. For \(k=4\), the logarithm requires \(4x-x^2=x(4-x)>0\). Thus the maximal domain is \((0, 4)\). 4. As \(x\to0^+\) or \(x\to4^-\), the logarithm's argument approaches \(0^+\), so the function approaches \(-\infty\). The vertical asymptotes are \(x=0\) and \(x=4\). 5. For \(k=4\), \(f_4''(2)=-\frac{1}{2}<0\). Therefore, the point is a local maximum.

a) \(k=4\) b) Domain: \((0, 4)\); vertical asymptotes: \(x=0\) and \(x=4\) c) The point is a local maximum.

- Find the first derivative. - Factor the derivative to find all critical numbers. - Use the second derivative at each critical number. - Verify that every critical number is classified as a maximum or minimum.

1. Differentiate: \(g'(x) = x^3 - 5x^2 + 6x = x(x - 2)(x - 3)\). 2. The critical numbers are \(x = 0\), \(x = 2\), and \(x = 3\). 3. Use \(g''(x) = 3x^2 - 10x + 6\). Since \(g''(0) = 6 > 0\), \(x = 0\) gives a local minimum. Since \(g''(2) = -2 < 0\), \(x = 2\) gives a local maximum. Since \(g''(3) = 3 > 0\), \(x = 3\) gives a local minimum. 4. Each critical number produces a local extremum, so the function has exactly three local extrema.

\(g\) has local minima at \(x = 0\) and \(x = 3\), and a local maximum at \(x = 2\). Therefore, it has exactly three local extrema.

- Rewrite \(\frac{1}{x}\) as \(x^{-1}\) before differentiating. - Set the derivative equal to zero and clear the denominator carefully. - Explain why the derivative cannot be zero when \(x < 0\). - Use the second derivative to classify the critical point.

1. Differentiate: \(f'(x) = \frac{2}{3}x - \frac{18}{x^2}\). 2. Solve \(f'(x) = 0\): \(\frac{2}{3}x = \frac{18}{x^2}\), so \(x^3 = 27\) and \(x = 3\). 3. For \(x < 0\), both terms in \(f'(x) = \frac{2}{3}x - \frac{18}{x^2}\) are negative, so there are no critical numbers on the negative part of the domain. 4. At \(x = 3\), \(f''(3) = \frac{2}{3} + \frac{36}{27} = 2 > 0\), so \(f\) has a local minimum there. 5. Since \(f(3) = 9\), the only local extremum is \((3, 9)\).

The function has exactly one local extremum, a local minimum at \((3, 9)\).

- “Sufficient” means the conditions guarantee the conclusion. - A horizontal tangent need not be a peak or valley. - Use a cubic function as a counterexample.

1. The conditions \(f'(x_0)=0\) and \(f''(x_0)<0\) are sufficient for a local maximum at \(x_0\). 2. A zero derivative only guarantees a horizontal tangent. It does not guarantee a change from increasing to decreasing or vice versa. 3. For \(f(x)=x^3\) at \(x_0=0\), \(f'(0)=0\), but the function has a stationary inflection point rather than a local extremum.

a) The conditions \(f'(x_0)=0\) and \(f''(x_0)<0\) are sufficient for a local maximum at \(x_0\). b) They are not sufficient. For \(f(x)=x^3\), \(f'(0)=0\), but no local extremum occurs at \(x_0=0\).

- A horizontal tangent occurs where the first derivative is \(0\). - Check the derivative's sign on both sides of \(5\). - Compare the second derivative result with the actual behavior of the function. - Separate necessary conditions from sufficient tests.

1. \(f'(x)=4(x-5)^3\), so \(f'(x)=0\) at \(x_0=5\). 2. For \(x<5\), \(f'(x)<0\); for \(x>5\), \(f'(x)>0\). Therefore, \(f\) has a local minimum at \(x=5\). 3. \(f''(x)=12(x-5)^2\), so \(f''(5)=0\). 4. Since a local minimum exists even though the second derivative is \(0\), the condition \(f''(x_0)\neq0\) is not necessary.

a) \(f'(x)=4(x-5)^3\), and \(x_0=5\) b) \(f'\) changes from negative to positive, so there is a local minimum. c) \(f''(5)=0\) d) No. A local extremum can occur when the second derivative equals \(0\).

- Differentiate and set the first derivative equal to zero. - Use exact unit-circle values for cosine. - Evaluate the second derivative at each critical number.

1. Differentiate: \(f'(x)=1-2\cos x\). 2. Set the derivative equal to zero: \(1-2\cos x=0\), so \(\cos x=\frac{1}{2}\). In \((0, 2\pi)\), the critical numbers are \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\). 3. Differentiate again: \(f''(x)=2\sin x\). 4. At \(x=\frac{\pi}{3}\), \(f''\left(\frac{\pi}{3}\right)=\sqrt{3}>0\), so \(f\) has a local minimum. At \(x=\frac{5\pi}{3}\), \(f''\left(\frac{5\pi}{3}\right)=-\sqrt{3}<0\), so \(f\) has a local maximum.

Local minimum at \(x=\frac{\pi}{3}\); local maximum at \(x=\frac{5\pi}{3}\)

- Factor the polynomial to find its zeros. - Use the first and second derivatives for local extrema. - Use the second and third derivatives for the inflection point. - To test point symmetry, recenter the function at the suspected center.

1. The function is neither even nor odd, so the graph is not symmetric about the y-axis or the origin. Rewriting around \(x=2\) gives \(f(2+u)-2=u^3-3u\), an odd expression in \(u\). Therefore, the graph has rotational symmetry about \((2, 2)\). 2. Factor: \(f(x)=x(x-3)^2\). The x-intercepts are \((0, 0)\) and \((3, 0)\), with \(x=3\) a double zero. The y-intercept is \((0, 0)\). 3. The derivative is \(f'(x)=3(x-1)(x-3)\), so the critical values are \(x=1\) and \(x=3\). Since \(f''(x)=6x-12\), \((1, 4)\) is a local maximum and \((3, 0)\) is a local minimum. 4. The second derivative is zero at \(x=2\), and the third derivative is \(6\ne0\). Thus, the inflection point is \((2, 2)\).

Symmetry: rotational symmetry about \((2, 2)\), but no y-axis or origin symmetry Intercepts: \((0, 0)\) and \((3, 0)\) Local maximum: \((1, 4)\) Local minimum: \((3, 0)\) Inflection point: \((2, 2)\)

- Use the powers of \(x\) to test symmetry. - Factor out \(x\) to find the zeros. - Use the first and second derivatives to find and classify extrema. - Use the cubic end behavior to determine the range.

1. Since \(g(-x)=-g(x)\), the graph is symmetric about the origin. 2. The leading term is \(\frac{1}{4}x^3\), so \(g(x)\to\infty\) as \(x\to\infty\), and \(g(x)\to-\infty\) as \(x\to-\infty\). 3. Factoring gives \(g(x)=x(\frac{1}{4}x^2-3)\). The intercepts are \((0, 0)\) and \((\pm2\sqrt{3}, 0)\). 4. The derivative is \(g'(x)=\frac{3}{4}x^2-3\), so the critical values are \(x=\pm2\). Since \(g''(x)=\frac{3}{2}x\), \((-2, 4)\) is a local maximum and \((2, -4)\) is a local minimum. 5. The second derivative is zero at \(x=0\), and the third derivative is nonzero. Thus, the inflection point is \((0, 0)\). 6. The cubic is continuous and has opposite infinite end behavior, so its range is \(\mathbb{R}\).

Symmetry: about the origin End behavior: \(g(x)\to\infty\) as \(x\to\infty\), and \(g(x)\to-\infty\) as \(x\to-\infty\) Intercepts: \((0, 0)\), \((\pm2\sqrt{3}, 0)\) Local maximum: \((-2, 4)\); local minimum: \((2, -4)\) Inflection point: \((0, 0)\) Range: \(\mathbb{R}\)

- Check whether the function is even or odd. - Use the leading term for end behavior. - Read the zeros from the factored form. - Apply the derivative tests for extrema and inflection points.

1. Expanding gives \(f(x)=3x^2-\frac{1}{2}x^4\), an even function. Therefore, the graph is symmetric about the y-axis. 2. The leading term is \(-\frac{1}{2}x^4\), so \(f(x)\to-\infty\) as \(x\to\pm\infty\). 3. From the factored form, the zeros are \(x=0\), with multiplicity \(2\), and \(x=\pm\sqrt{6}\). 4. The derivative is \(f'(x)=2x(3-x^2)\), so the critical values are \(x=0\) and \(x=\pm\sqrt{3}\). Since \(f''(x)=6-6x^2\), \((0, 0)\) is a local minimum and \((\pm\sqrt{3}, \frac{9}{2})\) are local maxima. 5. The second derivative is zero at \(x=\pm1\), and the third derivative is nonzero there. The inflection points are \((\pm1, \frac{5}{2})\).

Symmetry: about the y-axis End behavior: \(f(x)\to-\infty\) as \(x\to\pm\infty\) Zeros: \(x=0\) with multiplicity \(2\), and \(x=\pm\sqrt{6}\) Local minimum: \((0, 0)\); local maxima: \((\pm\sqrt{3}, \frac{9}{2})\) Inflection points: \((\pm1, \frac{5}{2})\)

- Read zeros and multiplicities from the factored form. - Expand the function to identify the leading term. - Use the first and second derivatives for the extrema. - A cubic's inflection value lies midway between its two critical values.

1. The factored form gives a zero at \(x=-2\) with multiplicity \(2\) and a zero at \(x=1\) with multiplicity \(1\). 2. Expanding gives \(f(x)=x^3+3x^2-4\). Because this is a cubic with positive leading coefficient, \(f(x)\to-\infty\) as \(x\to-\infty\), and \(f(x)\to\infty\) as \(x\to\infty\). 3. The derivative is \(f'(x)=3x(x+2)\), so the critical values are \(x=-2\) and \(x=0\). Since \(f''(x)=6x+6\), \((-2, 0)\) is a local maximum and \((0, -4)\) is a local minimum. 4. The second derivative is zero at \(x=-1\), and the third derivative is nonzero. Therefore, the inflection point is \((-1, -2)\).

Zeros: \(x=-2\), multiplicity \(2\); \(x=1\), multiplicity \(1\) End behavior: \(f(x)\to-\infty\) as \(x\to-\infty\), and \(f(x)\to\infty\) as \(x\to\infty\) Local maximum: \((-2, 0)\); local minimum: \((0, -4)\) Inflection point: \((-1, -2)\)

- Try factoring the derivative by grouping. - After finding a factor, continue factoring completely. - Use the second derivative to classify the critical numbers. - Check signs carefully when evaluating the original function.

1. Differentiate: \(f'(x) = x^3 - 3x^2 - 4x + 12\). 2. Factor by grouping: \(f'(x) = x^2(x - 3) - 4(x - 3) = (x - 3)(x - 2)(x + 2)\). The critical numbers are \(x = -2\), \(x = 2\), and \(x = 3\). 3. Use \(f''(x) = 3x^2 - 6x - 4\). Since \(f''(-2) = 20 > 0\), \((-2, -20)\) is a local minimum. 4. Since \(f''(2) = -4 < 0\), \((2, 12)\) is a local maximum. 5. Since \(f''(3) = 5 > 0\), \((3, \frac{45}{4})\) is a local minimum.

Local minima: \((-2, -20)\) and \((3, \frac{45}{4})\) Local maximum: \((2, 12)\)

- Check whether the function is even or odd. - Use the leading term for end behavior. - Use the first and second derivatives to find and classify extrema. - Substitute the inflection value into the original function.

1. The function is neither even nor odd, so it has no y-axis or origin symmetry. The leading term is \(\frac{1}{3}x^3\), so \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\). 2. The derivative is \(f'(x)=x^2-2x-3=(x-3)(x+1)\). The critical values are \(x=-1\) and \(x=3\). Since \(f''(x)=2x-2\), \((-1, \frac{16}{3})\) is a local maximum and \((3, -\frac{16}{3})\) is a local minimum. 3. The second derivative is zero at \(x=1\), and the third derivative is \(2\ne0\). Since \(f(1)=0\), the inflection point is \((1, 0)\).

1. No y-axis or origin symmetry; \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\) 2. Local maximum: \((-1, \frac{16}{3})\); local minimum: \((3, -\frac{16}{3})\) 3. Inflection point: \((1, 0)\)

- Use even powers to identify symmetry. - Substitute \(u=x^2\) for the quartic equation. - Use the first and second derivatives for extrema. - Verify the second-derivative zeros with the third derivative.

1. Only even powers occur, so the graph is symmetric about the y-axis. 2. The y-intercept is \((0, -2)\). To find the x-intercepts, let \(u=x^2\). The equation \(-\frac{1}{8}x^4+x^2-2=0\) becomes \(u^2-8u+16=(u-4)^2=0\), so \(x=\pm2\). The x-intercepts are \((-2, 0)\) and \((2, 0)\). 3. The derivative is \(g'(x)=-\frac{1}{2}x(x^2-4)\), so the critical values are \(x=-2, 0, 2\). Since \(g''(x)=-\frac{3}{2}x^2+2\), \((0, -2)\) is a local minimum, and \((-2, 0)\) and \((2, 0)\) are local maxima. 4. Solving \(g''(x)=0\) gives \(x=\pm\frac{2}{\sqrt{3}}\). The third derivative is nonzero at both values, and \(g(\pm\frac{2}{\sqrt{3}})=-\frac{8}{9}\). Therefore, the inflection points are \((\pm\frac{2}{\sqrt{3}}, -\frac{8}{9})\).

Symmetry: about the y-axis Intercepts: \((0, -2)\), \((-2, 0)\), \((2, 0)\) Local minimum: \((0, -2)\); local maxima: \((-2, 0)\), \((2, 0)\) Inflection points: \((\pm\frac{2}{\sqrt{3}}, -\frac{8}{9})\)

- Use the leading term for end behavior. - Test whether the function is even. - Factor out \(x^2\) to find the zeros. - Compare the degrees of the first and second derivatives with the maximum numbers of their real zeros.

1. The degree is even and the leading coefficient is positive, so \(f(x)\to\infty\) as \(x\to\pm\infty\). 2. Since \(f(-x)=f(x)\), the graph is symmetric about the y-axis. 3. Factor: \(f(x)=x^2(\frac{1}{8}x^2-1)\). The zeros are \(x=0\), with multiplicity \(2\), and \(x=\pm2\sqrt{2}\). 4. A fourth-degree polynomial can have at most \(3\) local extrema because its first derivative has degree \(3\), and at most \(2\) inflection values because its second derivative has degree \(2\). Here, \(f'(x)=\frac{1}{2}x(x^2-4)\), so the function has \(3\) local extrema at \(x=-2, 0, 2\). Also, \(f''(x)=\frac{3}{2}x^2-2\), so it has \(2\) inflection values at \(x=\pm\frac{2}{\sqrt{3}}\). The function reaches both maximum possible counts.

1. \(f(x)\to\infty\) as \(x\to\pm\infty\) 2. Symmetric about the y-axis 3. \(x=0\), multiplicity \(2\), and \(x=\pm2\sqrt{2}\) 4. Maximum possible: \(3\) local extrema and \(2\) inflection values. This function has exactly \(3\) local extrema and \(2\) inflection values.

- Factor the polynomial to find its zeros. - Use the first and second derivatives for local extrema. - What degree is the second derivative of a cubic? - Substitute each x-coordinate into the original function.

1. Factor: \(f(x)=\frac{1}{3}(x-3)^2(x+3)\). Thus, \(x=3\) is a double zero and \(x=-3\) is a simple zero. 2. The derivative is \(f'(x)=x^2-2x-3=(x-3)(x+1)\), so the critical values are \(x=-1\) and \(x=3\). Since \(f''(x)=2x-2\), \((-1, \frac{32}{3})\) is a local maximum and \((3, 0)\) is a local minimum. 3. Solving \(f''(x)=0\) gives \(x=1\), and \(f(1)=\frac{16}{3}\). Thus, the inflection point is \((1, \frac{16}{3})\). For any polynomial of degree exactly \(3\), the second derivative is a nonconstant linear function. It has exactly one real zero and changes sign there, so the cubic has exactly one inflection point.

1. \(x=-3\), multiplicity \(1\); \(x=3\), multiplicity \(2\) 2. Local maximum: \((-1, \frac{32}{3})\); local minimum: \((3, 0)\) 3. Inflection point: \((1, \frac{16}{3})\). A cubic's second derivative is a nonconstant linear function, so it has exactly one sign-changing zero.

- Check whether the function is even. - Recognize the polynomial as a perfect square. - Use the first and second derivatives for extrema. - Verify the second-derivative zeros with the third derivative.

1. Since \(f(-x)=f(x)\), the graph is symmetric about the y-axis. 2. The y-intercept is \((0, 81)\). Also, \(f(x)=(x^2-9)^2\), so the x-intercepts are \((-3, 0)\) and \((3, 0)\), both corresponding to double zeros. 3. The derivative is \(f'(x)=4x(x^2-9)\), so the critical values are \(x=-3, 0, 3\). Since \(f''(x)=12x^2-36\), \((0, 81)\) is a local maximum and \((-3, 0)\) and \((3, 0)\) are local minima. 4. Solving \(f''(x)=0\) gives \(x=\pm\sqrt{3}\). The third derivative is nonzero there, and \(f(\pm\sqrt{3})=36\). Thus, the inflection points are \((\pm\sqrt{3}, 36)\).

Symmetry: about the y-axis Intercepts: \((0, 81)\), \((-3, 0)\), \((3, 0)\) Local maximum: \((0, 81)\); local minima: \((-3, 0)\), \((3, 0)\) Inflection points: \((\pm\sqrt{3}, 36)\)

- Check whether the function is even or odd. - Use the leading term to determine the end behavior. - Substitute \(u=x^2\) when solving for the zeros. - Use the first three derivatives to find and classify extrema and inflection points.

1. Since \(f(-x)=f(x)\), the graph is symmetric about the y-axis. Because the leading term has even degree and a positive coefficient, \(f(x)\to\infty\) as \(x\to\pm\infty\). 2. To find the x-intercepts, solve \(\frac{1}{4}x^4-\frac{5}{2}x^2+4=0\). Let \(u=x^2\). Then \(u^2-10u+16=0\), so \(u=2\) or \(u=8\). Therefore, the x-intercepts are \((\pm\sqrt{2}, 0)\) and \((\pm2\sqrt{2}, 0)\). The y-intercept is \((0, 4)\). 3. The derivative is \(f'(x)=x^3-5x=x(x^2-5)\), so the critical values are \(x=0\) and \(x=\pm\sqrt{5}\). Since \(f''(x)=3x^2-5\), \((0, 4)\) is a local maximum and \((\pm\sqrt{5}, -\frac{9}{4})\) are local minima. 4. Solving \(f''(x)=0\) gives \(x=\pm\sqrt{\frac{5}{3}}\). Because \(f'''(x)=6x\) is nonzero at both values, the inflection points are \((\pm\sqrt{\frac{5}{3}}, \frac{19}{36})\).

Symmetry: about the y-axis End behavior: \(f(x)\to\infty\) as \(x\to\pm\infty\) Intercepts: \((0, 4)\), \((\pm\sqrt{2}, 0)\), and \((\pm2\sqrt{2}, 0)\) Local maximum: \((0, 4)\); local minima: \((\pm\sqrt{5}, -\frac{9}{4})\) Inflection points: \((\pm\sqrt{\frac{5}{3}}, \frac{19}{36})\)

- Compare \(f(-x)\) with \(f(x)\) and \(-f(x)\). - Use the leading term to determine the end behavior. - Apply the derivative tests to find extrema and the inflection point. - Use digital graphing or algebra technology when the zeros do not factor conveniently.

1. The function is neither even nor odd, so its graph has neither y-axis symmetry nor origin symmetry. Since the leading term is \(\frac{1}{3}x^3\), \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\). 2. The derivative is \(f'(x)=x^2-x-2=(x-2)(x+1)\). The critical values are \(x=-1\) and \(x=2\). Since \(f''(x)=2x-1\), \((-1, \frac{13}{6})\) is a local maximum and \((2, -\frac{7}{3})\) is a local minimum. 3. Solving \(f''(x)=0\) gives \(x=\frac{1}{2}\). Because \(f'''(x)=2\ne0\), the inflection point is \((\frac{1}{2}, -\frac{1}{12})\). 4. Graphing technology or a computer algebra system gives the zeros \(x\approx-2.08\), \(x\approx0.46\), and \(x\approx3.12\).

Symmetry: neither y-axis symmetry nor origin symmetry End behavior: \(f(x)\to\infty\) as \(x\to\infty\), and \(f(x)\to-\infty\) as \(x\to-\infty\) Local maximum: \((-1, \frac{13}{6})\); local minimum: \((2, -\frac{7}{3})\) Inflection point: \((\frac{1}{2}, -\frac{1}{12})\) Zeros: \(x\approx-2.08\), \(x\approx0.46\), and \(x\approx3.12\)

- Complete the square in terms of \(x^2\). - First find critical numbers from the first derivative. - Use the second derivative test separately to classify each critical number. - A positive factor does not change a derivative's sign.

1. Complete the square: \(x^4-8x^2+25=(x^2-4)^2+9\ge9\). Thus, \(f(x)\ge3>0\). 2. For \(g\), \(g'(x)=4x^3-16x=4x(x^2-4)\), so the critical numbers are \(x=-2,0,2\). 3. The second derivative is \(g''(x)=12x^2-16\). Since \(g''(0)=-16<0\), \(g\) has a local maximum at \((0,25)\). Since \(g''(\pm2)=32>0\), \(g\) has local minima at \((-2,9)\) and \((2,9)\). 4. By the chain rule, \(f'(x)=\frac{g'(x)}{2\sqrt{g(x)}}\). The denominator is always positive, so \(f'\) and \(g'\) have the same zeros and signs. Therefore, their local extrema occur at the same x-values and have the same types.

1. \(x^4-8x^2+25=(x^2-4)^2+9\ge9\), so \(f(x)>0\). 2. Local maximum at \((0,25)\); local minima at \((-2,9)\) and \((2,9)\) 3. Since \(f'(x)=\frac{g'(x)}{2f(x)}\) and \(f(x)>0\), the derivatives have the same zeros and signs.

- Use only even powers in the general quartic form. - Use the y-intercept and the given point to form two equations. - At a local extremum, the first derivative is \(0\). - Verify the local minimum with the second derivative.

1. Because \(f\) is an even quartic polynomial, let \(f(x)=ax^4+cx^2+e\). Then \(f'(x)=4ax^3+2cx\). 2. The y-intercept gives \(f(0)=5\), so \(e=5\). The point \(P(1, -2)\) gives \(a+c=-7\). 3. Since \(x=2\) is a local minimum, \(f'(2)=0\). Thus, \(32a+4c=0\), or \(8a+c=0\). 4. Solving \(a+c=-7\) and \(8a+c=0\) gives \(a=1\) and \(c=-8\). Therefore, \(f(x)=x^4-8x^2+5\). 5. The second derivative is \(f''(x)=12x^2-16\), and \(f''(2)=32>0\), confirming that \(x=2\) is a local minimum.

\(f(x)=x^4-8x^2+5\)

- Begin with the general form of a cubic polynomial. - Translate the local-minimum coordinates into conditions on \(f\) and \(f'\). - Translate the inflection-point coordinates into conditions on \(f\) and \(f''\). - Verify both point types after finding the coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Then \(f'(x)=3ax^2+2bx+c\) and \(f''(x)=6ax+2b\). 2. The local minimum at \(T(0, 2)\) gives \(f(0)=2\) and \(f'(0)=0\). Therefore, \(d=2\) and \(c=0\). 3. The inflection point at \(W(1, 4)\) gives \(f(1)=4\) and \(f''(1)=0\). Thus, \(a+b=2\) and \(6a+2b=0\). 4. Solving the system gives \(a=-1\) and \(b=3\). Therefore, \(f(x)=-x^3+3x^2+2\). 5. Since \(f''(0)=6>0\), \(T(0, 2)\) is a local minimum. Also, the third derivative is \(-6\ne0\), confirming the inflection point at \(W(1, 4)\).

\(f(x)=-x^3+3x^2+2\)

- An even quartic contains only even powers of \(x\). - A point on the graph gives an equation involving \(f\). - A local extremum gives both a function-value condition and a first-derivative condition.

1. Because the polynomial is even, let \(f(x)=ax^4+cx^2+e\). 2. Since \(P(0, 4)\) lies on the graph, \(e=4\). The point \(E(2, 0)\) gives \(f(2)=0\), so \(16a+4c+4=0\), or \(4a+c=-1\). 3. Because \(E(2, 0)\) is a local extremum, \(f'(2)=0\). Since \(f'(x)=4ax^3+2cx\), this gives \(32a+4c=0\), or \(8a+c=0\). 4. Solving the system gives \(a=\frac{1}{4}\) and \(c=-2\). Therefore, \(f(x)=\frac{1}{4}x^4-2x^2+4\). 5. Since \(f''(x)=3x^2-4\) and \(f''(2)=8>0\), \(E(2,0)\) is indeed a local minimum.

\(f(x)=\frac{1}{4}x^4-2x^2+4\)

- Begin with the general form of a cubic polynomial. - An inflection point gives conditions on the function and second derivative. - A local minimum gives conditions on the function and first derivative. - Verify the point classifications after finding the coefficients.

1. Let \(f(x)=ax^3+bx^2+cx+d\). The inflection point at \(W(0, 1)\) gives \(f(0)=1\) and \(f''(0)=0\), so \(d=1\) and \(b=0\). 2. The local minimum at \(T(1, 0)\) gives \(f(1)=0\) and \(f'(1)=0\). Therefore, \(a+c=-1\) and \(3a+c=0\). 3. Solving the system gives \(a=\frac{1}{2}\) and \(c=-\frac{3}{2}\). Thus, \(f(x)=\frac{1}{2}x^3-\frac{3}{2}x+1\). 4. Since \(f''(1)=3>0\), \(T(1, 0)\) is a local minimum. The nonzero third derivative confirms the inflection point.

\(f(x)=\frac{1}{2}x^3-\frac{3}{2}x+1\)

- Use only even powers for an even quartic polynomial. - Use the y-intercept and the given point to form equations. - At a local minimum, the first derivative is \(0\). - Verify the classification with the second derivative.

1. Because the polynomial is even, let \(f(x)=ax^4+bx^2+c\). The y-intercept gives \(c=2\). 2. The point \(P(2, -6)\) gives \(16a+4b+2=-6\), or \(16a+4b=-8\). 3. Since \(x=2\) is a local minimum, \(f'(2)=0\). Because \(f'(x)=4ax^3+2bx\), this gives \(32a+4b=0\). 4. Solving the system gives \(a=\frac{1}{2}\) and \(b=-4\). Therefore, \(f(x)=\frac{1}{2}x^4-4x^2+2\). 5. Since \(f''(x)=6x^2-8\), \(f''(2)=16>0\), confirming the local minimum.

\(f(x)=\frac{1}{2}x^4-4x^2+2\)

- Find the positive local-minimum x-coordinate and the positive inflection-point x-coordinate separately. - Form the ratio only after expressing both coordinates in terms of \(a\). - Simplify the radicals.

1. The derivatives are \(g_a'(x)=4ax^3-4x=4x(ax^2-1)\) and \(g_a''(x)=12ax^2-4\). 2. The positive critical number is \(x_{\min}=\frac{1}{\sqrt{a}}\). Because \(g_a''\left(\frac{1}{\sqrt{a}}\right)=8>0\), it is the x-coordinate of a local minimum. 3. The positive solution of \(g_a''(x)=0\) is \(x_{\mathrm{infl}}=\frac{1}{\sqrt{3a}}\). The second derivative changes sign there, so it is an inflection-point x-coordinate. 4. Therefore, \(\frac{x_{\min}}{x_{\mathrm{infl}}}=\frac{1/\sqrt{a}}{1/\sqrt{3a}}=\sqrt{3}\), which is independent of \(a\).

\(\frac{x_{\min}}{x_{\mathrm{infl}}}=\sqrt{3}\)

- Compare \(g_a(-x)\) with \(g_a(x)\). - Find the critical numbers and use the second derivative test. - Substitute the critical numbers into the original function. - Apply the restriction \(a > 0\) when solving for \(a\).

1. Evaluate \(g_a(-x)\): \((-x)^4 - 2a(-x)^2 = x^4 - 2ax^2 = g_a(x)\). Therefore, every graph is symmetric about the \(y\)-axis. 2. Differentiate: \(g_a'(x) = 4x^3 - 4ax = 4x(x^2 - a)\). For \(a > 0\), the critical numbers are \(x = 0\) and \(x = \pm\sqrt{a}\). 3. The second derivative is \(g_a''(x) = 12x^2 - 4a\). At \(x = \pm\sqrt{a}\), \(g_a''(x) = 8a > 0\), so both points are local minima. 4. Their function value is \(g_a(\pm\sqrt{a}) = a^2 - 2a^2 = -a^2\). Thus, the local minima are \(\left(-\sqrt{a}, -a^2\right)\) and \(\left(\sqrt{a}, -a^2\right)\). 5. Set \(-a^2 = -25\). Since \(a > 0\), \(a = 5\).

a) \(g_a(-x) = g_a(x)\), so every graph is symmetric about the \(y\)-axis. b) \(\left(-\sqrt{a}, -a^2\right)\) and \(\left(\sqrt{a}, -a^2\right)\) c) \(a = 5\)

- Begin with the general cubic form. - Translate the point, extremum, and inflection conditions into equations. - Use the second-derivative condition first to simplify the system. - Classify the extremum after finding the polynomial.

1. Let \(f(x)=ax^3+bx^2+cx+d\). Passing through the origin gives \(d=0\). 2. The local extremum at \(E(2, -4)\) gives \(f(2)=-4\) and \(f'(2)=0\). The inflection condition gives \(f''(1)=0\). 3. From \(f''(1)=6a+2b=0\), \(b=-3a\). Substituting into \(f'(2)=12a+4b+c=0\) gives \(c=0\). 4. Using \(f(2)=-4\) gives \(8a+4(-3a)=-4\), so \(a=1\) and \(b=-3\). Therefore, \(f(x)=x^3-3x^2\). 5. Since \(f''(2)=6>0\), the extremum at \(E(2, -4)\) is a local minimum. Also, \(f'''(x)=6\ne0\), confirming the inflection point at \(x=1\).

\(f(x)=x^3-3x^2\)

- Test the endpoint \(x=0\) for a common point. - Differentiate for \(x>0\) and solve the critical-point equation. - Eliminate \(k\) from the minimum coordinates. - Include the attainable x-values in the locus.

1. Since \(f_k(0)=0\), every graph passes through \((0, 0)\). 2. For \(x>0\), \(f_k'(x)=1-\frac{k}{2\sqrt{x}}\). Setting this equal to zero gives \(\sqrt{x}=\frac{k}{2}\), so \(x=\frac{k^2}{4}\). 3. The second derivative is \(f_k''(x)=\frac{k}{4x^{3/2}}>0\), so this critical point is the unique local minimum. Its y-coordinate is \(-\frac{k^2}{4}\). 4. The minimum coordinates satisfy \(y=-x\). Since \(k>0\), \(x=\frac{k^2}{4}>0\), so the locus is \(y=-x\) for \(x>0\).

a) \((0, 0)\) b) \(\left(\frac{k^2}{4}, -\frac{k^2}{4}\right)\) c) \(y=-x\) for \(x>0\)

- Find and classify both critical points. - Substitute the minimum x-coordinate into the function. - Eliminate the parameter and preserve its domain restriction.

1. The derivatives are \(f_a'(x)=3x^2-6ax=3x(x-2a)\) and \(f_a''(x)=6x-6a\). 2. The critical points are \(x=0\) and \(x=2a\). Since \(a>0\), \(f_a''(0)=-6a<0\) and \(f_a''(2a)=6a>0\). Thus the local minimum occurs at \(x=2a\). 3. Its y-coordinate is \(f_a(2a)=-4a^3+4\), so the minimum is \((2a, -4a^3+4)\). 4. Since \(a=\frac{x}{2}\), the locus is \(y=-4\left(\frac{x}{2}\right)^3+4=-\frac{1}{2}x^3+4\). Because \(a>0\), the locus has \(x>0\).

a) \((2a, -4a^3+4)\) b) \(y=-\frac{1}{2}x^3+4\) for \(x>0\)

- Find both critical points. - Express the parameter using the moving critical-point x-coordinate. - Check that the fixed extremum also lies on the resulting curve.

1. The derivative is \(f_a'(x)=3x(x-2a)\), so the critical points occur at \(x=0\) and \(x=2a\). The second derivative is nonzero at both points because \(a\ne0\), so both are local extrema. 2. The fixed extremum is \((0, 2)\). For the moving extremum, \(x=2a\), so \(a=\frac{x}{2}\). 3. Substituting into the family gives \(y=x^3-3\left(\frac{x}{2}\right)x^2+2=-\frac{1}{2}x^3+2\). 4. The fixed extremum \((0, 2)\) also lies on this curve. Therefore, all extrema lie on \(y=-\frac{1}{2}x^3+2\).

\(y=-\frac{1}{2}x^3+2\)

- Set the first derivative equal to zero and use the critical point that is not the origin. - Eliminate the parameter between the extremum coordinates. - Use the second derivative to find the inflection point. - Simplify the ratio and check whether the parameter remains.

1. Differentiate: \(f_k'(x)=2kx-3x^2=x(2k-3x)\). The critical point away from the origin has x-coordinate \(x_{\mathrm{ext}}=\frac{2k}{3}\). Its y-coordinate is \(f_k\left(\frac{2k}{3}\right)=\frac{4k^3}{27}\), so the point is \(\left(\frac{2k}{3}, \frac{4k^3}{27}\right)\). Since \(f_k''\left(\frac{2k}{3}\right)=-2k\), it is a local maximum for \(k>0\) and a local minimum for \(k<0\). 2. From \(x=\frac{2k}{3}\), obtain \(k=\frac{3x}{2}\). Substituting into \(y=\frac{4k^3}{27}\) gives \(y=\frac{1}{2}x^3\). Because \(k\ne0\), the locus has \(x\ne0\). 3. Since \(f_k''(x)=2k-6x\), the inflection-point x-coordinate is \(x_{\mathrm{infl}}=\frac{k}{3}\). Also, \(f_k'''(x)=-6\neq0\), so this is an inflection point. Therefore, \(\frac{x_{\mathrm{ext}}}{x_{\mathrm{infl}}}=\frac{2k/3}{k/3}=2\), independent of \(k\).

1. \(\left(\frac{2k}{3}, \frac{4k^3}{27}\right)\); local maximum for \(k>0\), local minimum for \(k<0\) 2. \(y=\frac{1}{2}x^3\) for \(x\ne0\) 3. \(x_{\mathrm{infl}}=\frac{k}{3}\) and \(\frac{x_{\mathrm{ext}}}{x_{\mathrm{infl}}}=2\)

- Check how the parameter term behaves at a common point. - Factor the function to analyze its zeros. - Use the first and second derivatives to classify the critical points. - Eliminate \(k\) between the coordinates of the local minimum.

1. Since \(f_k(0)=0\), every graph passes through the origin. If a point with x-coordinate \(x\) were common to all graphs, the term \(-kx\) would have to be independent of \(k\), which occurs only when \(x=0\). Thus there is no other common point. 2. Factoring gives \(f_k(x)=x(x^2-k)\). There are three distinct real zeros, \(0\) and \(\pm\sqrt{k}\), exactly when \(k>0\). 3. Since \(f_k'(x)=3x^2-k\), the critical points are \(x=\pm\sqrt{\frac{k}{3}}\). Because \(f_k''(x)=6x\), the positive critical point is a local minimum. Therefore, the local minimum is \(\left(\sqrt{\frac{k}{3}}, -\frac{2k}{3}\sqrt{\frac{k}{3}}\right)\). 4. At the local minimum, \(x=\sqrt{\frac{k}{3}}\), so \(k=3x^2\). Substitution gives \(y=-\frac{2(3x^2)}{3}x=-2x^3\). Because \(k>0\), the minimum has \(x>0\).

a) The origin is the only common point. b) \(k>0\) c) \(\left(\sqrt{\frac{k}{3}}, -\frac{2k}{3}\sqrt{\frac{k}{3}}\right)\) d) \(y=-2x^3\) for \(x>0\)

- Factor the first derivative to find the critical points. - Treat the cases \(k>0\), \(k<0\), and \(k=0\) separately. - Eliminate \(k\) using the critical-point equation. - Check directly what happens when the second derivative test is inconclusive.

1. Differentiate: \(f_k'(x)=x^3-2kx=x(x^2-2k)\) and \(f_k''(x)=3x^2-2k\). 2. For \(k>0\), the nonzero critical points are \(x=\pm\sqrt{2k}\). At either point, \(f_k''(x)=4k>0\), so both are local minima. Since \(k=\frac{x^2}{2}\), their y-coordinate is \(y=\frac{1}{4}x^4-\frac{x^2}{2}x^2=-\frac{1}{4}x^4\). 3. For \(k<0\), \(f_k(x)=\frac{1}{4}x^4+|k|x^2\geq 0\), with a local minimum at the origin. For \(k=0\), \(f_0(x)=\frac{1}{4}x^4\geq 0\), so the origin is also a local minimum even though the second derivative test is inconclusive there. 4. The origin lies on \(y=-\frac{1}{4}x^4\), and every nonzero x-value on this curve is attained by choosing \(k=\frac{x^2}{2}>0\). Therefore, the locus is \(y=-\frac{1}{4}x^4\).

\(y=-\frac{1}{4}x^4\)

- A graph is tangent to the x-axis when both the function value and derivative are zero. - Find both critical points and evaluate the function at each one. - Eliminate the parameter using the x-coordinate of the second extremum.

1. Differentiate: \(f_a'(x)=3x^2-3a^2=3(x-a)(x+a)\). The critical points occur at \(x=a\) and \(x=-a\). Since \(f_a''(x)=6x\) and \(a\neq0\), both are local extrema. 2. At \(x=a\), \(f_a(a)=0\) and \(f_a'(a)=0\). Therefore, the graph is tangent to the x-axis at \((a, 0)\). 3. The other extremum occurs at \(x=-a\), with \(f_a(-a)=4a^3\). Thus its coordinates are \((-a, 4a^3)\). 4. Since \(x=-a\), substitute \(a=-x\) to obtain \(y=4(-x)^3=-4x^3\). Because \(a\ne0\), the locus has \(x\ne0\).

a) The graph is tangent to the x-axis at \((a, 0)\). b) The other extremum is \((-a, 4a^3)\), and its locus is \(y=-4x^3\) for \(x\ne0\).

- Rewrite \(4^x\) as a power with base \(2\). - Factor out \(2^x\) when solving the equations. - Use the derivative rule for \(a^x\). - Compare the two exponential terms for very large positive and negative values of \(x\).

1. Rewrite \(4^x=(2^x)^2\). Then \(f(x)=2^x(2^x-1)\). Since \(2^x>0\), the only zero satisfies \(2^x=1\), so \(x=0\). 2. Differentiate: \(f'(x)=\ln(2)2^x(2\cdot2^x-1)\). The only critical number satisfies \(2\cdot2^x=1\), so \(x=-1\). 3. The second derivative is \(f''(x)=(\ln(2))^2 2^x(4\cdot2^x-1)\). Thus \(f''(-1)=\frac{1}{2}(\ln(2))^2>0\), so the critical point is a local minimum. Since \(f(-1)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\), the minimum is \(\left(-1,-\frac{1}{4}\right)\). 4. As \(x\to\infty\), the term \(4^x\) dominates, so \(f(x)\to\infty\). As \(x\to-\infty\), both exponential terms approach \(0\), so \(f(x)\to0\).

a) The only zero is \(x=0\), giving the intercept \((0,0)\). b) Local minimum at \(\left(-1,-\frac{1}{4}\right)\). c) \(\lim_{x\to\infty}f(x)=\infty\) and \(\lim_{x\to-\infty}f(x)=0\).

- Recall the range of an exponential function. - Rewrite the critical-point equation so both sides use base \(2\). - Apply the chain rule to the term with the negative exponent. - Consider which exponential term grows in each direction.

1. Both \(2^x\) and \(4\cdot2^{-x}\) are positive for every real \(x\). Their sum is therefore always positive, so \(g\) has no zeros. 2. Differentiate: \(g'(x)=\ln(2)(2^x-4\cdot2^{-x})\). Setting this equal to zero gives \(2^x=4\cdot2^{-x}\). Multiplying by \(2^x\) gives \(2^{2x}=4=2^2\), so \(x=1\). 3. The second derivative is \(g''(x)=(\ln(2))^2(2^x+4\cdot2^{-x})\), which is positive for every real \(x\). Therefore, the critical point is a local minimum. Since \(g(1)=2+2=4\), the minimum is \((1,4)\). 4. As \(x\to\infty\), \(2^x\to\infty\) and \(4\cdot2^{-x}\to0\), so \(g(x)\to\infty\). As \(x\to-\infty\), \(2^x\to0\) and \(4\cdot2^{-x}\to\infty\), so \(g(x)\to\infty\).

a) Both terms are positive for every real \(x\), so \(g(x)>0\) and there are no zeros. b) Local minimum at \((1,4)\). c) \(\lim_{x\to\infty}g(x)=\infty\) and \(\lim_{x\to-\infty}g(x)=\infty\).

- Factor out the common exponential factor. - Solve equations of the form \(e^x=c\) with logarithms. - Use the second derivative test to classify the critical point. - Substitute the critical x-coordinate into the original function.

1. Factor: \(f_a(x)=e^x(e^x-a)\). Since \(e^x\ne0\), the zero satisfies \(e^x=a\), so \(x=\ln(a)\). The x-intercept is \((\ln(a), 0)\). 2. The first derivative is \(f_a'(x)=e^x(2e^x-a)\). A critical point satisfies \(e^x=\frac{a}{2}\), so \(x=\ln\left(\frac{a}{2}\right)\). 3. The second derivative is \(f_a''(x)=4e^{2x}-ae^x\). At the critical point, \(f_a''\left(\ln\left(\frac{a}{2}\right)\right)=\frac{a^2}{2}>0\), so the point is a local minimum. 4. Its y-coordinate is \(\left(\frac{a}{2}\right)^2-a\left(\frac{a}{2}\right)=-\frac{a^2}{4}\).

Zero: \((\ln(a), 0)\) Local minimum: \(\left(\ln\left(\frac{a}{2}\right), -\frac{a^2}{4}\right)\)

- Solve the first-derivative equation on the stated domain. - Use the sign of the second derivative to classify the critical point. - A point on the x-axis has y-coordinate zero. - Use logarithm properties to simplify \(\ln\left(\frac{1}{a}\right)\).

1. The first derivative is \(f_a'(x)=\frac{1}{x}-a\). It equals zero only at \(x=\frac{1}{a}\), which is in the domain because \(a>0\). 2. The second derivative is \(f_a''(x)=-\frac{1}{x^2}<0\) for every \(x>0\). Therefore, the critical point is the unique local maximum. 3. For the maximum to lie on the x-axis, require \(f_a\left(\frac{1}{a}\right)=0\). This gives \(\ln\left(\frac{1}{a}\right)-1=0\), or \(-\ln(a)-1=0\). 4. Thus \(\ln(a)=-1\), so \(a=e^{-1}=\frac{1}{e}\).

a) The unique local maximum occurs at \(x=\frac{1}{a}\). b) \(a=\frac{1}{e}\)

- Solve the first-derivative equation using the range of \(e^x\). - For \(u^2=v^2\), consider both \(u=v\) and \(u=-v\). - A horizontal tangent has slope zero.

1. The derivative is \(f_a'(x)=-2e^x(a-e^x)\). Since \(e^x>0\), a critical point requires \(e^x=a\), which is possible exactly when \(a>0\). Then \(x=\ln(a)\). 2. The second derivative is \(f_a''(x)=4e^{2x}-2ae^x\). At \(x=\ln(a)\), it equals \(2a^2>0\), so the point is a local minimum. Its y-coordinate is \(0\), giving \((\ln(a), 0)\). 3. For an intersection, \((a_1-e^x)^2=(a_2-e^x)^2\). Because \(a_1\ne a_2\), the equal-base case is impossible, so \(a_1-e^x=-(a_2-e^x)\). Thus \(e^x=\frac{a_1+a_2}{2}\). An intersection exists exactly when \(a_1+a_2>0\), and then \(x=\ln\left(\frac{a_1+a_2}{2}\right)\). 4. A horizontal tangent at \(x=0\) requires \(f_a'(0)=-2(a-1)=0\), so \(a=1\).

1. For \(a>0\), the local minimum is \((\ln(a), 0)\). 2. An intersection exists when \(a_1+a_2>0\), at \(x=\ln\left(\frac{a_1+a_2}{2}\right)\). 3. \(a=1\)

- Solve the first-derivative equation by combining the exponential terms. - Check when the resulting right side is positive. - Use the second derivative test for the extremum. - For inflection points, solve the second-derivative equation and verify with the third derivative.

1. The first two derivatives are \(f_a'(x)=e^x-ae^{-x}\) and \(f_a''(x)=e^x+ae^{-x}\). 2. Critical points satisfy \(e^{2x}=a\). If \(a>0\), there is exactly one solution, \(x=\frac{1}{2}\ln(a)\). If \(a<0\), there are no critical points. 3. For \(a>0\), the second derivative at this critical point is \(2\sqrt{a}>0\), so the point is a local minimum. Its y-coordinate is also \(2\sqrt{a}\), giving \(\left(\frac{1}{2}\ln(a), 2\sqrt{a}\right)\). 4. For \(a<0\), inflection points satisfy \(f_a''(x)=0\), or \(e^{2x}=-a\). This has exactly one solution, \(x=\frac{1}{2}\ln(-a)\). 5. The third derivative is \(f_a'''(x)=e^x-ae^{-x}\), and at this x-coordinate it equals \(2\sqrt{-a}\ne0\). Therefore, the graph has exactly one inflection point.

1. \(a>0\): one local extremum; \(a<0\): no local extrema 2. For \(a>0\), the point \(\left(\frac{1}{2}\ln(a), 2\sqrt{a}\right)\) is a local minimum. 3. For \(a<0\), there is exactly one inflection point at \(x=\frac{1}{2}\ln(-a)\).

- Rewrite the rational expression before differentiating. - Use the first and second derivative tests. - Eliminate \(a\) from each extremum’s coordinates. - Keep track of the sign restrictions implied by \(a>0\).

1. Rewrite the function as \(f_a(x)=x+2a+\frac{4a^2}{x}\). Then \(f_a'(x)=1-\frac{4a^2}{x^2}\) and \(f_a''(x)=\frac{8a^2}{x^3}\). 2. The critical-point equation gives \(x^2=4a^2\), so \(x=\pm2a\). Since \(a>0\), \(f_a''(2a)=\frac{1}{a}>0\) and \(f_a''(-2a)=-\frac{1}{a}<0\). 3. Therefore, the local minimum is \((2a, 6a)\), and the local maximum is \((-2a, -2a)\). 4. For the minima, \(x=2a>0\), so \(y=6a=3x\). Their locus is \(y=3x\) for \(x>0\). 5. For the maxima, \(x=-2a<0\), and \(y=-2a=x\). Their locus is \(y=x\) for \(x<0\).

a) Local minimum: \((2a, 6a)\); local maximum: \((-2a, -2a)\) b) \(y=3x\) for \(x>0\) c) \(y=x\) for \(x<0\)

- Find and classify both critical points. - Use \(k>0\) when applying the second derivative test. - Replace \(k\) using the maximum x-coordinate. - Include the x-values implied by \(k>0\).

1. Differentiate: \(f_k'(x)=x^2-k^2\) and \(f_k''(x)=2x\). The critical points are \(x=\pm k\). 2. Since \(f_k''(-k)=-2k<0\), the point at \(x=-k\) is the unique local maximum. Its y-coordinate is \(f_k(-k)=-\frac{k^3}{3}+k^3+k=\frac{2}{3}k^3+k\). Thus the local maximum is \(\left(-k, \frac{2}{3}k^3+k\right)\). 3. At the maximum, \(x=-k<0\), so \(k=-x\). Substitution gives \(y=\frac{2}{3}(-x)^3-x=-\frac{2}{3}x^3-x\). Therefore, the locus is \(y=-\frac{2}{3}x^3-x\) for \(x<0\).

a) \(\left(-k, \frac{2}{3}k^3+k\right)\) b) \(y=-\frac{2}{3}x^3-x\) for \(x<0\)

- Use a double-angle identity for the product \(\sin(x)\cos(x)\). - A product is zero when at least one factor is zero. - Rewrite the derivative equation as a quadratic equation in \(\cos(x)\). - Use higher derivatives when the second derivative test is inconclusive.

1. Expand the product: \(f(x)=\sin(x)+\sin(x)\cos(x)\). Since \(\sin(2x)=2\sin(x)\cos(x)\), this becomes \(f(x)=\sin(x)+\frac{1}{2}\sin(2x)\). 2. The product is zero when \(\sin(x)=0\) or \(1+\cos(x)=0\). On \([0,2\pi]\), the zeros are \(x=0\), \(x=\pi\), and \(x=2\pi\). 3. Differentiate: \(f'(x)=\cos(x)+\cos(2x)\). Using \(\cos(2x)=2\cos^2(x)-1\), the critical-point equation becomes \(2\cos^2(x)+\cos(x)-1=0\). Thus \(\cos(x)=\frac{1}{2}\) or \(\cos(x)=-1\), giving \(x=\frac{\pi}{3}\), \(x=\pi\), and \(x=\frac{5\pi}{3}\). 4. The second derivative is \(f''(x)=-\sin(x)-2\sin(2x)\). At \(x=\frac{\pi}{3}\), it equals \(-\frac{3\sqrt{3}}{2}<0\), so there is a local maximum at \(\left(\frac{\pi}{3},\frac{3\sqrt{3}}{4}\right)\). At \(x=\frac{5\pi}{3}\), it equals \(\frac{3\sqrt{3}}{2}>0\), so there is a local minimum at \(\left(\frac{5\pi}{3},-\frac{3\sqrt{3}}{4}\right)\). 5. At \(x=\pi\), \(f'(\pi)=0\) and \(f''(\pi)=0\), but \(f'''(\pi)=-3\ne0\). The concavity changes there, so \((\pi,0)\) is a stationary inflection point, not a local extremum.

a) \(f(x)=\sin(x)+\frac{1}{2}\sin(2x)\). b) \(x=0,\pi,2\pi\). c) Local maximum: \(\left(\frac{\pi}{3},\frac{3\sqrt{3}}{4}\right)\). Local minimum: \(\left(\frac{5\pi}{3},-\frac{3\sqrt{3}}{4}\right)\). Stationary inflection point: \((\pi,0)\).

- Use the required function value and the condition that the first derivative is zero. - Solve the resulting system for the two parameters. - Use the second derivative test to classify the critical point.

1. The point \((-2, -1)\) must satisfy \(f_{a,b}(-2)=-1\), and a local extremum requires \(f_{a,b}^{\prime}(-2)=0\). 2. From the function value, \(\frac{-2a+b}{5}=-1\), so \(-2a+b=-5\) and \(b=2a-5\). 3. Differentiate: \(f_{a,b}^{\prime}(x)=\frac{a(x^2+1)-2x(ax+b)}{(x^2+1)^2}=\frac{-ax^2-2bx+a}{(x^2+1)^2}\). 4. Applying \(f_{a,b}^{\prime}(-2)=0\) gives \(-3a+4b=0\). Substitute \(b=2a-5\): \(-3a+4(2a-5)=0\), so \(5a=20\) and \(a=4\). Then \(b=3\). 5. For \(a=4\) and \(b=3\), the numerator of \(f^{\prime}\) is \(N(x)=-4x^2-6x+4\). Since \(N(-2)=0\), \(f^{\prime\prime}(-2)=\frac{N^{\prime}(-2)}{25}=\frac{10}{25}=0.4>0\). Therefore, \((-2, -1)\) is a local minimum.

\(a=4\) and \(b=3\). Since \(f^{\prime\prime}(-2)=0.4>0\), the point \((-2, -1)\) is a local minimum.

- A horizontal tangent occurs where the first derivative is zero. - Use the unit circle to solve \(\sin x=\cos x\). - Classify each critical number using the sign of the second derivative.

1. Differentiate twice: \(h'(x)=\cos x-\sin x\) and \(h''(x)=-\sin x-\cos x\). 2. A horizontal tangent requires \(\cos x-\sin x=0\). Thus, \(\tan x=1\), which gives \(x=\frac{\pi}{4}\) and \(x=\frac{5\pi}{4}\) on \([0, 2\pi]\). 3. At \(x=\frac{\pi}{4}\), \(h''\left(\frac{\pi}{4}\right)=-\sqrt{2}<0\), so there is a local maximum. At \(x=\frac{5\pi}{4}\), \(h''\left(\frac{5\pi}{4}\right)=\sqrt{2}>0\), so there is a local minimum.

\(x=\frac{\pi}{4}\): local maximum \(x=\frac{5\pi}{4}\): local minimum

- Find the critical points and classify them with the second derivative. - Changing \(k\) shifts the graph vertically. - Tangency to the x-axis at a minimum requires the minimum value to be zero. - Use the relative heights of the maximum and minima to count zeros.

1. The first derivative is \(f_k'(x)=4x(x^2-5)\), so the critical points are \(x=0\) and \(x=\pm\sqrt{5}\). 2. The second derivative is \(f_k''(x)=12x^2-20\). Since \(f_k''(0)<0\), \((0, k)\) is a local maximum. Since \(f_k''(\pm\sqrt{5})>0\), \((-\sqrt{5}, k-25)\) and \((\sqrt{5}, k-25)\) are local minima. 3. Tangency at both minima requires \(k-25=0\), so \(k=25\). 4. Exactly three distinct zeros occur when the central local maximum lies on the x-axis while both minima lie below it. This gives \(k=0\), with zeros \(x=0\) and \(x=\pm\sqrt{10}\). 5. There are no real zeros when the absolute minima are above the x-axis: \(k-25>0\), so \(k>25\).

a) Local maximum: \((0, k)\); local minima: \((-\sqrt{5}, k-25)\) and \((\sqrt{5}, k-25)\) b) \(k=25\) c) \(k=0\) d) \(k>25\)

- Compare \(f_a(-x)\) with \(f_a(x)\). - Factor the first derivative and separate the cases \(a\leq0\) and \(a>0\). - Use the second derivative to classify the critical points. - Eliminate \(a\) from the minimum coordinates.

1. Since \(f_a(-x)=f_a(x)\), every graph is symmetric about the y-axis. 2. Differentiate: \(f_a'(x)=4x(ax^2-1)\) and \(f_a''(x)=12ax^2-4\). 3. If \(a\leq0\), the only critical point is \(x=0\). Since \(f_a''(0)=-4<0\), the origin is the only local extremum and is a local maximum. 4. If \(a>0\), the critical points are \(x=0\) and \(x=\pm\frac{1}{\sqrt{a}}\). The origin is a local maximum, while \(f_a''\left(\pm\frac{1}{\sqrt{a}}\right)=8>0\), so the other two points are local minima. 5. At a local minimum, \(x^2=\frac{1}{a}\), so \(a=\frac{1}{x^2}\). Substituting gives \(y=\frac{x^4}{x^2}-2x^2=-x^2\). Because these minima have \(x\neq0\), the attained locus is \(y=-x^2\) for \(x\neq0\).

a) Every graph is symmetric about the y-axis. b) For \(a\leq0\), there is one local maximum at \((0, 0)\). For \(a>0\), there is a local maximum at \((0, 0)\) and two local minima at \(\left(\pm\frac{1}{\sqrt{a}}, -\frac{1}{a}\right)\). c) \(y=-x^2\) for \(x\neq0\)

- First locate the zeros of the first derivative. - Then read the second derivative at each candidate. - A negative second derivative indicates a local maximum. - A positive second derivative indicates a local minimum.

1. Local extrema can occur where \(f'(x)=0\). From the orange graph, the zeros are \(x=-1\) and \(x=3\). 2. From the blue graph, \(f''(-1)=-4<0\), so \(f\) has a local maximum at \(x=-1\). 3. Also, \(f''(3)=4>0\), so \(f\) has a local minimum at \(x=3\).

There is a local maximum at \(x=-1\) and a local minimum at \(x=3\).

- Use the labels and turning points shown on the coordinate plane. - At an interior local extremum, the first derivative is zero. - The sign of the second derivative classifies the critical point. - A horizontal tangent has slope zero.

1. From the graph, Cards A and B match \(f\), while Card C matches \(g\). 2. Differentiate \(f\): \(f'(x) = 0.75x^2 - 3\) and \(f''(x) = 1.5x\). Solving \(f'(x) = 0\) gives \(x = -2\) and \(x = 2\). 3. Since \(f''(-2) = -3 < 0\), \(f\) has a local maximum at \(x = -2\). Evaluate \(f(-2) = 6\), giving \((-2, 6)\). 4. Since \(f''(2) = 3 > 0\), \(f\) has a local minimum at \(x = 2\). Evaluate \(f(2) = -2\), giving \((2, -2)\). 5. Differentiate \(g\): \(g'(x) = 2(x - 1)\). Then \(g'(1) = 0\), so the tangent is horizontal at \(x = 1\).

1. Card A matches \(f\); Card B matches \(f\); Card C matches \(g\). 2. The local extrema of \(f\) are the local maximum \((-2, 6)\) and the local minimum \((2, -2)\). 3. \(g'(1) = 0\), so \(g\) has a horizontal tangent at \(x = 1\).

- Candidates occur at zeros of the first derivative. - Read the second derivative at each candidate. - Positive second derivative means a local minimum. - Negative second derivative means a local maximum.

1. The orange graph has zeros at \(x=-3\), \(x=1\), and \(x=3\). These are the critical-value candidates. 2. At \(x=-3\), \(f''(-3)=6>0\), so there is a local minimum. 3. At \(x=1\), \(f''(1)=-2<0\), so there is a local maximum. 4. At \(x=3\), \(f''(3)=3>0\), so there is a local minimum.

a) \(x=-3\), \(x=1\), and \(x=3\) b) Local minima at \(x=-3\) and \(x=3\); local maximum at \(x=1\)

- Which derivatives are needed for the second derivative test? - Solve the equation where the first derivative equals zero. - Use the second derivative to distinguish a maximum from a minimum. - Evaluate the original function for the exact coordinates. - Compare the locations of the turning points with the three graphs.

1. Differentiate: \(f'(x) = \frac{1}{2}x^2 - 2\) and \(f''(x) = x\). 2. Solve \(f'(x) = 0\): \(\frac{1}{2}x^2 - 2 = 0\), so \(x = -2\) and \(x = 2\). 3. Since \(f''(-2) = -2 < 0\), \(f\) has a local maximum at \(x = -2\). Evaluate \(f(-2) = \frac{8}{3}\), giving \((-2, \frac{8}{3})\). 4. Since \(f''(2) = 2 > 0\), \(f\) has a local minimum at \(x = 2\). Evaluate \(f(2) = -\frac{8}{3}\), giving \((2, -\frac{8}{3})\). 5. Only Graph C has a local maximum at \(x = -2\) and a local minimum at \(x = 2\), so Graph C represents \(f\).

1. Local maximum: \((-2, \frac{8}{3})\); local minimum: \((2, -\frac{8}{3})\) 2. Graph C

- Compare symmetry, intercepts, end behavior, or extrema for part a. - Use the leading term to determine the end behavior. - Substitute \(u=x^2\) to solve the quartic equation. - Use the first and second derivatives to find and classify extrema.

1. The graph shown cannot represent \(f\). For example, \(f\) is even and therefore symmetric about the y-axis, while the displayed graph is symmetric about \(x=1\). Also, \(f(0)=6\), but the displayed graph has y-intercept \((0, 1)\). 2. The positive leading coefficient and even degree give \(f(x)\to\infty\) as \(x\to\pm\infty\). 3. Solving \(\frac{1}{2}x^4-4x^2+6=0\) and substituting \(u=x^2\) gives \(u^2-8u+12=0\). Thus, \(u=2\) or \(u=6\), so the zeros are \(x=\pm\sqrt{2}\approx\pm1.41\) and \(x=\pm\sqrt{6}\approx\pm2.45\). 4. The derivative is \(f'(x)=2x(x^2-4)\), so the critical values are \(x=-2, 0, 2\). Since \(f''(x)=6x^2-8\), \((0, 6)\) is a local maximum and \((-2, -2)\) and \((2, -2)\) are local minima.

a) Possible reasons include: \(f\) is symmetric about the y-axis, but the displayed graph is symmetric about \(x=1\); and \(f(0)=6\), but the displayed graph has y-intercept \((0, 1)\). b) 1. \(f(x)\to\infty\) as \(x\to\pm\infty\) 2. \(x=\pm\sqrt{2}\approx\pm1.41\) and \(x=\pm\sqrt{6}\approx\pm2.45\) 3. Local maximum: \((0, 6)\); local minima: \((-2, -2)\) and \((2, -2)\)

- Compare the turning point of one graph with the zero of the other. - Local extrema of \(f\) can occur where \(f'(x) = 0\). - Use the sign of \(f''\) at each critical point to apply the second derivative test.

1. Since \(f''\) is the derivative of \(f'\), the graph of \(f''\) must be zero where the graph of \(f'\) has a horizontal tangent. The parabola \(g\) has its minimum at \(x = 1\), and the line \(h\) is zero there. The signs of \(h\) also match the decreasing and increasing behavior of \(g\). Therefore, \(g = f'\) and \(h = f''\). 2. Local extrema of \(f\) occur where \(f'(x) = 0\). The graph of \(g\) is zero at \(x = -1\) and \(x = 3\). 3. Since \(f''(-1) = -4 < 0\), \(f\) has a local maximum at \(x = -1\). Since \(f''(3) = 4 > 0\), \(f\) has a local minimum at \(x = 3\).

a) The blue graph \(g\) represents \(f'\), and the red graph \(h\) represents \(f''\). b) \(f\) has a local maximum at \(x = -1\) and a local minimum at \(x = 3\).

- Find the first three derivatives. - Use the first and second derivatives to find and classify extrema. - Use the second and third derivatives to find the inflection point. - Compare the exact points with the displayed graph. - Check the axis scales carefully.

1. The derivatives are \(f'(x)=-\frac{3}{4}x^2+3\), \(f''(x)=-\frac{3}{2}x\), and \(f'''(x)=-\frac{3}{2}\). 2. Solving \(f'(x)=0\) gives \(x=\pm2\). The second derivative shows that \((-2, -4)\) is a local minimum and \((2, 4)\) is a local maximum. 3. Solving \(f''(x)=0\) gives \(x=0\). Since the third derivative is nonzero, the inflection point is \((0, 0)\). 4. The displayed graph has extrema at approximately \((-2, -3)\) and \((2, 3)\), not at \((-2, -4)\) and \((2, 4)\). Therefore, it does not represent \(f\) correctly.

a) Local minimum: \((-2, -4)\); local maximum: \((2, 4)\); inflection point: \((0, 0)\) b) No. The displayed graph has extremum y-values \(-3\) and \(3\), but \(f\) has extremum y-values \(-4\) and \(4\).

- Set the first derivative equal to \(0\) to find critical values. - Use the second derivative to classify each critical point. - Set the second derivative equal to \(0\) to find possible concavity changes. - Verify each value with the third derivative or a sign change.

1. The derivatives are \(f'(x)=-\frac{2}{5}x^3+\frac{8}{5}x\), \(f''(x)=-\frac{6}{5}x^2+\frac{8}{5}\), and \(f'''(x)=-\frac{12}{5}x\). 2. Solving \(f'(x)=0\) gives \(x=0\) and \(x=\pm2\). The second derivative gives a local minimum at \((0, 1)\) and local maxima at \((-2, \frac{13}{5})\) and \((2, \frac{13}{5})\). 3. Solving \(f''(x)=0\) gives \(x=\pm\frac{2\sqrt{3}}{3}\). The third derivative is nonzero at both values, so these are the concavity-change x-values.

a) Local maxima: \((-2, \frac{13}{5})\) and \((2, \frac{13}{5})\); local minimum: \((0, 1)\) b) \(x=\pm\frac{2\sqrt{3}}{3}\)

- First locate the zeros of \(f'\) from the graph. - Use the sign of \(f''\) at each critical value to apply the second derivative test. - Compare what a necessary condition guarantees with what a sufficient condition guarantees. - Inflection points occur where \(f''\) changes sign.

1. From the blue graph, the zeros of \(f'\) occur at about \(x=-2.5\), \(x=1\), and \(x=4.5\). 2. At the two outer zeros, the orange graph is above the x-axis, so \(f''>0\) and \(f\) has local minima. At \(x=1\), the orange graph is below the x-axis, so \(f''<0\) and \(f\) has a local maximum. 3. For a differentiable function, \(f'(x_0)=0\) is necessary at an interior local extremum because the tangent must be horizontal. It is not sufficient because a horizontal tangent can occur without an extremum, as for \(g(x)=x^3\) at \(x=0\). A sign change in \(f'\), or the second derivative test when \(f''(x_0)\ne0\), supplies a sufficient condition. 4. The orange graph crosses the x-axis at \(x=-1\) and \(x=3\). At \(x=-1\), \(f''\) changes from positive to negative, so \(f\) changes from concave up to concave down. At \(x=3\), \(f''\) changes from negative to positive, so \(f\) changes from concave down to concave up.

a) Local minima at about \(x=-2.5\) and \(x=4.5\); local maximum at \(x=1\) b) \(f'(x_0)=0\) identifies a candidate but does not guarantee an extremum; an additional sufficient test is required. c) Inflection point at \(x=-1\): concave up to concave down. Inflection point at \(x=3\): concave down to concave up.

- Factor the function to find its zeros. - Compare \(f_a(-x)\) with \(f_a(x)\). - Use the first and second derivatives to locate and classify extrema. - Determine when \(\pm\sqrt{a}\) are distinct from zero.

1. Factor \(f_a(x)=x(x^2-a)\). Graph p has only the zero \(x=0\), so \(a=0\). Graph q has zeros \(-1, 0, 1\), so \(a=1\). Graph r has zeros \(-2, 0, 2\), so \(a=4\). 2. Since \(f_a(-x)=-f_a(x)\), every graph is symmetric about the origin. 3. The first derivative is \(f_a'(x)=3x^2-a\), so the critical points are \(x=\pm\sqrt{\frac{a}{3}}\). The second derivative is \(f_a''(x)=6x\). 4. The local maximum is \(\left(-\sqrt{\frac{a}{3}}, \frac{2a}{3}\sqrt{\frac{a}{3}}\right)\), and the local minimum is \(\left(\sqrt{\frac{a}{3}}, -\frac{2a}{3}\sqrt{\frac{a}{3}}\right)\). 5. The zeros are \(0\) and \(\pm\sqrt{a}\). They are three distinct real numbers exactly when \(a>0\).

a) p: \(a=0\); q: \(a=1\); r: \(a=4\) b) Symmetric about the origin c) Local maximum: \(\left(-\sqrt{\frac{a}{3}}, \frac{2a}{3}\sqrt{\frac{a}{3}}\right)\); local minimum: \(\left(\sqrt{\frac{a}{3}}, -\frac{2a}{3}\sqrt{\frac{a}{3}}\right)\) d) \(a>0\)

- Factor the first derivative. - Notice which factors can never be zero. - Substitute the maximum x-coordinate into the original function. - Use point-slope form for the tangent line.

1. The first derivative is \(f_a'(x)=ax(2-x)e^{-x}\). Since \(a>0\) and the exponential is positive, the critical points are \(x=0\) and \(x=2\), independent of \(a\). 2. The second derivative is \(f_a''(x)=a(x^2-4x+2)e^{-x}\). At \(x=0\), it is positive, so there is a local minimum. At \(x=2\), it is negative, so there is a local maximum. 3. The maximum value is \(f_a(2)=4ae^{-2}\). Setting this equal to \(4\) gives \(a=e^2\). 4. For \(a=10\), \(f_{10}(1)=\frac{10}{e}\) and \(f_{10}'(1)=\frac{10}{e}\). Thus the tangent line is \(y-\frac{10}{e}=\frac{10}{e}(x-1)\), or \(y=\frac{10}{e}x\). It passes through the origin.

a) \(x=0\): local minimum; \(x=2\): local maximum b) \(a=e^2\) c) \(y=\frac{10}{e}x\); it passes through the origin.

- Find the critical numbers from the first derivative. - Use the second derivative to classify them. - Compare the locations and heights of the extrema with the three graphs.

1. Differentiate: \(f'(x) = x^2 - 2x - 3 = (x - 3)(x + 1)\), and \(f''(x) = 2x - 2\). 2. The critical numbers are \(x = -1\) and \(x = 3\). 3. Since \(f''(-1) = -4 < 0\), \(f\) has a local maximum at \(x = -1\). Evaluate \(f(-1) = \frac{5}{3}\), giving \((-1, \frac{5}{3})\). 4. Since \(f''(3) = 4 > 0\), \(f\) has a local minimum at \(x = 3\). Evaluate \(f(3) = -9\), giving \((3, -9)\). 5. Graph A shows a local maximum near \(x = -1\) and a local minimum at \(x = 3\), so Graph A represents \(f\).

Local maximum: \((-1, \frac{5}{3})\) Local minimum: \((3, -9)\) The correct graph is A.

- Set the first derivative equal to \(0\). - Use the second derivative to classify each critical point. - Substitute the critical values into the original function. - For the tangent line, find both the point and the derivative at \(x=1\).

1. The derivatives are \(p'(x)=-4x^3+8x=-4x(x^2-2)\) and \(p''(x)=-12x^2+8\). 2. The critical values are \(x=0\) and \(x=\pm\sqrt{2}\). Since \(p''(0)=8>0\), \((0, 0)\) is a local minimum. Since \(p''(\pm\sqrt{2})=-16<0\), \((\pm\sqrt{2}, 4)\) are local maxima. These locations agree with the graph. 3. At \(x=1\), \(p(1)=3\) and \(p'(1)=4\). Using point-slope form, \(y-3=4(x-1)\), so the tangent line is \(y=4x-1\).

a) Local maxima: \((\pm\sqrt{2}, 4)\); local minimum: \((0, 0)\) b) The computed points agree with the graph. c) \(y=4x-1\)

- Factor the polynomial to find its zeros. - Use the first and second derivatives to find the local maximum. - Set the second derivative equal to \(0\) for possible inflection points. - Use the graph's symmetry to check paired answers.

1. Since \(f(x)=\frac{1}{4}(x^2-4)^2\), the zeros are \(x=-2\) and \(x=2\), each with multiplicity \(2\). Thus, \(N_1=(-2, 0)\) and \(N_2=(2, 0)\). 2. The derivatives are \(f'(x)=x^3-4x\), \(f''(x)=3x^2-4\), and \(f'''(x)=6x\). The second-derivative test gives a local maximum at \(H=(0, 4)\). 3. Solving \(f''(x)=0\) gives \(x=\pm\frac{2\sqrt{3}}{3}\). At either value, \(f(x)=\frac{16}{9}\), and the third derivative is nonzero. Therefore, \(W_1=(-\frac{2\sqrt{3}}{3}, \frac{16}{9})\) and \(W_2=(\frac{2\sqrt{3}}{3}, \frac{16}{9})\).

\(N_1=(-2, 0)\), \(N_2=(2, 0)\), \(H=(0, 4)\) \(W_1=(-\frac{2\sqrt{3}}{3}, \frac{16}{9})\), \(W_2=(\frac{2\sqrt{3}}{3}, \frac{16}{9})\)

- Factor the polynomial to find zeros and multiplicities. - Use the first and second derivatives for extrema. - Use the second derivative for the inflection point. - Compare each coordinate with the graph.

1. Factoring gives \(f(x)=x(x-3)^2\). Therefore, \(x=0\) is a simple zero and \(x=3\) is a double zero. 2. The derivatives are \(f'(x)=3(x-1)(x-3)\), \(f''(x)=6x-12\), and \(f'''(x)=6\). The second-derivative test gives a local maximum at \(H=(1, 4)\) and a local minimum at \(T=(3, 0)\). 3. Solving \(f''(x)=0\) gives \(x=2\), and \(f(2)=2\). Thus, \(W=(2, 2)\). All points agree with the graph.

Zeros: \((0, 0)\), simple; \((3, 0)\), double Local maximum: \(H=(1, 4)\); local minimum: \(T=(3, 0)\) Inflection point: \(W=(2, 2)\)

- Compare the polynomial degrees and the locations of local extrema and zeros. - Horizontal tangents of \(f\) occur where \(f' = 0\). - Use the sign of \(f''\) for the second derivative test.

1. The blue graph is cubic and the red graph is quadratic. The red graph is the derivative of the blue graph: its zeros occur at the local extrema of the blue graph. Therefore, \(g_1 = f'\) and \(g_2 = f''\). 2. Horizontal tangents of \(f\) occur where \(f'(x) = 0\). The blue graph is zero at \(x = -3\), \(x = 0\), and \(x = 3\). 3. At \(x = -3\), \(f''(-3) > 0\), so \(f\) has a local minimum. At \(x = 0\), \(f''(0) < 0\), so \(f\) has a local maximum. At \(x = 3\), \(f''(3) > 0\), so \(f\) has a local minimum.

a) \(g_1 = f'\) and \(g_2 = f''\) b) Local minima at \(x = -3\) and \(x = 3\); local maximum at \(x = 0\)

- Use the x-intercept shown in the graph. - Apply the product rule. - The exponential factor is always positive. - Use the second derivative test.

1. The graph crosses the x-axis at \(x=2\). Since \((2-a)e^2=0\), \(a=2\). 2. The first derivative is \(g_a'(x)=(x-a+1)e^x\). Since the exponential factor is positive, the only critical point is \(x=a-1\). 3. The second derivative is \(g_a''(x)=(x-a+2)e^x\), so \(g_a''(a-1)=e^{a-1}>0\). Therefore, this point is a local minimum. 4. Its y-coordinate is \(g_a(a-1)=-e^{a-1}\).

a) \(a=2\) b) Local minimum: \((a-1, -e^{a-1})\)

- Factor the polynomial to find x-intercepts and multiplicities. - Use the first and second derivatives to find and classify extrema. - Use the second derivative to find the inflection point. - Compare the exact coordinates with the graph.

1. Since \(f(x)=x^2(x-3)\), the x-intercepts are \((0, 0)\), with multiplicity \(2\), and \((3, 0)\), with multiplicity \(1\). The y-intercept is also \((0, 0)\). 2. The derivatives are \(f'(x)=3x(x-2)\), \(f''(x)=6x-6\), and \(f'''(x)=6\). The second-derivative test gives a local maximum at \((0, 0)\) and a local minimum at \((2, -4)\). 3. The inflection point occurs at \(x=1\), where \(f(1)=-2\). Thus, the inflection point is \((1, -2)\). 4. These exact points agree with the diagram, so the graph is correct.

Yes, the graph is correct. Intercepts: \((0, 0)\), a double x-intercept and the y-intercept; \((3, 0)\), a simple x-intercept Local maximum: \((0, 0)\); local minimum: \((2, -4)\) Inflection point: \((1, -2)\)

- Evaluate the function at \(x=0\). - Factor the cubic to find zeros and multiplicities. - Use the first and second derivatives to find and classify extrema. - Compare the exact values with the graph's scale.

1. The y-intercept is \((0, 2)\). Factoring gives \(f(x)=-(x+1)^2(x-2)\), so the x-intercepts are \((-1, 0)\), with multiplicity \(2\), and \((2, 0)\), with multiplicity \(1\). 2. The derivatives are \(f'(x)=-3x^2+3\) and \(f''(x)=-6x\). The critical values are \(x=-1\) and \(x=1\). The second-derivative test gives a local minimum at \((-1, 0)\) and a local maximum at \((1, 4)\). 3. The displayed graph has y-intercept \((0, 1.6)\) and a local maximum near \((1, 3.2)\). These do not match the exact values, so the graph is incorrect.

No. Intercepts: \((0, 2)\), \((-1, 0)\) with multiplicity \(2\), and \((2, 0)\) Local minimum: \((-1, 0)\); local maximum: \((1, 4)\)

- Use even symmetry to check paired points. - Factor out \(x^2\) to find the zeros. - Set the first derivative equal to \(0\). - Use the second derivative to classify the critical points.

1. The function is even, so its graph is symmetric about the y-axis. Factoring gives \(f(x)=\frac{1}{2}x^2(x^2-4)\). Thus, the x-intercepts are \((0, 0)\), with multiplicity \(2\), and \((\pm2, 0)\). The y-intercept is \((0, 0)\). 2. The derivatives are \(f'(x)=2x(x^2-2)\) and \(f''(x)=6x^2-4\). The critical values are \(x=0\) and \(x=\pm\sqrt{2}\). The second-derivative test gives a local maximum at \((0, 0)\) and local minima at \((\pm\sqrt{2}, -2)\). 3. These exact features agree with the displayed graph, so the graph is correct.

Yes. Intercepts: \((0, 0)\), a double x-intercept and the y-intercept; \((\pm2, 0)\) Local maximum: \((0, 0)\); local minima: \((\pm\sqrt{2}, -2)\)

- Check \(f(0)\) before doing further calculations. - Find the zeros of the first derivative; if the first and second derivatives are both \(0\), check the third derivative. - Compute the exact stationary-point coordinates and compare them with the graph.

1. The y-intercept is \((0, 2)\), but the displayed graph has y-intercept \((0, 4)\). This already shows that the graph is incorrect. 2. The derivative is \(f'(x)=x^2(x-3)\). Thus, the critical values are \(x=0\) and \(x=3\). 3. At \(x=3\), \(f''(3)=9>0\), so there is a local minimum at \((3, -\frac{19}{4})\). At \(x=0\), the first and second derivatives are \(0\), while \(f'''(0)=-6\ne0\), so \((0, 2)\) is a stationary inflection point. 4. These points do not match the displayed graph, so the graph is incorrect.

No. Y-intercept and stationary inflection point: \((0, 2)\) Local minimum: \((3, -\frac{19}{4})\)

- Use the function's even symmetry and substitute \(u=x^2\) to find the zeros. - Set the first derivative equal to \(0\). - Use the second derivative to classify the critical points.

1. The function is even, so the graph is symmetric about the y-axis. The y-intercept is \((0, -5)\). 2. Let \(u=x^2\). Solving \(-u^2+6u-5=0\) gives \(u=1\) or \(u=5\). Therefore, the x-intercepts are \((\pm1, 0)\) and \((\pm\sqrt{5}, 0)\). 3. The derivatives are \(f'(x)=-4x(x^2-3)\) and \(f''(x)=-12x^2+12\). The second-derivative test gives a local minimum at \((0, -5)\) and local maxima at \((\pm\sqrt{3}, 4)\). 4. These exact features agree with the graph, so it is correct.

Yes. Intercepts: \((0, -5)\), \((\pm1, 0)\), and \((\pm\sqrt{5}, 0)\) Local minimum: \((0, -5)\); local maxima: \((\pm\sqrt{3}, 4)\)

- Find the critical values from the first derivative and classify them with the second derivative. - A local minimum on the x-axis is a multiple zero. - Set the second derivative equal to \(0\) to locate the inflection point. - Compare the exact critical and inflection values with the graph.

1. The y-intercept is \((0, 4)\). The derivative is \(f'(x)=3x(x-2)\), so the critical values are \(x=0\) and \(x=2\). 2. Since \(f''(0)=-6<0\), \((0, 4)\) is a local maximum. Since \(f''(2)=6>0\), \((2, 0)\) is a local minimum. 3. Because the local minimum lies on the x-axis, \(x=2\) is a double zero. Factoring gives \(f(x)=(x-2)^2(x+1)\), so the other x-intercept is \((-1, 0)\). 4. Solving \(f''(x)=6x-6=0\) gives \(x=1\). Since \(f(1)=2\), the inflection point is \((1,2)\). 5. The displayed graph places the local minimum at \(x=3\), not \(x=2\), and its inflection point is also misplaced. Therefore, it is incorrect.

No. Intercepts: \((0, 4)\), \((-1, 0)\), and \((2, 0)\), where \(x=2\) is a double zero Local maximum: \((0, 4)\); local minimum: \((2, 0)\) Inflection point: \((1,2)\)

- Factor the first derivative before solving for critical values. - Use the second derivative to classify the local minimum. - At the special inflection point, check the first three derivatives. - A horizontal tangent has derivative \(0\).

1. The derivatives are \(f'(x)=x(x-3)^2\), \(f''(x)=3(x-1)(x-3)\), and \(f'''(x)=6x-12\). 2. At \(x=0\), \(f'(0)=0\) and \(f''(0)=9>0\), so the local minimum is \((0, 2)\). 3. At \(x=3\), \(f'(3)=0\), \(f''(3)=0\), and \(f'''(3)=6\ne0\). Also, \(f(3)=\frac{35}{4}\). Therefore, \((3, \frac{35}{4})\) is a stationary inflection point. 4. A stationary inflection point additionally satisfies \(f'(x_0)=0\), meaning its tangent line is horizontal. An inflection point with a nonhorizontal tangent has \(f'(x_0)\ne0\).

a) Local minimum: \((0, 2)\) b) Stationary inflection point: \((3, \frac{35}{4})\) c) It must satisfy \(f'(x_0)=0\).

- A flat-looking graph does not determine the point type. - Check the first derivative at the origin. - Use the second derivative to classify the critical point. - Consider how the vertical scale affects the graph's appearance.

1. The derivatives are \(f'(x)=-4x^3+30x^2+\frac{1}{5}x\) and \(f''(x)=-12x^2+60x+\frac{1}{5}\). 2. At \(x=0\), \(f'(0)=0\), so the origin is a critical point. However, \(f''(0)=\frac{1}{5}>0\), so the origin is a local minimum, not a stationary inflection point. 3. The viewing scale makes the shallow minimum look nearly flat compared with the much larger function values elsewhere in the window.

a) The claim is false. The wide viewing window makes the graph appear flatter than it is near the origin. b) Since \(f'(0)=0\) and \(f''(0)=\frac{1}{5}>0\), the origin is a local minimum, not a stationary inflection point.

- Factor out \(x\) to find the zeros. - Use the first three derivatives to find and classify extrema and inflection points. - Give both coordinates for every point.

1. Factoring gives \(f(x)=\frac{1}{2}x(x^2-12)\). Therefore, the x-intercepts are \((0, 0)\) and \((\pm2\sqrt{3}, 0)\). The y-intercept is \((0, 0)\). 2. The derivatives are \(f'(x)=\frac{3}{2}x^2-6\), \(f''(x)=3x\), and \(f'''(x)=3\). The critical values are \(x=\pm2\). The second-derivative test gives a local maximum at \((-2, 8)\) and a local minimum at \((2, -8)\). 3. The second derivative is \(0\) at \(x=0\), and the third derivative is nonzero. Thus, the inflection point is \((0, 0)\).

Intercepts: \((0, 0)\) and \((\pm2\sqrt{3}, 0)\) Local maximum: \((-2, 8)\); local minimum: \((2, -8)\) Inflection point: \((0, 0)\)

- Factor out \(x^3\) to find the zeros. - Use the leading term for end behavior. - Use derivative tests for local extrema, and remember that a stationary inflection point has zero slope.

1. Factoring gives \(f(x)=x^3(x-4)\). Thus, \(x=0\) is a zero of multiplicity \(3\), and \(x=4\) is a simple zero. Since the leading term is \(x^4\), \(f(x)\to\infty\) as \(x\to\pm\infty\). 2. The derivatives are \(f'(x)=4x^2(x-3)\), \(f''(x)=12x(x-2)\), and \(f'''(x)=24x-24\). At \(x=3\), the second derivative is positive and \(f(3)=-27\), so \((3, -27)\) is a local minimum. The critical value \(x=0\) is not an extremum. 3. The second derivative is \(0\) at \(x=0\) and \(x=2\). The third derivative is nonzero at both values, so the inflection points are \((0, 0)\) and \((2, -16)\). The point \((0, 0)\) is a stationary inflection point.

a) Zeros: \(x=0\), multiplicity \(3\); \(x=4\), multiplicity \(1\). End behavior: \(f(x)\to\infty\) as \(x\to\pm\infty\). b) Local minimum: \((3, -27)\); no other local extrema c) Inflection points: \((0, 0)\) and \((2, -16)\)

- Find the zeros of \(f'\). - Use the sign of \(f''\) at each critical value. - The function decreases where \(f'<0\).

1. The zeros of \(f'\) are \(x=-4\), \(x=-1\), and \(x=2\). 2. At \(x=-4\), the graph of \(f''\) is above the x-axis, so \(f''(-4)>0\) and \(f\) has a local minimum. At \(x=-1\), the graph of \(f''\) is below the x-axis, so \(f''(-1)<0\) and \(f\) has a local maximum. At \(x=2\), the graph of \(f''\) is above the x-axis, so \(f''(2)>0\) and \(f\) has a local minimum. 3. The graph of \(f'\) is below the x-axis for \(-5<x<-4\) and \(-1<x<2\). Therefore, \(f\) is strictly decreasing on \([-5,-4]\) and \([-1,2]\).

a) \(x=-4\), \(x=-1\), and \(x=2\) b) Local minima at \(x=-4\) and \(x=2\); local maximum at \(x=-1\) c) Strictly decreasing on \([-5,-4]\) and \([-1,2]\)

- The leading term determines polynomial end behavior. - A local maximum must satisfy a first-derivative condition. - Use the second derivative to confirm the classification.

1. The leading term is \(-\frac{1}{4}x^4\). Therefore, \(f_k(x)\to-\infty\) as \(x\to\pm\infty\). Figure 1 has this end behavior, while Figure 2 does not. 2. The first derivative is \(f_k'(x)=2kx-x^3\). Requiring \(f_k'(\sqrt{6})=0\) gives \(2k\sqrt{6}-6\sqrt{6}=0\), so \(k=3\). 3. The second derivative is \(f_k''(x)=2k-3x^2\). For \(k=3\), \(f_3''(\sqrt{6})=-12<0\), confirming a local maximum.

a) Figure 1 b) \(k=3\)

- Set the first derivative equal to zero and use the second derivative to classify the points. - Eliminate \(k\) from each extremum’s coordinates. - Factor the original function to find its zeros. - Look for an x-value that makes the parameter term vanish.

1. Differentiate: \(g_k'(x)=x^2-k^2\). The critical points are \(x=\pm k\). Since \(g_k''(x)=2x\), the local maximum is \(\left(-k, \frac{2}{3}k^3\right)\), and the local minimum is \(\left(k, -\frac{2}{3}k^3\right)\). 2. For the minimum, \(x=k\), giving \(y=-\frac{2}{3}x^3\). For the maximum, \(x=-k\), so \(k=-x\), and the same equation results. Thus the locus is \(y=-\frac{2}{3}x^3\), with \(x\neq0\). 3. Factor: \(g_k(x)=x\left(\frac{1}{3}x^2-k^2\right)\). The three distinct zeros are \(x=0\) and \(x=\pm\sqrt{3}k\). 4. Since \(g_k(0)=0\) for every \(k\), all graphs pass through the origin. No other point is common to all graphs because the coefficient of \(k^2\) is \(-x\), which vanishes only at \(x=0\).

a) \(\left(-k, \frac{2}{3}k^3\right)\) and \(\left(k, -\frac{2}{3}k^3\right)\) b) \(y=-\frac{2}{3}x^3\) for \(x\neq0\) c) Three zeros: \(x=0\) and \(x=\pm\sqrt{3}k\) d) \((0, 0)\)

- Translate “slope \(0\)” and “local maximum” into derivative conditions. - Begin with the general form of a cubic polynomial. - Use the fact that the graph passes through the origin to simplify the polynomial. - Substitute the known point and slope conditions into the function and its derivative.

1. Start with \(f(x)=ax^3+bx^2+cx+d\). 2. Since the road begins at \(O(0, 0)\), \(f(0)=0\), so \(d=0\). Since its slope there is \(0\), \(f'(0)=0\), so \(c=0\). Thus, \(f(x)=ax^3+bx^2\). 3. The local maximum at \((2, 6)\) gives \(f(2)=6\) and \(f'(2)=0\). Therefore, \(8a+4b=6\) and \(12a+4b=0\). 4. Subtracting the first equation from the second gives \(4a=-6\), so \(a=-\frac{3}{2}\). Substitution gives \(b=\frac{9}{2}\). 5. Therefore, \(f(x)=-\frac{3}{2}x^3+\frac{9}{2}x^2\). Also, \(f''(2)=-9<0\), confirming a local maximum at \(x=2\).

\(f(x)=-\frac{3}{2}x^3+\frac{9}{2}x^2\)

- A horizontal tangent gives a condition on the first derivative. - Use the inflection point to obtain conditions on the function and its second derivative. - Reflecting across \(y=1\) sends an output \(y\) to \(2-y\). - Find critical values from the first derivative and classify them with the second derivative.

1. Write \(f(x)=ax^3+bx^2+cx+d\). 2. Since \(W(1, 1)\) is the inflection point, \(f(1)=1\) and \(f''(1)=0\). From \(f''(x)=6ax+2b\), \(6a+2b=0\), so \(b=-3a\). 3. The horizontal tangent at \(x=0\) gives \(f'(0)=0\). Since \(f'(x)=3ax^2+2bx+c\), \(c=0\). 4. Using \(A(3, 5)\), \(5=f(3)=27a+9b+d=d\), so \(d=5\). Then \(f(1)=1\) gives \(a-3a+5=1\), so \(a=2\) and \(b=-6\). Thus, \(f(x)=2x^3-6x^2+5\). 5. Reflection across \(y=1\) changes each output \(y\) to \(2-y\). Therefore, \(g(x)=2-f(x)=-2x^3+6x^2-3\). 6. Since \(f'(x)=6x^2-12x=6x(x-2)\), the critical values are \(x=0\) and \(x=2\). The second derivative is \(f''(x)=12x-12\). Thus, \(f''(0)<0\), so \((0, 5)\) is a local maximum, and \(f''(2)>0\), so \((2, -3)\) is a local minimum.

a) \(f(x)=2x^3-6x^2+5\) b) Reflection across \(y=1\) gives \(g(x)=2-f(x)=-2x^3+6x^2-3\). c) Local maximum: \((0, 5)\) Local minimum: \((2, -3)\)

- Find the x-intercept algebraically. - Examine the exponential factor as \(x\to-\infty\). - Use the product rule to find the critical number. - Use the second derivative to classify the critical point.

1. The function \(p\) is zero when \(x-1=0\), so its x-intercept is \(x=1\). The graph labeled \(f\) has this intercept. Also, as \(x\to-\infty\), \(e^x\to 0\) quickly enough that \((x-1)e^x\to 0\) from below. This also matches the graph labeled \(f\). 2. Differentiate using the product rule: \(p'(x)=e^x+(x-1)e^x=xe^x\). Since \(e^x>0\), the only critical number is \(x=0\). 3. The second derivative is \(p''(x)=(x+1)e^x\). Since \(p''(0)=1>0\), \(x=0\) gives a local minimum. Finally, \(p(0)=-1\). Therefore, the local minimum is \((0, -1)\).

a) The graph labeled \(f\) represents \(p\). b) Local minimum: \((0, -1)\)

- Factor the function to find its zeros. - Use the first and second derivatives to find and classify the critical points. - Eliminate \(a\) from the maximum coordinates. - Compare the positive x-intercept of each graph.

1. Factor: \(g_a(x)=x^2(a-x)\). The zeros are \(x=0\), with multiplicity two, and \(x=a\). 2. Differentiate: \(g_a'(x)=x(2a-3x)\) and \(g_a''(x)=2a-6x\). The critical points are \(x=0\) and \(x=\frac{2a}{3}\). Since \(g_a''(0)=2a>0\), the origin is a local minimum. Since \(g_a''\left(\frac{2a}{3}\right)=-2a<0\), the local maximum is \(M_a\left(\frac{2a}{3},\frac{4a^3}{27}\right)\). 3. At the maximum, \(a=\frac{3x}{2}\). Substitution gives \(y=\frac{4}{27}\left(\frac{3x}{2}\right)^3=\frac{1}{2}x^3\). Since \(a>0\), the locus has \(x>0\). 4. Setting \(\frac{4a^3}{27}=4\) gives \(a^3=27\), so \(a=3\). 5. The positive zero occurs at \(x=a\). Graph p has its positive zero at \(x=1.5\), so it corresponds to \(a=1.5\). Graph q has its positive zero at \(x=3\), so it corresponds to \(a=3\).

a) Zeros: \(x=0\) with multiplicity two and \(x=a\); local maximum: \(M_a\left(\frac{2a}{3},\frac{4a^3}{27}\right)\) b) \(y=\frac{1}{2}x^3\) for \(x>0\) c) \(a=3\) d) Graph p: \(a=1.5\); graph q: \(a=3\)

- Symmetry about the y-axis restricts the powers that may appear. - An inflection point on the x-axis gives a function-value condition and a second-derivative condition. - Use the y-intercept to determine the constant term. - Solve the resulting two-equation system for the remaining coefficients.

1. Because \(f\) is even, write \(f(x)=ax^4+bx^2+c\). 2. The point \(M(0, 4)\) gives \(c=4\). At the inflection point with \(x=1\), both \(f(1)=0\) and \(f''(1)=0\). The first condition gives \(a+b=-4\). Since \(f''(x)=12ax^2+2b\), the second gives \(12a+2b=0\), or \(6a+b=0\). Solving the system yields \(a=\frac{4}{5}\) and \(b=-\frac{24}{5}\). Therefore, \(f(x)=\frac{4}{5}x^4-\frac{24}{5}x^2+4\). 3. Since \(f''(0)=-\frac{48}{5}<0\), \(M(0,4)\) is a local maximum. Also, \(f'''(\pm1)=\pm\frac{96}{5}\ne0\), so \(x=-1\) and \(x=1\) are indeed inflection values.

1. \(f(x)=ax^4+bx^2+c\) 2. \(f(x)=\frac{4}{5}x^4-\frac{24}{5}x^2+4\)

- Use the y-axis symmetry to reduce the number of coefficients. - A marked inflection point gives conditions on the function and its second derivative. - Find the critical values from the first derivative, then classify them. - Keep \(\sqrt{3}\) as an exact value.

1. Because \(g\) is even, write \(g(x)=ax^4+bx^2+c\). The graph passes through the origin, so \(c=0\). The marked inflection point \(I_2(1, \frac{5}{2})\) gives \(g(1)=\frac{5}{2}\) and \(g''(1)=0\). Therefore, \(a+b=\frac{5}{2}\) and \(12a+2b=0\). Solving gives \(a=-\frac{1}{2}\) and \(b=3\), so \(g(x)=-\frac{1}{2}x^4+3x^2\). Since \(g'''(\pm1)=\mp12\ne0\), both marked points are indeed inflection points. 2. The derivative is \(g'(x)=-2x^3+6x=-2x(x^2-3)\). Thus, the critical values are \(x=0\) and \(x=\pm\sqrt{3}\). Since \(g''(0)=6>0\), \(x=0\) is a local minimum. At \(x=\pm\sqrt{3}\), \(g''(x)=-6x^2+6=-12<0\), so both are local maxima. Their common function value is \(g(\pm\sqrt{3})=-\frac{1}{2}(9)+3(3)=\frac{9}{2}\).

1. \(g(x)=-\frac{1}{2}x^4+3x^2\) 2. Local maxima: \((-\sqrt{3}, \frac{9}{2})\) and \((\sqrt{3}, \frac{9}{2})\)

- Use the y-axis symmetry to simplify the polynomial form. - Translate “meets the ground with a horizontal tangent” into a function-value condition and a derivative condition. - Use the center height to determine the constant term. - Solve the resulting system for the remaining coefficients.

1. Because the arch is symmetric about the y-axis, write \(f(x)=ax^4+bx^2+c\). 2. The center point \((0, 20)\) gives \(c=20\). 3. The endpoint \((20, 0)\) gives \(160{,}000a+400b+20=0\). 4. A horizontal tangent at \(x=20\) requires \(f'(20)=0\). Since \(f'(x)=4ax^3+2bx\), \(32{,}000a+40b=0\), so \(b=-800a\). 5. Substitution into the endpoint equation gives \(-160{,}000a=-20\), so \(a=\frac{1}{8000}\) and \(b=-\frac{1}{10}\). 6. Therefore, \(f(x)=\frac{1}{8000}x^4-\frac{1}{10}x^2+20\), for \(-20\le x\le20\). Also, \(f''(0)=-\frac{1}{5}<0\), confirming the maximum at the center.

\(f(x)=\frac{1}{8000}x^4-\frac{1}{10}x^2+20\), for \(-20\le x\le20\)

- What powers can appear in a polynomial whose graph is symmetric about the y-axis? - Use the derivative tests for a local maximum and for inflection points. - Choose simple coefficients that make the second derivative have two distinct real zeros. - Verify every condition after choosing the function.

1. Symmetry about the y-axis requires an even polynomial, so begin with \(f(x)=ax^4+bx^2+c\). 2. Choose \(f(x)=x^4-6x^2\). This is a fourth-degree even polynomial, so its graph is symmetric about the y-axis. 3. The derivatives are \(f'(x)=4x^3-12x\) and \(f''(x)=12x^2-12\). Since \(f'(0)=0\) and \(f''(0)=-12<0\), the function has a local maximum at \(x=0\). 4. Solving \(f''(x)=0\) gives \(x=\pm1\). The second derivative changes sign at both values, so the function has exactly two inflection values.

One possible function is \(f(x)=x^4-6x^2\).

- Substitute the given point into the function to determine \(k\). - Factor the function to find its zeros and their multiplicities. - Find the critical numbers from the first derivative. - Use the second derivative and the sign of \(k\) to classify each critical point.

1. Substitute \((2, 4)\): \(4k - 8 = 4\), so \(k = 3\). 2. Factor the function: \(f_k(x) = x^2(k - x)\). Thus, \(x = 0\) is a zero of multiplicity \(2\), and \(x = k\) is a simple zero. 3. Differentiate: \(f_k'(x) = 2kx - 3x^2 = x(2k - 3x)\). The critical numbers are \(x = 0\) and \(x = \frac{2k}{3}\). 4. The second derivative is \(f_k''(x) = 2k - 6x\). Therefore, \(f_k''(0) = 2k\) and \(f_k''\left(\frac{2k}{3}\right) = -2k\). 5. Also, \(f_k(0) = 0\) and \(f_k\left(\frac{2k}{3}\right) = \frac{4k^3}{27}\). 6. If \(k > 0\), then \((0, 0)\) is a local minimum and \(\left(\frac{2k}{3}, \frac{4k^3}{27}\right)\) is a local maximum. 7. If \(k < 0\), then \((0, 0)\) is a local maximum and \(\left(\frac{2k}{3}, \frac{4k^3}{27}\right)\) is a local minimum.

a) \(k = 3\) b) \(x = 0\), with multiplicity \(2\), and \(x = k\), with multiplicity \(1\) c) For \(k > 0\): local minimum at \((0, 0)\) and local maximum at \(\left(\frac{2k}{3}, \frac{4k^3}{27}\right)\) For \(k < 0\): local maximum at \((0, 0)\) and local minimum at \(\left(\frac{2k}{3}, \frac{4k^3}{27}\right)\)

- Factor the function to find its zeros. - Use the first derivative to find critical numbers and the second derivative to classify them. - Eliminate \(t\) from the coordinates of the local minimum. - Write and solve an inequality using the local maximum value.

1. Factor: \(f_t(x) = x\left(\frac{1}{3}x^2 - t^2\right)\). The zeros are \(x = 0\) and \(x = \pm t\sqrt{3}\). 2. Differentiate: \(f_t'(x) = x^2 - t^2\). The critical numbers are \(x = -t\) and \(x = t\). 3. Since \(f_t''(x) = 2x\), \(f_t''(-t) < 0\) and \(f_t''(t) > 0\). Therefore, the local maximum is \(\left(-t, \frac{2}{3}t^3\right)\), and the local minimum is \(\left(t, -\frac{2}{3}t^3\right)\). 4. For the local minima, \(x = t\). Eliminating the parameter gives \(y = -\frac{2}{3}x^3\), with \(x > 0\). 5. The local maximum value is greater than \(18\) when \(\frac{2}{3}t^3 > 18\). This simplifies to \(t^3 > 27\), so \(t > 3\).

a) \(x = -t\sqrt{3}\), \(x = 0\), and \(x = t\sqrt{3}\) b) Local maximum at \(\left(-t, \frac{2}{3}t^3\right)\); local minimum at \(\left(t, -\frac{2}{3}t^3\right)\) c) \(y = -\frac{2}{3}x^3\), for \(x > 0\) d) \(t > 3\)

- Factor both functions and compare their zeros. - Find the critical numbers of \(q_a\) and use the second derivative test. - Eliminate \(a\) from the coordinates of the local minimum. - Remember to reverse the inequality when multiplying or dividing by a negative quantity.

1. Factor each function: \(p_a(x) = x(x - 2a)\) and \(q_a(x) = x^2\left(\frac{x}{a} - 2\right)\). Both functions have zeros at \(x = 0\) and \(x = 2a\). 2. Differentiate \(q_a\): \(q_a'(x) = \frac{3}{a}x^2 - 4x = x\left(\frac{3}{a}x - 4\right)\). The critical numbers are \(x = 0\) and \(x = \frac{4a}{3}\). 3. Since \(q_a''(x) = \frac{6}{a}x - 4\), we have \(q_a''\left(\frac{4a}{3}\right) = 4 > 0\). Thus, the local minimum is \(\left(\frac{4a}{3}, -\frac{32a^2}{27}\right)\). 4. From \(x = \frac{4a}{3}\), obtain \(a = \frac{3x}{4}\). Substitute into the \(y\)-coordinate: \(y = -\frac{32}{27}\left(\frac{3x}{4}\right)^2 = -\frac{2}{3}x^2\). Because \(a > 0\), the locus has \(x > 0\). 5. Solve \(-\frac{32a^2}{27} < -32\). Dividing by a negative number reverses the inequality, giving \(a^2 > 27\). Since \(a > 0\), \(a > 3\sqrt{3}\).

a) Both functions have zeros at \(x = 0\) and \(x = 2a\). b) Local minimum at \(\left(\frac{4a}{3}, -\frac{32a^2}{27}\right)\) c) \(y = -\frac{2}{3}x^2\), for \(x > 0\) d) \(a > 3\sqrt{3}\)

- Find the critical numbers in terms of \(a\). - Compute the corresponding function values. - Eliminate \(a\) in both cases and compare the resulting equations. - Apply the restriction \(a \ne 0\).

1. Differentiate: \(g_a'(x) = 3x^2 - 3a^2\). The critical numbers satisfy \(x^2 = a^2\), so \(x = a\) or \(x = -a\). 2. Since \(g_a''(x) = 6x\) and \(a \ne 0\), the second derivative is nonzero at both critical numbers. Thus, both points are local extrema. 3. At \(x = a\), the function value is \(g_a(a) = -2a^3 + 2\). Replacing \(a\) with \(x\) gives \(y = -2x^3 + 2\). 4. At \(x = -a\), the function value is \(g_a(-a) = 2a^3 + 2\). Since \(a = -x\), this also gives \(y = -2x^3 + 2\). 5. Because \(a \ne 0\), neither critical number can be zero. Therefore, the locus is \(y = -2x^3 + 2\), for \(x \ne 0\).

\(y = -2x^3 + 2\), for \(x \in \mathbb{R} \setminus \{0\}\)

- Factor out \(x^2\) before finding the zeros. - Separate the cases \(a>0\) and \(a\leq0\) when square roots appear. - Use the first and second derivatives to find and classify critical points. - Set the second derivative equal to \(0\), check whether the candidates are real, and verify a change in concavity.

1. Factor \(f_a(x)=x^2(x^2-2a)\). The value \(x=0\) is always a zero. When \(a>0\), the additional zeros are \(x=\pm\sqrt{2a}\); when \(a\leq0\), there are no additional real zeros. 2. Differentiate: \(f_a'(x)=4x(x^2-a)\) and \(f_a''(x)=4(3x^2-a)\). 3. If \(a>0\), the critical numbers are \(0\) and \(\pm\sqrt{a}\). Since \(f_a''(0)=-4a<0\), \((0, 0)\) is a local maximum. Since \(f_a''(\pm\sqrt{a})=8a>0\), \((\pm\sqrt{a}, -a^2)\) are local minima. 4. If \(a\leq0\), \(x=0\) is the only critical number. The first derivative changes from negative to positive there, so \((0, 0)\) is a local minimum. 5. Inflection-point candidates satisfy \(f_a''(x)=0\), so \(x=\pm\sqrt{a/3}\). These are real only when \(a>0\). Because \(f_a''(x)=4(3x^2-a)\) changes sign at both values, they are inflection points. Thus, there are two inflection points for \(a>0\) and none for \(a\leq0\).

Zeros: \(x=0\); when \(a>0\), also \(x=\pm\sqrt{2a}\). For \(a>0\): local maximum \((0, 0)\); local minima \((\pm\sqrt{a}, -a^2)\). For \(a\leq0\): local minimum \((0, 0)\). Inflection-point x-coordinates: \(x=\pm\sqrt{a/3}\) for \(a>0\); none for \(a\leq0\).

- Factor out common factors after using the product rule. - The exponential factor is never zero. - Set each remaining factor of the first derivative equal to \(0\). - Solve the quadratic from the second derivative, then verify that its sign changes at each root.

1. Since \(e^x>0\), the only zero is \(x=k\), with multiplicity \(2\). 2. Use the product rule: \(g_k'(x)=(x-k)(x-k+2)e^x\). 3. The critical numbers are \(x=k\) and \(x=k-2\). 4. The second derivative is \(g_k''(x)=\bigl(x^2+(4-2k)x+k^2-4k+2\bigr)e^x\). 5. At \(x=k\), \(g_k''(k)=2e^k>0\), so \((k, 0)\) is a local minimum. At \(x=k-2\), \(g_k''(k-2)=-2e^{k-2}<0\), so \((k-2, 4e^{k-2})\) is a local maximum. 6. Set the second derivative equal to \(0\). Because \(e^x>0\), solve \(x^2+(4-2k)x+k^2-4k+2=0\), which gives \(x=k-2\pm\sqrt{2}\). The quadratic factor changes sign at both distinct roots, so both values are x-coordinates of inflection points.

Zero: \(x=k\), with multiplicity \(2\). Local minimum: \((k, 0)\). Local maximum: \((k-2, 4e^{k-2})\). Inflection-point x-coordinates: \(x=k-2\pm\sqrt{2}\).

- Apply the chain rule to \((g(x))^2\). - Use the product rule to find the second derivative. - A graph is concave up where its second derivative is positive. - Substitute \(g(x_0)=0\) into both derivative formulas. - Recall the second derivative test for a local minimum.

1. Apply the chain rule: \(f'(x)=2g(x)g'(x)\). 2. Differentiate using the product rule: \(f''(x)=2[(g'(x))^2+g(x)g''(x)]\) \(=2(g'(x))^2+2g(x)g''(x)\). 3. Under the assumptions in b), \((g'(x))^2\ge 0\) and \(g(x)g''(x)>0\). Therefore, \(f''(x)>0\) for every real \(x\), so the graph of \(f\) is concave up everywhere. 4. At \(x=x_0\), \(g(x_0)=0\), so \(f'(x_0)=2g(x_0)g'(x_0)=0\). Also, \(f''(x_0)=2(g'(x_0))^2+2g(x_0)g''(x_0)=2(g'(x_0))^2>0\). By the second derivative test, \(f\) has a local minimum at \(x_0\).

a) \(f'(x)=2g(x)g'(x)\) and \(f''(x)=2(g'(x))^2+2g(x)g''(x)\) b) \(f''(x)>0\) everywhere, so the graph of \(f\) is concave up everywhere. c) \(f'(x_0)=0\) and \(f''(x_0)=2(g'(x_0))^2>0\), so \(f\) has a local minimum at \(x_0\).

- Find the first and second derivatives. - Use the domain \(x>0\) when solving the critical-point equation. - A positive second derivative throughout the domain establishes strict convexity. - A point on the x-axis has y-coordinate \(0\).

1. Differentiate: \(g_k'(x)=x-\frac{k}{x}\) and \(g_k''(x)=1+\frac{k}{x^2}\). 2. The critical-point equation is \(x-\frac{k}{x}=0\), so \(x^2=k\). Because \(x>0\), the only critical point is \(x=\sqrt k\). 3. Since \(g_k''(x)>0\) for all \(x>0\), the function is strictly convex and this critical point is the unique minimum. Its y-coordinate is \(g_k(\sqrt k)=\frac{k}{2}(1-\ln k)\). Thus, the minimum point is \(T=\left(\sqrt k, \frac{k}{2}(1-\ln k)\right)\). 4. For the point to lie on the x-axis, \(\frac{k}{2}(1-\ln k)=0\). Since \(k>0\), \(\ln k=1\), so \(k=e\).

a) \(T=\left(\sqrt k, \frac{k}{2}(1-\ln k)\right)\) b) \(k=e\)

- Recall the first- and second-derivative conditions at a local minimum when the second derivative is positive. - Apply the chain rule to the square root and the squared inner function. - When evaluating the second derivative at \(x_0\), identify terms containing \(g'(x_0)\). - Use the signs of all factors in the final expression.

1. By the chain rule, \(h'(x)=\frac{g(x)g'(x)}{\sqrt{(g(x))^2+1}}\). 2. Since \(g\) has a local minimum at \(x_0\), \(g'(x_0)=0\). Therefore, \(h'(x_0)=0\). 3. Differentiate \(h'\). At \(x_0\), every term containing \(g'(x_0)\) vanishes, leaving \(h''(x_0)=\frac{g(x_0)g''(x_0)}{\sqrt{(g(x_0))^2+1}}\). 4. The numerator factors \(g(x_0)\) and \(g''(x_0)\) are positive, and the denominator is positive. Hence, \(h''(x_0)>0\). 5. By the second derivative test, \(h\) has a local minimum at \(x_0\).

\(h'(x_0)=0\) and \(h''(x_0)=\frac{g(x_0)g''(x_0)}{\sqrt{(g(x_0))^2+1}}>0\). Therefore, \(h\) has a local minimum at \(x_0\).

- A product is zero when at least one factor is zero. - Use the product and chain rules. - Show that the first derivative has only one zero by analyzing the second derivative. - Examine the logarithm near the left endpoint of the domain.

1. The product is zero when \(x-2=0\) or \(\ln(x-1)=0\). Both conditions give \(x=2\), so this is the only zero. 2. Differentiate using the product rule: \(g'(x)=\ln(x-1)+\frac{x-2}{x-1}\). At \(x=2\), both terms are zero, so \(x=2\) is a critical number. 3. The second derivative is \(g''(x)=\frac{1}{x-1}+\frac{1}{(x-1)^2}>0\) for every \(x>1\). Therefore, \(g'\) is strictly increasing and can have at most one zero. Thus \(x=2\) is the only critical number, and \((2,0)\) is a local minimum. 4. As \(x\to1^+\), \(x-2\to-1\) while \(\ln(x-1)\to-\infty\), so \(g(x)\to\infty\). Hence \(x=1\) is a vertical asymptote. As \(x\to\infty\), both factors grow without bound, so \(g(x)\to\infty\). There is no horizontal asymptote.

a) Zero: \(x=2\). Local minimum: \((2,0)\). b) \(\lim_{x\to1^+}g(x)=\infty\), and \(\lim_{x\to\infty}g(x)=\infty\). Vertical asymptote: \(x=1\).

- Use the point and the horizontal-tangent condition to write two equations in \(a\) and \(b\). - Recall that \(\ln(e)=1\). - Compare the growth of a logarithm with the growth of a quadratic term. - Use the second derivative test.

1. Since \(P(e,1)\) is on the graph, \(a+be^2=1\). 2. The derivative is \(g'(x)=\frac{a}{x}+2bx\). The horizontal tangent at \(x=e\) gives \(\frac{a}{e}+2be=0\), or \(a+2be^2=0\). 3. Subtracting the first equation from the second gives \(be^2=-1\), so \(b=-\frac{1}{e^2}\). Substitution gives \(a=2\). 4. Therefore, \(g(x)=2\ln(x)-\frac{x^2}{e^2}\). As \(x\to0^+\), the logarithmic term approaches \(-\infty\), so \(g(x)\to-\infty\). As \(x\to\infty\), the negative quadratic term dominates, so \(g(x)\to-\infty\). 5. The second derivative is \(g''(x)=-\frac{2}{x^2}-\frac{2}{e^2}\). Thus \(g''(e)=-\frac{4}{e^2}<0\), so \(P\) is a local maximum.

a) \(a=2\), \(b=-\frac{1}{e^2}\) b) \(\lim_{x\to0^+}g(x)=-\infty\) and \(\lim_{x\to\infty}g(x)=-\infty\) c) Since \(g'(e)=0\) and \(g''(e)=-\frac{4}{e^2}<0\), \(P\) is a local maximum.

- Set the first derivative equal to zero. - Factor out a common factor first. - For an equation involving \(x^4\) and \(x^2\), substitute \(u = x^2\). - Use the second derivative to classify every critical number.

1. Differentiate: \(g'(x) = x^5 - 6x^3 + 8x = x(x^4 - 6x^2 + 8)\). 2. One solution is \(x = 0\). For the quartic factor, let \(u = x^2\). Then \(u^2 - 6u + 8 = 0\), so \(u = 2\) or \(u = 4\). 3. Therefore, the critical numbers are \(x = 0\), \(x = \pm\sqrt{2}\), and \(x = \pm 2\). 4. Use \(g''(x) = 5x^4 - 18x^2 + 8\). At \(x = 0\) and \(x = \pm 2\), the second derivative is positive, so these are local minima. At \(x = \pm\sqrt{2}\), the second derivative is negative, so these are local maxima.

Local minima occur at \(x = -2\), \(x = 0\), and \(x = 2\). Local maxima occur at \(x = -\sqrt{2}\) and \(x = \sqrt{2}\).

- Recenter the expression around \(x=1\) to identify symmetry. - Factor to find the zeros. - Use the leading term to determine end behavior. - Use the first and second derivatives for extrema and inflection points. - Use the absolute minimum and end behavior to determine the range.

1. Factor and recenter: \(f(x)=x^2(x-2)^2=((x-1)^2-1)^2\). Thus, \(f(1+u)=f(1-u)\), so the graph is symmetric about \(x=1\). 2. The leading term is \(x^4\), so \(f(x)\to\infty\) as \(x\to\pm\infty\). 3. The zeros are \(x=0\) and \(x=2\), both with multiplicity \(2\). The y-intercept is \((0, 0)\). 4. The derivative is \(f'(x)=4x(x-1)(x-2)\). The second derivative is \(f''(x)=12x^2-24x+8\). Therefore, \((0, 0)\) and \((2, 0)\) are local minima, and \((1, 1)\) is a local maximum. 5. Solving \(f''(x)=0\) gives \(x=1\pm\frac{\sqrt{3}}{3}\). At both values, \(f(x)=\frac{4}{9}\). Thus, the inflection points are \((1\pm\frac{\sqrt{3}}{3}, \frac{4}{9})\). 6. The absolute minimum value is \(0\), and the function is unbounded above. Therefore, the range is \([0, \infty)\).

1) Symmetric about \(x=1\) 2) \(f(x)\to\infty\) as \(x\to\pm\infty\) 3) Intercepts: \((0, 0)\) and \((2, 0)\) 4) Local minima: \((0, 0)\), \((2, 0)\); local maximum: \((1, 1)\) 5) \((1\pm\frac{\sqrt{3}}{3}, \frac{4}{9})\) 6) \([0, \infty)\)

- Find the critical numbers from the first derivative. - Use the second derivative to classify them. - Analyze the sign of the first derivative between its zeros. - Use the parameter restriction \(t > 0\) in part c.

1. Differentiate: \(f_t'(x) = x^2 - t^2\) and \(f_t''(x) = 2x\). 2. The critical numbers satisfy \(x^2 - t^2 = 0\), so \(x = -t\) and \(x = t\). 3. Since \(f_t''(-t) = -2t < 0\), there is a local maximum at \((-t, 5 + \frac{2}{3}t^3)\). Since \(f_t''(t) = 2t > 0\), there is a local minimum at \((t, 5 - \frac{2}{3}t^3)\). 4. The derivative \((x - t)(x + t)\) is positive on \((-\infty, -t)\) and \((t, \infty)\), and negative on \((-t, t)\). Thus, \(f_t\) is increasing on the first and third intervals and decreasing on the middle interval. 5. A local extremum at \(x = 4\) requires \(4 = t\) or \(4 = -t\). Because \(t > 0\), \(t = 4\).

a) Local maximum at \((-t, 5 + \frac{2}{3}t^3)\); local minimum at \((t, 5 - \frac{2}{3}t^3)\) b) Increasing on \((-\infty, -t)\) and \((t, \infty)\); decreasing on \((-t, t)\) c) \(t = 4\)

- Use the zero-product property for the x-intercepts. - Expand or use the product rule to differentiate. - A double zero often corresponds to a horizontal tangent. - Use the second derivative to classify the critical points.

1. The y-intercept is \(f(0)=-4\), so it is \((0, -4)\). From the factored form, the x-intercepts are \((-2, 0)\), \((1, 0)\), and \((2, 0)\); \(x=1\) is a double zero. 2. Expanding gives \(f(x)=x^4-2x^3-3x^2+8x-4\). The derivative factors as \(f'(x)=(x-1)(4x^2-2x-8)\). Thus, the critical values are \(x=1\) and \(x=\frac{1\pm\sqrt{33}}{4}\). 3. Since \(f''(1)=-6<0\), \((1, 0)\) is a local maximum. The other two critical points are local minima. Their exact coordinates are \((\frac{1-\sqrt{33}}{4}, -\frac{207+33\sqrt{33}}{32})\) and \((\frac{1+\sqrt{33}}{4}, -\frac{207-33\sqrt{33}}{32})\). Approximately, these are \((-1.19, -12.39)\) and \((1.69, -0.54)\).

1. Intercepts: \((0, -4)\), \((-2, 0)\), \((1, 0)\), \((2, 0)\) 2. Local maximum: \((1, 0)\) Local minima: \((\frac{1-\sqrt{33}}{4}, -\frac{207+33\sqrt{33}}{32})\) and \((\frac{1+\sqrt{33}}{4}, -\frac{207-33\sqrt{33}}{32})\)

- Test whether the function is odd. - Factor the polynomial completely to find zeros and multiplicities. - Substitute \(u=x^2\) when solving the first-derivative equation. - Use origin symmetry to check paired extrema.

1. Since \(f(-x)=-f(x)\), the graph is symmetric about the origin. Factoring gives \(f(x)=x(x^2-1)^2=x(x-1)^2(x+1)^2\). Thus, \(x=0\) is a simple zero, and \(x=\pm1\) are double zeros. 2. The derivative is \(f'(x)=5x^4-6x^2+1=(x^2-1)(5x^2-1)\). The critical values are \(x=\pm1\) and \(x=\pm\frac{\sqrt{5}}{5}\). 3. The second derivative is \(f''(x)=20x^3-12x\). It classifies \((-1, 0)\) and \((\frac{\sqrt{5}}{5}, \frac{16\sqrt{5}}{125})\) as local maxima, and \((1, 0)\) and \((-\frac{\sqrt{5}}{5}, -\frac{16\sqrt{5}}{125})\) as local minima.

a) Symmetric about the origin. Zeros: \(x=0\), multiplicity \(1\); \(x=\pm1\), multiplicity \(2\). b) Local maxima: \((-1, 0)\) and \((\frac{\sqrt{5}}{5}, \frac{16\sqrt{5}}{125})\) Local minima: \((1, 0)\) and \((-\frac{\sqrt{5}}{5}, -\frac{16\sqrt{5}}{125})\)

- Differentiate the original function first. - Differentiate the square using the chain rule or expansion. - Compare the squares of nearby negative numbers. - Consider the monotonicity of \(y=u^2\) when \(u<0\).

1. Since \(f'(x)=2x\) and \(f''(x)=2>0\), \(f\) has a local minimum at \(x=0\), where \(f(0)=-5\). 2. The squared function is \(g(x)=(x^2-5)^2\). Its second derivative is \(g''(x)=12x^2-20\), so \(g''(0)=-20<0\). Therefore, \(g\) has a local maximum at \(x=0\). 3. Near \(x=0\), the values of \(f\) are negative. On negative inputs, squaring reverses order by magnitude: values closer to \(-5\) produce larger squares than nearby values closer to \(0\). Thus the minimum becomes a maximum.

1. \(f\) has a local minimum at \(x=0\). 2. \(g\) has a local maximum at \(x=0\). 3. Because the nearby values of \(f\) are negative, squaring reverses their order and changes the minimum into a maximum.

- Begin with the general form of a cubic polynomial and its first two derivatives. - Use the inflection point at the origin to determine two coefficients. - Use the tangent line to determine the slope at the origin. - Apply the first-derivative condition at \(x=1\). - Use the second derivative to classify the resulting critical point.

1. Let \(f(x)=ax^3+bx^2+cx+d\). An inflection point at the origin gives \(f(0)=0\) and \(f''(0)=0\), so \(d=0\) and \(b=0\). 2. The tangent line \(y=x\) has slope \(1\), so \(f'(0)=1\). Therefore, \(c=1\), and \(f(x)=ax^3+x\). 3. A local extremum at \(x=1\) requires \(f'(1)=0\). Since \(f'(x)=3ax^2+1\), this gives \(3a+1=0\), so \(a=-\frac{1}{3}\). 4. Then \(f''(x)=6ax=-2x\), so \(f''(1)=-2<0\). Thus, the critical point at \(x=1\) is a local maximum, not a local minimum. Therefore, no such cubic polynomial exists.

No such cubic polynomial exists. The conditions force \(f(x)=-\frac{1}{3}x^3+x\), which has a local maximum rather than a local minimum at \(x=1\).

- Translate each point and derivative condition into an equation. - First find the only cubic that satisfies the point, critical-point, and inflection conditions. - Use the second derivative to classify the critical point at \(x=2\). - A contradiction in the classification proves nonexistence.

1. Let \(f(x)=ax^3+bx^2+cx+d\). The conditions \(f(0)=0\) and \(f''(3)=0\) give \(d=0\) and \(b=-9a\). 2. The point \(T(2, 4)\) and the necessary extremum condition give \(f(2)=4\) and \(f'(2)=0\). Substituting \(b=-9a\) yields \(-28a+2c=4\) and \(-24a+c=0\). 3. Solving gives \(a=\frac{1}{5}\), \(b=-\frac{9}{5}\), and \(c=\frac{24}{5}\). This is the only cubic satisfying the point, critical-point, and inflection conditions. 4. Its second derivative at \(x=2\) is \(f''(2)=-\frac{6}{5}<0\). Therefore, \(T(2, 4)\) would be a local maximum, not a local minimum. No cubic polynomial satisfies all the stated conditions.

No such cubic polynomial exists. The conditions force \(f(x)=\frac{1}{5}x^3-\frac{9}{5}x^2+\frac{24}{5}x\), and \(f''(2)=-\frac{6}{5}<0\), so \(T(2, 4)\) is a local maximum.

- Treat the leading coefficient as a free parameter. - Express the remaining coefficients in terms of that parameter. - Use the second-derivative test at \(x=2\). - Verify that the concavity actually changes at \(x=3\).

1. The conditions \(g(0)=0\), \(g''(3)=0\), \(g(2)=4\), and \(g'(2)=0\) give \(e=0\) and a system for the other coefficients. 2. Solving in terms of the free parameter \(a\) gives \(b=\frac{1}{5}-\frac{42}{5}a\), \(c=\frac{108}{5}a-\frac{9}{5}\), and \(d=\frac{24}{5}-\frac{88}{5}a\). 3. For \(T(2, 4)\) to be a local minimum, \(g''(2)>0\). Substitution gives \(g''(2)=-\frac{48}{5}a-\frac{6}{5}\). Thus, \(-\frac{48}{5}a-\frac{6}{5}>0\), which is equivalent to \(a<-\frac{1}{8}\). 4. For this range, \(g'''(3)=\frac{108}{5}a+\frac{6}{5}\ne0\), so \(x=3\) is indeed an inflection point. Therefore, infinitely many such quartic polynomials exist.

Yes. Infinitely many such quartic polynomials exist, and the leading coefficient must satisfy \(a<-\frac{1}{8}\).

- Factor the first derivative to find the critical numbers. - Use the sign of the second derivative and the sign of \(k\) to classify the extrema. - Find the inflection point from the second derivative. - Check separately what happens when \(k = 0\).

1. Differentiate: \(f_k'(x) = 3x^2 - 12kx + 9k^2 = 3(x - k)(x - 3k)\), and \(f_k''(x) = 6x - 12k\). 2. The critical numbers are \(x = k\) and \(x = 3k\). Their function values are \(f_k(k) = 4k^3\) and \(f_k(3k) = 0\). 3. If \(k > 0\), then \(f_k''(k) = -6k < 0\) and \(f_k''(3k) = 6k > 0\). Thus, \((k, 4k^3)\) is a local maximum and \((3k, 0)\) is a local minimum. 4. If \(k < 0\), the signs reverse. Thus, \((k, 4k^3)\) is a local minimum and \((3k, 0)\) is a local maximum. 5. If \(k = 0\), the two critical numbers coincide at \(x = 0\). The function is \(f_0(x) = x^3\), so there is no local extremum; the origin is a stationary inflection point. 6. An inflection point satisfies \(f_k''(x) = 0\), so \(x = 2k\). Since \(f_k'''(x) = 6 \ne 0\), this is an inflection point. Its value is \(f_k(2k) = 2k^3\), so the inflection point is \((2k, 2k^3)\).

For \(k > 0\): local maximum at \((k, 4k^3)\); local minimum at \((3k, 0)\) For \(k < 0\): local minimum at \((k, 4k^3)\); local maximum at \((3k, 0)\) For \(k = 0\): no local extrema; stationary inflection point at \((0, 0)\) Inflection point for all \(k\): \((2k, 2k^3)\)

- Find the two critical numbers in terms of \(a\). - Use the second derivative and consider the sign of \(a\). - Determine when the two critical numbers coincide. - Use the second and third derivatives to locate the inflection point.

1. Differentiate: \(g_a'(x) = \frac{3}{2}x^2 - 3ax = \frac{3}{2}x(x - 2a)\), \(g_a''(x) = 3x - 3a\), and \(g_a'''(x) = 3\). 2. The critical numbers are \(x = 0\) and \(x = 2a\). Their function values are \(g_a(0) = 2\) and \(g_a(2a) = 2 - 2a^3\). 3. If \(a > 0\), then \(g_a''(0) = -3a < 0\) and \(g_a''(2a) = 3a > 0\). Thus, \((0, 2)\) is a local maximum and \((2a, 2 - 2a^3)\) is a local minimum. 4. If \(a < 0\), the classifications reverse: \((0, 2)\) is a local minimum and \((2a, 2 - 2a^3)\) is a local maximum. 5. If \(a = 0\), the critical numbers coincide at \(x = 0\). Since \(g_0''(0) = 0\) and \(g_0'''(0) = 3 \ne 0\), the point \((0, 2)\) is a stationary inflection point, not a local extremum. 6. Inflection points satisfy \(g_a''(x) = 0\), so \(x = a\). The function value is \(g_a(a) = 2 - a^3\). Therefore, the inflection point is \((a, 2 - a^3)\).

For \(a > 0\): local maximum at \((0, 2)\); local minimum at \((2a, 2 - 2a^3)\) For \(a < 0\): local minimum at \((0, 2)\); local maximum at \((2a, 2 - 2a^3)\) For \(a = 0\): stationary inflection point at \((0, 2)\), with no local extrema Inflection point: \((a, 2 - a^3)\) The point on the \(y\)-axis is a local maximum when \(a > 0\).

- Use symmetry when finding the two local maxima. - Eliminate \(a\) between the coordinates of a maximum. - Use the second derivative to locate the inflection points. - Simplify the ratio and check whether \(a\) cancels.

1. Differentiate: \(g_a'(x)=ax-\frac{1}{3}x^3=x\left(a-\frac{x^2}{3}\right)\). The nonzero critical points are \(x=\pm\sqrt{3a}\). Since \(g_a''(x)=a-x^2\), we have \(g_a''(\pm\sqrt{3a})=-2a<0\), so both are local maxima. Their y-coordinate is \(\frac{3a^2}{4}\). Thus the local maxima are \(\left(\pm\sqrt{3a}, \frac{3a^2}{4}\right)\). 2. At a local maximum, \(x^2=3a\), so \(a=\frac{x^2}{3}\). Substituting into \(y=\frac{3a^2}{4}\) gives \(y=\frac{x^4}{12}\). Because \(a>0\), the locus has \(x\ne0\). 3. Inflection-point candidates satisfy \(g_a''(x)=a-x^2=0\), so \(x=\pm\sqrt{a}\). The second derivative changes sign at both values, so they are inflection points. Their y-coordinate is \(y_{\mathrm{infl}}=\frac{1}{2}a^2-\frac{1}{12}a^2=\frac{5a^2}{12}\). Therefore, \(\frac{y_{\max}}{y_{\mathrm{infl}}}=\frac{3a^2/4}{5a^2/12}=\frac{9}{5}\), independent of \(a\).

1. \(\left(\pm\sqrt{3a}, \frac{3a^2}{4}\right)\) 2. \(y=\frac{1}{12}x^4\) for \(x\ne0\) 3. \(y_{\mathrm{infl}}=\frac{5a^2}{12}\) and \(\frac{y_{\max}}{y_{\mathrm{infl}}}=\frac{9}{5}\)

- Relate the multiplicity of a zero to how the graph meets the x-axis. - Use the product rule and factor the derivative. - Use the second derivative to classify each critical point. - A point on the y-axis has x-coordinate \(0\).

1. The factor \((x-a)^2\) shows that \(x=a\) is a zero of even multiplicity. Since \(a\neq-1\), the factor \(x+1\) is nonzero there, so the graph is tangent to the x-axis at \(x=a\). 2. Differentiate: \(g_a'(x)=2(x-a)(x+1)+(x-a)^2=(x-a)(3x+2-a)\). Thus the critical points have \(x=a\) and \(x=\frac{a-2}{3}\). 3. Their coordinates are \(P_1(a, 0)\) and \(P_2\left(\frac{a-2}{3}, \frac{4}{27}(a+1)^3\right)\). Since \(g_a''(x)=6x+2-4a\), we have \(g_a''(a)=2(a+1)\) and \(g_a''\left(\frac{a-2}{3}\right)=-2(a+1)\). For \(a>-1\), \(P_1\) is a local minimum and \(P_2\) is a local maximum. For \(a<-1\), the classifications are reversed. 4. When \(a>-1\), the local maximum is \(P_2\). It lies on the y-axis when \(\frac{a-2}{3}=0\), giving \(a=2\). When \(a<-1\), the local maximum is \(P_1\), whose x-coordinate is \(a\), so it cannot equal zero. Therefore, the only value is \(a=2\), and the maximum is \((0, 4)\).

a) \(x=a\) is a zero of even multiplicity, so the graph is tangent to the x-axis there. b) \(P_1(a, 0)\) and \(P_2\left(\frac{a-2}{3}, \frac{4}{27}(a+1)^3\right)\). For \(a>-1\), \(P_1\) is a local minimum and \(P_2\) is a local maximum; for \(a<-1\), the classifications are reversed. c) \(a=2\), giving the local maximum \((0, 4)\)

- Compare \(h_k(-x)\) with \(h_k(x)\). - Use the quotient rule and simplify the derivative's numerator. - Consider separately the value of \(k\) that makes the original numerator equal to the denominator. - Classify the critical point using the second derivative.

1. The domain excludes \(x = \pm 2\). Since \(h_k(-x) = h_k(x)\), every graph is symmetric about the \(y\)-axis. 2. Apply the quotient rule: \(h_k'(x) = \frac{2x(x^2 - 4) - 2x(x^2 + k)}{(x^2 - 4)^2} = \frac{-2x(k + 4)}{(x^2 - 4)^2}\). 3. If \(k = -4\), then \(h_{-4}(x) = 1\) throughout its domain. It is constant on each domain interval and has no isolated local extrema. 4. If \(k \ne -4\), the only critical number is \(x = 0\), and \(h_k(0) = -\frac{k}{4}\). 5. The second derivative at the critical number is \(h_k''(0) = -\frac{k + 4}{8}\). If \(k > -4\), this is negative, so \(\left(0, -\frac{k}{4}\right)\) is a local maximum. If \(k < -4\), it is positive, so the point is a local minimum.

a) Every graph is symmetric about the \(y\)-axis. b) For \(k > -4\): local maximum at \(\left(0, -\frac{k}{4}\right)\) For \(k < -4\): local minimum at \(\left(0, -\frac{k}{4}\right)\) For \(k = -4\): no isolated local extrema; \(h_{-4}(x) = 1\) on its domain

- Compare two different parameter values to find the common point. - Use the second derivative test when \(k\ne0\). - When the second derivative is zero, examine the sign of the first derivative and the concavity.

1. If two distinct parameter values \(k\) and \(m\) give the same function value, then \((k-m)x^2e^{-x}=0\). Since \(e^{-x}>0\) and \(k\ne m\), this requires \(x=0\). Also, \(g_k(0)=0\), so the only common point is \((0, 0)\). 2. The first derivative is \(g_k'(x)=\left(-2x^3+(6-k)x^2+2kx\right)e^{-x}\), and \(g_k''(0)=2k\). If \(k>0\), the second derivative is positive at \(0\), so \((0, 0)\) is a local minimum. If \(k<0\), it is a local maximum. If \(k=0\), the second derivative test is inconclusive. In that case, \(g_0'(x)=2x^2(3-x)e^{-x}\). This derivative is positive on both sides of \(0\), so the function has no local extremum there. The third derivative at \(0\) equals \(12\ne0\), so the point is a stationary inflection point.

1) \((0, 0)\) 2) Local minimum for \(k>0\); local maximum for \(k<0\); stationary inflection point for \(k=0\).

- Find all zeros of the first derivative. - Use the second derivative test at each critical value. - A stationary inflection point would require a critical value that is not an extremum. - Compare the coordinate differences between the two nearby extrema.

1. The zeros of \(f'\) are \(x=-2\), \(x=2\), and \(x=2.1\). 2. The second derivative is \(f''(x)=60x^2-84x-80\). It gives \(f''(-2)=328>0\), \(f''(2)=-8<0\), and \(f''(2.1)=8.2>0\). 3. Therefore, the graph has local minima at \((-2,-304)\) and \((2.1,143.9865)\), and a local maximum at \((2,144)\). There are no stationary inflection points. 4. The extrema at \(x=2\) and \(x=2.1\) are only \(0.1\) unit apart horizontally and \(0.0135\) unit apart vertically. At the displayed scale, this small change looks almost flat.

a) Local minima: \((-2,-304)\) and \((2.1,143.9865)\); local maximum: \((2,144)\); no stationary inflection points b) The two extrema near \(x=2\) are too close in both coordinates to be resolved clearly at the graph's scale.

- Compare the y-intercepts first. - Use the derivative to locate the two critical points. - Check whether the second extremum shown in a panel has the required y-coordinate. - Eliminate \(k\) from the inflection-point coordinates.

1. Every graph has y-intercept \(f_k(0)=4\), so panels (2) and (3), whose y-intercept is \(2\), cannot belong to the family. 2. The derivative is \(f_k'(x)=3x(x-2k)\), so the critical points are \(x=0\) and \(x=2k\). Panel (4) would require \(2k=1\), or \(k=\frac{1}{2}\), but then the second extremum would be \(\left(1, \frac{7}{2}\right)\), not \((1, 2)\). Thus panel (4) is excluded. Panel (1) has its second extremum at \(x=2\), giving \(k=1\), and \(f_1(2)=0\), which matches the graph. 3. Since \(f_k''(x)=6x-6k\), for \(k>0\) the point \((0, 4)\) is a local maximum and \((2k, 4-4k^3)\) is a local minimum. For \(k<0\), the classifications are reversed. 4. Inflection points satisfy \(f_k''(x)=0\), so \(x=k\). Since the third derivative is \(6\), this is an inflection point. Its y-coordinate is \(4-2k^3\). Replacing \(k\) with \(x\) gives \(y=-2x^3+4\). Because \(k\neq0\), the attained points have \(x\neq0\).

a) Only panel (1) belongs to the family, with \(k=1\). b) For \(k>0\): local maximum \((0, 4)\), local minimum \((2k, 4-4k^3)\). For \(k<0\): the classifications are reversed. c) \(y=-2x^3+4\), with attained x-values \(x\neq0\)

- Use both coordinates of the inflection point to obtain conditions on the polynomial. - A local extremum occurs where the first derivative is \(0\). - Use the second derivative to classify the extremum. - For part c, keep the leading coefficient as a parameter and determine what passing through \(Q\) would require.

1. Write \(f(x)=ax^3+bx^2+cx+d\). 2. Since \(W(0, 1)\) is an inflection point, \(f(0)=1\) and \(f''(0)=0\). Because \(f''(x)=6ax+2b\), these conditions give \(d=1\) and \(b=0\). 3. A local extremum at \(x=-1\) requires \(f'(-1)=0\). Since \(f'(x)=3ax^2+c\), \(3a+c=0\), so \(c=-3a\). 4. Using \(P(2, 0)\), \(0=f(2)=8a-6a+1=2a+1\). Thus, \(a=-\frac{1}{2}\) and \(c=\frac{3}{2}\). Therefore, \(f(x)=-\frac{1}{2}x^3+\frac{3}{2}x+1\). 5. Here \(f''(x)=-3x\), so \(f''(-1)=3>0\). The extremum is a local minimum. 6. For a cubic satisfying the conditions at \(W\) and having an extremum at \(x=-1\), the form is \(f(x)=ax^3-3ax+1\). Passing through \(Q(2, 3)\) would require \(3=2a+1\), so \(a=1\). Then \(f''(-1)=-6<0\), making the extremum a local maximum, not a local minimum. The claim is true.

a) \(f(x)=-\frac{1}{2}x^3+\frac{3}{2}x+1\) b) The extremum is a local minimum because \(f''(-1)=3>0\). c) The claim is true. Passing through \(Q(2, 3)\) would force \(a=1\), which gives \(f''(-1)=-6<0\) and therefore a local maximum at \(x=-1\).

- Relate a local extremum and a stationary inflection point of \(f\) to zeros of its first derivative. - Relate inflection values of \(f\) to extrema of its first derivative. - Write the first derivative in factored form from its known zeros and multiplicities. - Use the tangent slope to find the remaining factor, then use the point \(I(0, 2)\).

1. At the local minimum \(x=-\frac{3}{2}\), \(f'\) has a simple zero. At the stationary inflection point \(x=3\), \(f'\) has a double zero. Since \(f'\) is cubic, these are all its zeros. 2. The inflection values of \(f\) are \(x=0\) and \(x=3\), so these are the zeros of \(f''\). Each is simple because the concavity changes at both values. 3. At \(x=0\), \(f''\) changes from positive to negative, so \(f'\) has a local maximum. At \(x=3\), \(f''\) changes from negative to positive, so \(f'\) has a local minimum. 4. The zeros and multiplicities give \(f'(x)=k(x+\frac{3}{2})(x-3)^2\). The tangent slope at \(I\) gives \(f'(0)=2\), so \(2=k(\frac{3}{2})(9)\), which gives \(k=\frac{4}{27}\). Thus, \(f'\) is uniquely determined. Integrating gives \(f(x)=\frac{1}{27}x^4-\frac{2}{9}x^3+2x+C\). Since \(f(0)=2\), \(C=2\). Therefore, the information determines \(f\) uniquely.

a) Zeros of \(f'\): \(x=-\frac{3}{2}\), multiplicity \(1\); \(x=3\), multiplicity \(2\) Zeros of \(f''\): \(x=0\) and \(x=3\), each with multiplicity \(1\) b) \(f'\) has a local maximum at \(x=0\) and a local minimum at \(x=3\). c) Yes. The zero structure and \(f'(0)=2\) determine \(f'\), and \(f(0)=2\) determines the constant of integration. In fact, \(f(x)=\frac{1}{27}x^4-\frac{2}{9}x^3+2x+2\).