- Check the first-derivative condition before using the test. - What does a positive second derivative say about the graph near the critical number?

1. The condition \(f'(2)=0\) makes \(x=2\) a critical number with a horizontal tangent. 2. Since \(f''(2)=5>0\), the graph is concave up at that critical number. 3. By the second derivative test, \(f\) has a local minimum at \(x=2\).

Local minimum at \(x=2\)

- Before examining \(f''\), what condition must the first derivative satisfy? - Does concave up at one point automatically mean a local minimum?

1. The second derivative test requires a stationary critical point where \(f'(2)=0\). 2. Here \(f'(2)=3\ne0\), so the test's prerequisite fails. 3. The positive second derivative indicates increasing slopes, not a local minimum at this point.

No. The second derivative test requires \(f'(2)=0\), but \(f'(2)=3\).

- The derivative already identifies the stationary input. - What does a positive second derivative mean at that input? - Use the function value to name the point.

1. Differentiate: \(f'(x)=2(x-1)\) and \(f''(x)=2\). 2. At the stationary point, \(f''(1)=2>0\). 3. The second derivative test gives a local minimum at \((1,4)\).

\(f''(1)=2\); local minimum at \((1,4)\).

- Confirm that the first derivative is zero. - A negative second derivative means the curve bends downward near the point. - Which stationary extremum has that shape?

1. The zero first derivative means the second derivative test may be used at \(-2\). 2. The second derivative is negative there. 3. Hence the stationary point is a local maximum.

A local maximum at \(x=-2\).

- The test uses the sign of the second derivative at a stationary point. - Is zero positive or negative? - Does an inconclusive test rule out an extremum?

1. The first derivative vanishes, so \(0\) is a stationary candidate. 2. The second derivative is zero there, so neither the positive nor negative case of the test applies. 3. The test is inconclusive; another method is needed.

The second derivative test is inconclusive at \(x=0\).

- Check whether the second derivative exists at the point before applying the test. - A test that cannot be used does not mean the point is not a minimum. - Compare nearby curve heights with the height at the origin.

1. For \(x\ne0\), \(f'(x)=\frac{3}{2}\operatorname{sgn}(x)\sqrt{|x|}\). The derivative at \(0\) is \(0\). 2. The quotient \(\frac{f'(h)-f'(0)}{h}=\frac{3}{2\sqrt{|h|}}\) has no finite limit as \(h\to0\), so \(f''(0)\) does not exist. 3. The second derivative test cannot be used at \(0\). Nevertheless, \(f(x)\ge f(0)=0\), so \(0\) is a minimum.

No. \(f''(0)\) does not exist, so the second derivative test cannot be used. The point is still a minimum because \(f(x)\ge0=f(0)\).

- What extra condition does the second derivative test need? - Find the sign of the first derivative at any point. - Can a function that is always strictly increasing have an interior minimum?

1. Differentiate to get \(f'(x)=e^x>0\) for all real \(x\). 2. Because \(f\) is strictly increasing everywhere, it has no local minimum. 3. Positive \(f''\) alone does not establish a minimum; the second derivative test also requires a stationary point.

No. Since \(f'(x)=e^x>0\) for all \(x\), \(f\) is strictly increasing and has no local minimum.

- Check that the first derivative is zero. - Match the second derivative's sign to local graph shape. - Determine which turning point has that shape.

1. The first derivative is zero, so the test can be applied at this critical number. 2. The second derivative is positive, so the graph is concave up locally. 3. The point is a local minimum.

A local minimum at \(x=3\).

- 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: \(f'(x)=3x^2-3\). Its zeros are \(x=-1\) and \(x=1\), both in the interval. 2. The second derivative is \(f''(x)=6x\). Since \(f''(-1)=-6<0\), the second derivative test gives a local maximum at \(-1\). Since \(f''(1)=6>0\), it gives a local minimum at \(1\). 3. Therefore, the function has \(2\) local extrema in the interval.

\(2\)

- 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 numbers 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 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 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.

- 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: \(f'(x)=3x^2-6x-9=3(x+1)(x-3)\), so the critical numbers are \(x=-1\) and \(x=3\). 2. Since \(f''(x)=6x-6\), \(f''(-1)=-12<0\) gives a local maximum at \(-1\), and \(f''(3)=12>0\) gives a local minimum at \(3\). 3. Evaluating the function gives \(f(-1)=15\) and \(f(3)=-17\). The local maximum is \((-1,15)\), and the local minimum is \((3,-17)\).

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

- 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 local extremum at \(x=2\) requires \(f_k'(2)=0\), giving \(16-4k=0\). Hence \(k=4\). 3. For this value, \(f_4''(x)=6x^2-8\), so \(f_4''(2)=16>0\). By the second derivative test, \(x=2\) is a local minimum.

\(k=4\)

- Apply the chain rule to the exponential term. - Where is the tangent slope zero? - What does a positive second derivative mean at that critical number?

1. By the chain rule, \(f'(x)=e^{0.5x}-1\). 2. Solve \(f'(x)=0\): \(e^{0.5x}=1\), so \(x=0\). 3. The second derivative is \(f''(x)=0.5e^{0.5x}\), giving \(f''(0)=0.5>0\). Therefore \(x=0\) is a local minimum. 4. Since \(f(0)=2\), the point is \((0,2)\).

Local minimum: \((0,2)\).

- Which x-coordinates have horizontal tangents? - What do the signs of the second derivative say about those critical points? - Find the y-coordinates from the original function.

1. Differentiate: \(f'(x)=3x^2-9x+6=3(x-1)(x-2)\) and \(f''(x)=6x-9\). 2. The critical numbers are \(x=1\) and \(x=2\). 3. Since \(f''(1)=-3<0\), there is a local maximum at \(x=1\). Since \(f''(2)=3>0\), there is a local minimum at \(x=2\). 4. The corresponding function values are \(f(1)=4.5\) and \(f(2)=4\).

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

- Use the chain rule to differentiate the exponential. - Solve the first derivative equation with logarithms. - Evaluate the second derivative at the critical number, then find its height.

1. The first derivative is \(f'(x)=2e^{2x}-6\). 2. Setting \(f'(x)=0\) gives \(e^{2x}=3\), so the critical number is \(x=\frac{1}{2}\ln(3)\). 3. The second derivative is \(f''(x)=4e^{2x}\). At the critical number, \(f''\left(\frac{1}{2}\ln(3)\right)=12>0\). Thus the point is a local minimum. 4. The function value is \(f\left(\frac{1}{2}\ln(3)\right)=3-3\ln(3)\).

Local minimum: \(\left(\frac{1}{2}\ln(3),3-3\ln(3)\right)\), with x-coordinate approximately \(0.549\).

- Use the product rule and chain rule together. - Which part of the derivative can equal zero? - Evaluate the second derivative at the critical number.

1. By the product and chain rules, \(g'(x)=(-x-1)e^{-x}\). 2. Since \(e^{-x}>0\), the critical number is \(x=-1\). 3. The second derivative is \(g''(x)=xe^{-x}\), so \(g''(-1)=-e<0\). The point is a local maximum. 4. Its height is \(g(-1)=e\).

Local maximum: \((-1,e)\).

- 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, and there are no other critical numbers, so \(g\) 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 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 \((2,3)\).

- “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 \(0\).

- 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 \(f''(-1)=-3<0\), the local maximum is at \(\left(-1,\frac{31}{6}\right)\). 4. Since \(f''(2)=3>0\), the local minimum is at \(\left(2,\frac{2}{3}\right)\).

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

- 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

- First locate the zeros of the first derivative. - Then read the second derivative at each candidate. - How does the sign of the second derivative determine the type of extremum?

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 \(x=3\).

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

1. Differentiate: \(f'(x)=-3x^2+3\) and \(f''(x)=-6x\). 2. Solve \(f'(x)=0\), giving critical numbers \(x=-1\) and \(x=1\). 3. Since \(f''(-1)=6>0\), the local minimum is at \((-1,-1)\). 4. Since \(f''(1)=-6<0\), the local maximum is at \((1,3)\).

Local minimum: \((-1,-1)\); local maximum: \((1,3)\).

- Find the zeros of \(f'\). - At each critical number, read the sign of \(f''\). - How does the sign of \(f''\) classify the point?

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 number. - How do those signs determine the type of extremum?

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

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

- The critical numbers are already given. - Read the sign of \(f''\) at each critical number. - 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 \(x=4\)

- A horizontal tangent alone does not determine the type of critical point. - Compute the second derivative at the critical number. - If \(f''(c)>0\), the second derivative test gives a local minimum.

1. Differentiate: \(f'(x)=-4x^3+30x^2+\frac15x\), so \(f'(0)=0\). Thus \(x=0\) is a critical number. 2. Differentiate again: \(f''(x)=-12x^2+60x+\frac15\), so \(f''(0)=\frac15>0\). 3. Because \(f'(0)=0\) and \(f''(0)>0\), the second derivative test is conclusive and gives a local minimum at \(x=0\). 4. Since \(f(0)=0\), the point is \((0,0)\). Therefore Ema's stationary-inflection claim is false.

The claim is false. Since \(f'(0)=0\) and \(f''(0)=\frac15>0\), the second derivative test gives a local minimum at \((0,0)\).

- What conclusion, if any, does the second derivative test permit when the second derivative is zero? - Use another valid way to classify the critical number. - Compare the sign of the first derivative on the two sides of \(0\).

1. The second derivative test is inconclusive when \(f''(0)=0\); it does not imply that no extremum exists. 2. For \(x<0\), \(f'(x)=4x^3<0\), while for \(x>0\), \(f'(x)=4x^3>0\). 3. Thus, \(f\) changes from decreasing to increasing at \(x=0\), so \(x=0\) is a local minimum. In fact, since \(x^4\ge0=f(0)\) for all real \(x\), it is also an absolute minimum.

The error is treating an inconclusive second derivative test as a negative conclusion. \(x=0\) is a local and absolute minimum.

- Evaluate both derivatives at the proposed extremum. - When one test is inconclusive, examine the derivative signs on either side. - What does a change from rising to falling mean for the graph?

1. \(f'(x)=-4x^3\) and \(f''(x)=-12x^2\), so \(f'(0)=0\) and \(f''(0)=0\). 2. The second derivative test is inconclusive at \(0\). 3. For \(x<0\), \(f'(x)>0\); for \(x>0\), \(f'(x)<0\). The sign changes from positive to negative. 4. By the first derivative test, \(x=0\) is a local maximum.

The second derivative test is inconclusive because \(f''(0)=0\). The first derivative changes from positive to negative at \(0\), so \(f\) has a local maximum there.

- Which derivative must be zero before the second derivative test is used? - Check that condition at the student's chosen point. - Solve for stationary points before checking the second derivative.

1. Differentiate to get \(f'(x)=2x+2\). In particular, \(f'(0)=2\ne0\). 2. The second derivative test for a local extremum requires \(f'(x)=0\) at the tested point, so the student's argument fails. 3. Solve \(2x+2=0\) to obtain \(x=-1\). 4. Since \(f''(-1)=2>0\), the second derivative test confirms a local minimum at \(x=-1\).

The student skipped the condition \(f'(0)=0\); in fact \(f'(0)=2\). The local minimum occurs at \(x=-1\).

- Stationary points appear at zeros of the first derivative graph. - At each of those x-values, read the second derivative from the other graph. - Match the sign of the second derivative to the appropriate kind of extremum.

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

\(x=-1\): local maximum, since \(f''(-1)=-3\). \(x=2\): local minimum, since \(f''(2)=3\).

- Does a graph with a corner have a tangent slope at that point? - Check the direction the graph moves on each side of the corner. - Which derivative sign change signals a minimum?

1. The derivative of \(|x|\) is \(-1\) for \(x<0\) and \(1\) for \(x>0\), but it does not exist at \(0\). 2. Because the first derivative does not exist at \(0\), the second derivative test cannot be used there. 3. The derivative changes from negative to positive, so the first derivative test shows a local minimum at \(0\).

No. The second derivative test is not valid because \(f'(0)\) does not exist. The derivative changes from negative to positive, so the point is a local minimum.

- Start with the zeros of the first derivative. - Differentiate the given derivative to obtain the second derivative. - Use its sign at the critical number.

1. Solve \(6(x+1)=0\) to get the critical number \(x=-1\). 2. Differentiating again gives \(f''(x)=6\), so \(f''(-1)>0\). 3. The critical number is a local minimum.

Critical number \(x=-1\); local minimum.

- What does it mean when the second derivative test is inconclusive? - What other way could you compare the function value at the critical point with values nearby? - Can a conclusion about an extremum be supported without a nonzero second derivative?

1. The first derivative is \(f'(x)=-4x^3\), so \(f'(0)=0\). 2. The second derivative is \(f''(x)=-12x^2\), giving \(f''(0)=0\). The second derivative test is inconclusive. 3. Since \(-x^4\le0=f(0)\) for all \(x\), the function has a local (and absolute) maximum at \(0\).

\(f'(0)=f''(0)=0\), so the second derivative test is inconclusive. Nevertheless, \(f(x)\le f(0)\) nearby, so \(x=0\) is a local maximum.

- Find both derivatives at zero. - If the second derivative test cannot decide, inspect the sign of the first derivative. - Does the function reverse direction at zero?

1. We have \(g'(x)=3x^2\) and \(g''(x)=6x\), so both derivatives are zero at \(0\). 2. The second derivative test is inconclusive. 3. Since \(g'(x)>0\) on both sides of \(0\), the function keeps increasing; it has no local extremum there.

The second derivative test is inconclusive at \(x=0\). The first derivative test shows neither a local maximum nor a local minimum.

- The first derivative locates the critical numbers. - The second derivative classifies each one separately. - Be careful not to use the same second derivative sign for both points.

1. Solve \(x(x-2)=0\), giving critical numbers \(0\) and \(2\). 2. At \(0\), \(f''(0)=-2<0\), so there is a local maximum. 3. At \(2\), \(f''(2)=2>0\), so there is a local minimum.

Local maximum at \(x=0\); local minimum at \(x=2\).

- How does the curve behave just before and after the indicated input? - Which derivative facts would be necessary to apply the requested test? - How can the algebraic check support, rather than merely repeat, the visual prediction?

1. The graph rises toward \(x=1\) and falls afterward, so it suggests a local maximum there. 2. Differentiate the given equation: \(f'(x)=-2(x-1)\) and \(f''(x)=-2\). 3. At \(x=1\), \(f'(1)=0\) and \(f''(1)=-2<0\). The second derivative test confirms the local maximum.

a) Local maximum at \(x=1\). b) \(f'(1)=0\) and \(f''(1)=-2<0\), so the second derivative test confirms a local maximum.

- 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.

- 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 \((-1, 2)\)

- What does the second derivative need to equal at the minimum in each case? - Can two functions have a minimum at the same input but different second derivatives there? - In part b, how could the sign of the first derivative settle the question?

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 \(x=1\).

- 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

- 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. The first derivative is \(f'(x)=\cos(x)+\frac{1}{2}\). Solving \(f'(x)=0\) on \((0,2\pi)\) gives \(x=\frac{2\pi}{3}\) and \(x=\frac{4\pi}{3}\). 2. Since \(f''(x)=-\sin(x)\), \(f''(\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}<0\) gives a local maximum, and \(f''(\frac{4\pi}{3})=\frac{\sqrt{3}}{2}>0\) gives 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: \(g'(x)=\sqrt{2}\cos(x)-1\) and \(g''(x)=-\sqrt{2}\sin(x)\). 2. Solving \(g'(x)=0\) in \((0,2\pi)\) gives \(x=\frac{\pi}{4}\) and \(x=\frac{7\pi}{4}\). 3. The second derivative values are \(g''(\frac{\pi}{4})=-1<0\) and \(g''(\frac{7\pi}{4})=1>0\). Therefore, the first value gives a local maximum and the second gives a local minimum.

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

- Treat \(x\ln(x)\) as the product of two functions. - What equation describes a horizontal tangent? - Check the sign of the second derivative at that point.

1. By the product rule, \(g'(x)=1\cdot\ln(x)+x\cdot\frac{1}{x}=\ln(x)+1\). 2. Solve \(g'(x)=0\): \(\ln(x)+1=0\), so \(x=\frac{1}{e}\). 3. Since \(g''(x)=\frac{1}{x}\), \(g''\left(\frac{1}{e}\right)=e>0\). The second derivative test gives a local minimum. 4. Its coordinates are \(\left(\frac{1}{e},-\frac{1}{e}\right)\).

The product rule gives \(g'(x)=\ln(x)+1\). The critical point is a local minimum at \(\left(\frac{1}{e},-\frac{1}{e}\right)\).

- Combine the two exponential terms before finding derivatives. - The exponential factor cannot be zero. What must the other factor equal at a critical point? - Evaluate the second derivative at each critical number, then check the requested interval.

1. Combine terms: \(f(x)=(x^2+2x-1)e^x\). 2. The first derivative is \(f'(x)=(x^2+4x+1)e^x\). Since \(e^x>0\), the critical numbers are \(x=-2-\sqrt{3}\) and \(x=-2+\sqrt{3}\). 3. The second derivative is \(f''(x)=(x^2+6x+5)e^x\). At \(x=-2-\sqrt{3}\), \(f''(x)=-2\sqrt{3}e^{-2-\sqrt{3}}<0\), giving a local maximum. 4. At \(x=-2+\sqrt{3}\), \(f''(x)=2\sqrt{3}e^{-2+\sqrt{3}}>0\), giving a local minimum. 5. The respective function values are \((2+2\sqrt{3})e^{-2-\sqrt{3}}\approx0.131\) and \((2-2\sqrt{3})e^{-2+\sqrt{3}}\approx-1.120\).

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

- Put the entire inner expression into the exponential function. - Which factor of the derivative can equal zero? - Check whether the second derivative is positive or negative there. - What is the y-coordinate of any point on \(y=4\)?

1. Composing gives \(h_k(x)=e^{-x^2+k}\). 2. The chain rule gives \(h_k'(x)=-2xe^{-x^2+k}\). Its only critical number is \(x=0\). 3. Since \(h_k''(x)=e^{-x^2+k}(4x^2-2)\), we have \(h_k''(0)=-2e^k<0\). The second derivative test gives a local maximum. 4. Because \(-x^2+k\le k\), this is also the absolute maximum, at \((0,e^k)\). 5. Setting \(e^k=4\) gives \(k=\ln(4)\).

a) \(h_k(x)=e^{-x^2+k}\) b) Maximum point: \((0,e^k)\) c) \(k=\ln(4)\)

- Where does the sine function equal zero? - Use the chain rule, then check which derivative zeros lie in the interval. - Evaluate the second derivative at each critical number.

1. Solve \(\sin(x^2)=0\). Since \(0\le x^2\le4\), the solutions are \(x=0\) and \(x=\pm\sqrt{\pi}\). 2. The derivative is \(f'(x)=2x\cos(x^2)\). Its interior zeros are \(x=0\) 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. 4. At \(x=\pm\sqrt{\frac{\pi}{2}}\), \(f''(x)=-2\pi<0\). Thus both points are local maxima, with function value \(1\).

a) \(x=0\) and \(x=\pm\sqrt{\pi}\). b) Local minimum: \((0,0)\). Local maxima: \(\left(-\sqrt{\frac{\pi}{2}},1\right)\) and \(\left(\sqrt{\frac{\pi}{2}},1\right)\).

- What must the tangent slope be at an interior local maximum? - Which relationships involving the parameter follow from the proposed turning point? - How can you check that a derivative equation actually has a real solution? - What must be verified beyond the first-derivative condition?

1. The product and chain rules give \(g_k'(x)=e^{0.5x}(0.5k-0.5x-1)\). 2. At \(x=4\), the critical-point condition is \(g_k'(4)=e^2(0.5k-3)=0\), giving \(k=6\). 3. Since \(g_k''(x)=e^{0.5x}(0.25k-0.25x-1)\), \(g_6''(4)=-0.5e^2<0\). The second derivative test confirms a local maximum.

\(k=6\)

- Treat \(k\) as a constant when differentiating. - At what x-coordinate is the tangent horizontal? - Evaluate the second derivative at that specific x-coordinate. - How does the natural logarithm change as its positive input increases?

1. Differentiate: \(f_k'(x)=ke^x-1\). 2. Solve \(ke^x=1\). The critical number is \(x=-\ln(k)\). 3. Since \(f_k''(x)=ke^x\), at the critical number \(f_k''(-\ln(k))=ke^{-\ln(k)}=1>0\). The second derivative test confirms a local minimum. 4. As \(k\) increases, \(\ln(k)\) increases. Thus \(-\ln(k)\) decreases, and the minimum moves left.

1. \(f_k'(x)=ke^x-1\) 2. \(x=-\ln(k)\) 3. The x-coordinate decreases as \(k\) increases.

- Use the product and chain rules together. - Which factor of the first derivative can equal zero? - Evaluate the second derivative at the critical number.

1. Differentiate: \(f'(x)=(3.6-0.8x)e^{-0.2x}\). 2. The exponential factor is positive, so \(f'(x)=0\) when \(x=4.5\). 3. The second derivative is \(f''(x)=(0.16x-1.52)e^{-0.2x}\). At \(x=4.5\), \(f''(4.5)=-0.8e^{-0.9}<0\), so the second derivative test gives a local maximum. 4. Its height is \(f(4.5)=20e^{-0.9}\).

The local maximum is \(\left(4.5,20e^{-0.9}\right)\).

- Factor the first derivative using a common exponential term. - Use the relation between exponentials and logarithms to solve for the critical number. - Evaluate the second derivative at that number.

1. The derivative is \(f'(x)=e^{2x}-4e^x=e^x(e^x-4)\). 2. Since \(e^x>0\), the only critical number satisfies \(e^x=4\), giving \(x=\ln(4)\). 3. The second derivative is \(f''(x)=2e^{2x}-4e^x\), so \(f''(\ln(4))=2\cdot16-4\cdot4=16>0\). The second derivative test gives a local minimum. 4. Its height is \(f(\ln(4))=\frac{1}{2}\cdot16-4\cdot4=-8\).

Local minimum: \((\ln(4),-8)\).

- Use the product rule on the polynomial and the exponential. - Does the exponential factor ever equal zero? - Evaluate the second derivative at both critical numbers. - Substitute both x-coordinates into the original function.

1. Differentiate using the product rule: \(f'(x)=(x^2-3x+2)e^x=(x-1)(x-2)e^x\). 2. The exponential factor is positive, so the critical numbers are \(x=1\) and \(x=2\). 3. The second derivative is \(f''(x)=(x^2-x-1)e^x\). Since \(f''(1)=-e<0\), the point at \(x=1\) is a local maximum. Since \(f''(2)=e^2>0\), the point at \(x=2\) is a local minimum. 4. Substituting gives \(f(1)=3e\) and \(f(2)=e^2\).

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

- Use the product rule to differentiate \(f_a\). - The exponential factor is never zero. - Check the second derivative at the resulting critical number. - How does the y-coordinate of a point on \(y=2\) constrain \(a\)?

1. Differentiate: \(f_a'(x)=(0.5a-1-0.5x)e^{0.5x}\). Since the exponential factor is positive, the only critical number is \(x=a-2\). 2. The second derivative is \(f_a''(x)=(0.25a-1-0.25x)e^{0.5x}\). At the critical number, \(f_a''(a-2)=-0.5e^{0.5a-1}<0\), so the point is a local maximum. 3. Its y-coordinate is \(f_a(a-2)=2e^{0.5a-1}\), giving \((a-2,2e^{0.5a-1})\). 4. Setting \(2e^{0.5a-1}=2\) gives \(a=2\).

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

- Keep \(x>0\) in mind when solving the critical-point equation. - Evaluate the second derivative at the critical number, not just somewhere in the domain. - A point on the x-axis has height zero.

1. The derivatives are \(g_k'(x)=x-\frac{k}{x}\) and \(g_k''(x)=1+\frac{k}{x^2}\). 2. Solve \(g_k'(x)=0\): \(x^2=k\). Since \(x>0\), the only critical number is \(x=\sqrt{k}\). 3. At this point, \(g_k''(\sqrt{k})=1+\frac{k}{k}=2>0\). The second derivative test gives a local minimum. It is the only one because it is the only critical point. 4. The height is \(g_k(\sqrt{k})=\frac{k}{2}(1-\ln(k))\), so the minimum is \(\left(\sqrt{k},\frac{k}{2}(1-\ln(k))\right)\). 5. For this height to be zero, \(\frac{k}{2}(1-\ln(k))=0\). Since \(k>0\), \(\ln(k)=1\), giving \(k=e\).

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

- How are position, velocity, and acceleration related by derivatives? - What does instantaneous velocity mean in terms of \(s(t)\)? - At an interior minimum of velocity, what must the acceleration equal? - Track the units through each derivative.

1. Differentiate position to find velocity: \(v(t) = s'(t) = 0.6t^2 - 3t + 4\). 2. Evaluate \(v(5) = 0.6(5^2) - 3(5) + 4 = 4\,\text{m/s}\). 3. Differentiate velocity to find acceleration: \(a(t) = v'(t) = 1.2t - 3\). 4. A minimum of velocity occurs where \(a(t) = 0\). Solve \(1.2t - 3 = 0\) to get \(t = 2.5\,\text{s}\). Because \(v''(t) = 1.2 > 0\), this critical time gives a minimum. 5. At that time, \(a(2.5) = 0\,\text{m/s}^2\).

a) \(4\,\text{m/s}\) b) The velocity is minimized at \(t = 2.5\,\text{s}\), when the acceleration is \(0\,\text{m/s}^2\).

- How many distinct zeros can the quadratic first derivative have? - What does its discriminant tell you about that number? - For the specified parameter, test each critical number with the second derivative. - Evaluate the original function to find each height.

1. The first derivative is \(f_k'(x)=x^2-kx+4\). It has two distinct real zeros when its discriminant is positive: \(k^2-16>0\). 2. Thus, two local extrema occur for \(k<-4\) or \(k>4\). The two simple zeros of the quadratic derivative give opposite changes of sign. 3. For \(k=5\), \(f_5'(x)=(x-1)(x-4)\). The critical numbers are \(1\) and \(4\), and \(f_5''(x)=2x-5\). 4. Since \(f_5''(1)=-3<0\), the point at \(x=1\) is a local maximum, with height \(f_5(1)=\frac{11}{6}\). 5. Since \(f_5''(4)=3>0\), the point at \(x=4\) is a local minimum, with height \(f_5(4)=-\frac{8}{3}\).

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

- Use the quotient and chain rules. - Which factor of the numerator can equal zero? - Test the second derivative at each critical number. - Keep the excluded point \(x=0\) out of the domain.

1. The quotient and chain rules give \(f'(x)=\frac{(x^2-1)e^{x^2/2}}{x^2}\). 2. Since the exponential and denominator are positive on the domain, the critical numbers are \(x=-1\) and \(x=1\). 3. Differentiate again: \(f''(x)=\left(x-\frac{1}{x}+\frac{2}{x^3}\right)e^{x^2/2}\). 4. Since \(f''(1)=2\sqrt{e}>0\), there is a local minimum at \(x=1\). Since \(f''(-1)=-2\sqrt{e}<0\), there is a local maximum at \(x=-1\). 5. Their function values are \(f(1)=\sqrt{e}\) and \(f(-1)=-\sqrt{e}\).

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

- The first derivative is a difference of squares. - What happens to the critical numbers when \(k=0\)? - Check the second derivative first; if it is zero, inspect the first derivative on either side. - An extremum on the x-axis must have function value zero.

1. The derivatives are \(f_k'(x)=x^2-k^2\) and \(f_k''(x)=2x\). 2. For \(k=0\), the only critical number is \(x=0\), and \(f_0''(0)=0\), so the second derivative test is inconclusive. Since \(f_0'(x)=x^2>0\) on both sides, there is no local extremum. 3. For \(k\ne0\), the critical numbers are \(x=k\) and \(x=-k\), with function values \(\frac{2}{3}-\frac{2}{3}k^3\) and \(\frac{2}{3}+\frac{2}{3}k^3\). 4. Since \(f_k''(k)=2k\) and \(f_k''(-k)=-2k\), for \(k>0\) the point at \(-k\) is a local maximum and the point at \(k\) a local minimum. For \(k<0\), these classifications reverse. 5. Requiring either height to be zero gives \(k^3=1\) or \(k^3=-1\), so \(k=1\) or \(k=-1\). In both cases the local minimum is \((1,0)\).

1) For \(k=0\), no local extrema. For \(k>0\), maximum \(\left(-k,\frac{2}{3}+\frac{2}{3}k^3\right)\) and minimum \(\left(k,\frac{2}{3}-\frac{2}{3}k^3\right)\). For \(k<0\), the classifications reverse. 2) \(k=1\) or \(k=-1\); in both cases the minimum is \((1,0)\).

- What is the domain of the logarithm? - Which parameter values make the critical number lie in that domain? - Evaluate the second derivative at that number.

1. Differentiate: \(f_a'(x)=\frac{a}{x}-1\) and \(f_a''(x)=-\frac{a}{x^2}\). 2. The critical-point equation gives \(x=a\), which lies in the domain only if \(a>0\). 3. For \(a>0\), \(f_a''(a)=-\frac{1}{a}<0\), so the point is a local maximum at \(\left(a,a(\ln(a)-1)\right)\). 4. For \(a\le0\), there is no critical number in \(x>0\), and hence no local extremum.

A local maximum exists exactly when \(a>0\), at \(\left(a,a(\ln(a)-1)\right)\). For \(a\le0\), there is no local extremum.

- Think about which values an exponential expression can take. - Take logarithms only once the parameter conditions make that possible. - Evaluate the second derivative at the critical number, not just generally.

1. The derivatives are \(h_k'(x)=ke^{kx}-1\) and \(h_k''(x)=k^2e^{kx}\). 2. The critical-number equation is \(e^{kx}=\frac{1}{k}\). Because the exponential is positive, this has a real solution exactly when \(k>0\). 3. For \(k>0\), logarithms give \(x=-\frac{\ln(k)}{k}\). At this point, \(h_k''\left(-\frac{\ln(k)}{k}\right)=k^2\cdot\frac{1}{k}=k>0\), so it is a local minimum. 4. Its height is \(\frac{1}{k}+\frac{\ln(k)}{k}=\frac{1+\ln(k)}{k}\).

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

- Remember to exclude the zero of the denominator from the domain. - Use the quotient rule to locate horizontal tangents. - Evaluate the second derivative separately at both critical numbers.

1. By the quotient rule, \(f'(x)=\frac{(x-2)(x+4)}{(x+1)^2}\), for \(x\ne-1\). 2. The critical numbers are \(x=-4\) and \(x=2\). Their function values are \(f(-4)=-6\) and \(f(2)=6\). 3. The second derivative is \(f''(x)=\frac{18}{(x+1)^3}\). Thus \(f''(-4)=-\frac{2}{3}<0\) gives a local maximum at \((-4,-6)\), while \(f''(2)=\frac{2}{3}>0\) gives a local minimum at \((2,6)\).

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

- 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 extremum.

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 \((1,1)\)

- What equation must every interior critical number satisfy? - After finding one obvious factor, inspect the powers in the remaining expression. - Can a change of variable turn the remaining equation into a familiar type? - Apply the second derivative test at each solution.

1. Differentiate: \(g'(x)=x^5-6x^3+8x=x(x^4-6x^2+8)\). 2. One critical number is \(x=0\). For the quartic factor, let \(u=x^2\). Then \(u^2-6u+8=0\), so \(u=2\) or \(u=4\). 3. The critical numbers are \(x=0\), \(x=\pm\sqrt{2}\), and \(x=\pm2\). 4. Since \(g''(x)=5x^4-18x^2+8\), we have \(g''(0)=8>0\) and \(g''(\pm2)=16>0\), giving local minima. Also \(g''(\pm\sqrt{2})=-8<0\), giving local maxima.

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

- 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 extremum.

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. Since \(f''(x)=\frac{2}{3}+\frac{36}{x^3}\), we have \(f''(3)=2>0\); thus \(f\) has a local minimum at \(x=3\). 5. Since \(f(3)=9\), the only local extremum is \((3,9)\).

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

- 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. Setting \(f'(x)=0\) gives \(\cos(x)=\frac{1}{2}\). The interior critical numbers are \(x=\frac{\pi}{3}\) and \(x=\frac{5\pi}{3}\). 3. The second derivative is \(f''(x)=2\sin(x)\). 4. Since \(f''\left(\frac{\pi}{3}\right)=\sqrt{3}>0\), there is a local minimum at \(x=\frac{\pi}{3}\). Since \(f''\left(\frac{5\pi}{3}\right)=-\sqrt{3}<0\), there is a local maximum at \(x=\frac{5\pi}{3}\).

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

- 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. Since \(f''(x)=3x^2-6x-4\) and \(f''(-2)=20>0\), \((-2,-20)\) is a local minimum. 4. Since \(f''(2)=-4<0\), \((2,12)\) is a local maximum. 5. Since \(f''(3)=5>0\), \(\left(3,\frac{45}{4}\right)\) is a local minimum.

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

- 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.

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

1. Since \(f'(x)=2x\) and \(f''(0)=2>0\), \(f'(0)=0\) gives a local minimum at \(x=0\), where \(f(0)=-5\). 2. The squared function is \(g(x)=(x^2-5)^2\), with \(g'(x)=4x(x^2-5)\). At \(x=0\), \(g'(0)=0\) and \(g''(0)=12(0)^2-20=-20<0\), so the second derivative test gives a local maximum. 3. Near \(x=0\), the values of \(f\) are negative. Squaring reverses their order: negative values closer to \(-5\) give 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 nearby values of \(f\) are negative, squaring reverses their order and changes the minimum into a maximum.

- Solve the first-derivative equation on the stated domain. - Use the sign of the second derivative to classify the extremum. - 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 \(f_a'(x)=\frac{1}{x}-a\), which vanishes only at \(x=\frac{1}{a}\) in the domain. 2. The second derivative is \(f_a''(x)=-\frac{1}{x^2}\), so \(f_a''\left(\frac{1}{a}\right)=-a^2<0\). The second derivative test gives a unique local maximum. 3. For this maximum to lie on the x-axis, require \(\ln\left(\frac{1}{a}\right)-1=0\), giving \(-\ln(a)-1=0\). 4. Thus \(a=e^{-1}=\frac{1}{e}\).

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

- 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 \(x=\pm2\). Since \(h_k(-x)=h_k(x)\), every graph is symmetric about the y-axis. 2. The quotient rule gives \(h_k'(x)=\frac{-2x(k+4)}{(x^2-4)^2}\). 3. If \(k=-4\), then \(h_{-4}(x)=1\) throughout its domain. It is constant on each domain interval, with no isolated local extrema. 4. If \(k\ne-4\), the only critical number is \(x=0\), where \(h_k(0)=-\frac{k}{4}\). 5. Since \(h_k''(0)=-\frac{k+4}{8}\), \(k>-4\) gives a local maximum at \(\left(0,-\frac{k}{4}\right)\), and \(k<-4\) gives a local minimum there.

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.

- First solve \(f'(x)=0\) to find the critical numbers. - The second derivative test is conclusive when \(f''(c)\neq0\). - A negative second derivative gives a local maximum; a positive second derivative gives a local minimum.

1. Differentiate: \(f'(x)=0.75x^2-3=\frac34(x^2-4)\). Solving \(f'(x)=0\) gives the critical numbers \(x=-2\) and \(x=2\). 2. Differentiate again: \(f''(x)=1.5x=\frac32x\). 3. At \(x=-2\), \(f''(-2)=-3<0\). By the second derivative test, \(f\) has a local maximum there. \(f(-2)=6\), so the local maximum is \((-2,6)\). 4. At \(x=2\), \(f''(2)=3>0\). By the second derivative test, \(f\) has a local minimum there. \(f(2)=-2\), so the local minimum is \((2,-2)\).

Local maximum \((-2,6)\); local minimum \((2,-2)\), classified by the second derivative test.

- Candidates occur at zeros of the first derivative. - Read the second derivative at each candidate. - How can a positive or negative second derivative classify a critical point?

1. The graph of \(f'\) has zeros at \(x=-3\), \(x=1\), and \(x=3\). These are the critical-number 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

- 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 zero. - Use the second derivative to classify each extremum. - Check the locations against the graph. - For the tangent, find the point and derivative at \(x=1\).

1. The derivatives are \(p'(x)=-4x(x^2-2)\) and \(p''(x)=-12x^2+8\). 2. The critical numbers 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, agreeing with the graph. 3. At \(x=1\), \(p(1)=3\) and \(p'(1)=4\). The tangent is \(y-3=4(x-1)\), or \(y=4x-1\).

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

- Which feature of the displayed curve could identify its parameter? - Which derivative facts would establish that exactly one local extremum exists? - How would the classification change if the second derivative had the opposite sign?

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, its only zero is \(x=a-1\). 3. The second derivative is \(g_a''(x)=(x-a+2)e^x\), so \(g_a''(a-1)=e^{a-1}>0\), confirming a local minimum. 4. The y-coordinate is \(g_a(a-1)=-e^{a-1}\).

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

- Find the zeros of \(f'\). - Use the sign of \(f''\) at each critical number. - 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 \([-1, 2]\)

- Which term controls a quartic's end behavior? - A local maximum must satisfy a first-derivative condition. - Use the second derivative to confirm the classification.

1. The leading term is \(-\frac{1}{4}x^4\), so \(f_k(x)\to-\infty\) as \(x\to\pm\infty\). Figure 1 has this end behavior; 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) \(k=3\)

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

1. The function is zero at \(x=1\); the graph labeled \(f\) has this x-intercept. As \(x\to-\infty\), \((x-1)e^x\to0\) from below, also matching \(f\). 2. By 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 \(p''(x)=(x+1)e^x\), so \(p''(0)=1>0\). Therefore, the local minimum is \((0,-1)\).

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

- Differentiate the outer square before applying the chain rule. - The first derivative is a product. How do you differentiate it? - A square cannot be negative. - Substitute \(g(x_0)=0\) into both derivative formulas.

1. The chain rule gives \(f'(x)=2g(x)g'(x)\). 2. The product rule gives \(f''(x)=2(g'(x))^2+2g(x)g''(x)\). 3. Under the conditions in b), \((g'(x))^2\ge0\) and \(g(x)g''(x)>0\). Thus \(f''(x)>0\) everywhere, so \(f\) is concave up. 4. When \(g(x_0)=0\), we have \(f'(x_0)=0\) and \(f''(x_0)=2(g'(x_0))^2>0\), since \(g'(x_0)\ne0\). 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 \(f\) is concave up. c) \(f'(x_0)=0\) and \(f''(x_0)=2(g'(x_0))^2>0\), so \(f\) has a local minimum at \(x_0\).

- 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. Both numerator factors 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 \(x_0\).

- Find all zeros of the first derivative. - Compare second derivative signs at the critical numbers. - How would a stationary inflection differ from a local 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 \((2,144)\). Since every critical number is classified as an extremum by a nonzero second derivative, 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.