Free math worksheets

- Find every zero of \(f'\). - Check whether the derivative changes sign at each zero.

1. A local extremum of \(f\) occurs where \(f'\) has a zero and changes sign. 2. The graph of \(f'\) crosses the x-axis at \(x=-3\), \(x=1\), and \(x=4\). 3. The derivative changes sign at all three zeros, so \(f\) has three local extrema.

Three local extrema

- Translate each sign of \(f'\) into increasing or decreasing behavior. - Compare the direction of motion before and after the critical number.

1. To the left of \(2\), \(f'(x)>0\), so \(f\) is increasing. 2. To the right of \(2\), \(f'(x)<0\), so \(f\) is decreasing. 3. The function changes from increasing to decreasing at \(x=2\), so the first derivative test gives a local maximum there.

\(f\) has a local maximum at \(x=2\).

- Find and factor the first derivative. - Determine the derivative sign on every interval separated by the critical numbers. - Classify each critical number from the direction of the sign change. - Evaluate the original function after the classifications are known.

1. Differentiate: \(f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)\). 2. The critical numbers are \(x = 1\) and \(x = 3\). 3. The derivative is positive for \(x<1\), negative for \(1<x<3\), and positive for \(x>3\). Thus \(f'\) changes from positive to negative at \(x=1\), giving a local maximum, and from negative to positive at \(x=3\), giving a local minimum by the first derivative test. 4. Evaluate the function: \(f(1) = 6\) and \(f(3) = 2\). 5. The local maximum is \((1, 6)\), and the local minimum is \((3, 2)\).

Local maximum: \((1, 6)\) Local minimum: \((3, 2)\)

- Local extrema occur where the derivative changes sign. - Read the derivative's sign on each interval separated by its zeros. - Negative-to-positive means a local minimum. - Positive-to-negative means a local maximum.

1. The derivative is \(0\) at \(x=-3\), \(x=1\), and \(x=4\). 2. At \(x=-3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 3. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 4. At \(x=4\), \(f'\) changes from negative to positive, so \(f\) has a local minimum.

Local minima occur at \(x=-3\) and \(x=4\). A local maximum occurs at \(x=1\).

- Use the sign of the derivative to determine monotonicity. - Check the direction of each sign change at a zero of the derivative.

1. On \((1,4)\), the derivative is negative, so \(f\) is strictly decreasing there. Statement a) is true. 2. At \(x=-2\), the derivative changes from negative to positive. Therefore, \(f\) has a local minimum, not a local maximum. Statement b) is false. 3. On \((-1,1)\), the derivative is positive, so \(f\) is strictly increasing there. Statement c) is true.

a) True b) False c) True

- A zero derivative gives a critical number, not automatically an extremum. - Think of a function that flattens briefly but continues increasing. - Check whether the derivative changes sign at the critical number.

1. The statement is false. 2. The condition \(f'(x_0)=0\) identifies a critical number, but it does not guarantee a local extremum. 3. For example, let \(f(x)=x^3\). Then \(f'(x)=3x^2\), so \(f'(0)=0\). 4. However, \(f'(x)>0\) on both sides of \(x=0\). The function continues increasing through \(x=0\), so it has no local extremum there.

False. The function \(f(x)=x^3\) has \(f'(0)=0\), but \(f\) has no local extremum at \(x=0\) because \(f'\) does not change sign.

- A local extremum can occur where the derivative is \(0\). - A change from positive to negative indicates a local maximum. - A change from negative to positive indicates a local minimum.

1. Local extrema of \(h\) can occur where \(h^{\prime}=0\) and changes sign. The graph has zeros at \(x=-2\) and \(x=2\). 2. At \(x=-2\), \(h^{\prime}\) changes from positive to negative, so \(h\) has a local maximum. 3. At \(x=2\), \(h^{\prime}\) changes from negative to positive, so \(h\) has a local minimum.

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

- Use the derivative graph's position above or below the x-axis. - Classify each zero by the direction of the derivative's sign change.

1. The derivative is zero at \(x=-2\), \(x=1\), and \(x=3\). 2. It is positive on \((-3.5,-2)\) and \((1,3)\), so \(f\) is strictly increasing on those intervals. 3. It is negative on \((-2,1)\) and \((3,4.5)\), so \(f\) is strictly decreasing on those intervals. 4. At \(x=-2\) and \(x=3\), the derivative changes from positive to negative, so \(f\) has local maxima. At \(x=1\), it changes from negative to positive, so \(f\) has a local minimum.

a) Strictly increasing on \((-3.5,-2)\) and \((1,3)\); strictly decreasing on \((-2,1)\) and \((3,4.5)\). b) Local maxima at \(x=-2\) and \(x=3\); local minimum at \(x=1\).

- Find the zeros of \(f'\). - Check the sign of \(f'\) immediately before and after each zero. - Positive to negative gives a local maximum; negative to positive gives a local minimum.

1. The zeros of \(f'\) are \(x=-3\), \(x=1\), and \(x=4\). 2. At \(x=-3\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 3. At \(x=1\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 4. At \(x=4\), \(f'\) changes from positive to negative, so \(f\) has a local maximum.

a) \(x=-3\), \(x=1\), and \(x=4\) b) Local maxima at \(x=-3\) and \(x=4\); local minimum at \(x=1\)

- Read the derivative's sign from its position relative to the x-axis. - Use the direction of each sign change to classify the extremum.

1. The derivative is negative on \((-3,-1)\), positive on \((-1,3)\), and negative on \((3,6)\). 2. Therefore, \(f\) is strictly decreasing on \((-3,-1)\) and \((3,6)\), and strictly increasing on \((-1,3)\). 3. At \(x=-1\), the derivative changes from negative to positive, so \(f\) has a local minimum. At \(x=3\), it changes from positive to negative, so \(f\) has a local maximum.

Strictly increasing on \((-1,3)\); strictly decreasing on \((-3,-1)\) and \((3,6)\). Local minimum at \(x=-1\); local maximum at \(x=3\).

- Locate the x-intercepts of the derivative graph. - Check the sign of \(f'\) immediately on both sides of each intercept. - Classify each point from the direction of the sign change.

1. The zeros of \(f'\) are \(x=0\) and \(x=4\). 2. At \(x=0\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 3. At \(x=4\), \(f'\) changes from positive to negative, so \(f\) has a local maximum.

Local minimum at \(x=0\); local maximum at \(x=4\).

- A stationary inflection point has a horizontal tangent. - Compare the function's increasing or decreasing behavior on the two sides of the point. - A sign change in the first derivative would indicate a local extremum.

1. A stationary inflection point has a horizontal tangent, so \(f'(-1)=0\). 2. It is not a local extremum. The function continues increasing on both sides or continues decreasing on both sides, so \(f'\) keeps the same sign. 3. Therefore, statement c) is correct.

c) \(f'(-1)=0\), and \(f'\) does not change sign there.

- Find every zero of the derivative graph. - Check whether the graph crosses or merely touches the x-axis at each zero. - Classify each crossing from the derivative signs on its two sides.

1. The derivative zeros are \(x=-2,-0.5,1,2.5\). 2. The graph crosses the x-axis at every zero, so \(f'\) changes sign at all four. 3. From left to right, the derivative signs are positive, negative, positive, negative, positive. Therefore, the sign changes are positive-to-negative, negative-to-positive, positive-to-negative, and negative-to-positive. 4. Thus, \(f\) has four local extrema: local maximum, local minimum, local maximum, local minimum from left to right.

\(4\) local extrema: maximum, minimum, maximum, minimum from left to right.

- Locate the zeros of \(H'\). - Inspect the sign of the derivative on both sides of each zero. - Positive-to-negative gives a local maximum; negative-to-positive gives a local minimum.

1. Local extrema of \(H\) occur where \(H'\) changes sign. 2. At \(x=1\), \(H'\) changes from positive to negative, so \(H\) has a local maximum. 3. At \(x=5\), \(H'\) changes from negative to positive, so \(H\) has a local minimum.

Local maximum at \(x=1\); local minimum at \(x=5\).

- A local extremum requires the derivative to change sign. - Compare a derivative graph that only touches the x-axis with one that crosses it. - Check each zero of each derivative.

1. A local extremum can occur where the derivative is zero and changes sign. 2. The graph of \(f'\) touches the x-axis at \(x=1\) but remains nonnegative, so \(f'\) does not change sign. Therefore, \(f\) has no local extremum there. 3. The graph of \(g'\) changes from negative to positive at \(x=1\), so \(g\) has a local minimum. 4. The graph of \(h'\) changes sign at \(x=-2\) and \(x=2\), so \(h\) has local extrema at both values. 5. Therefore, only \(f\) has no local extrema in the displayed domain.

Only \(f\) has no local extrema in the displayed domain.

- A local minimum occurs where the derivative changes from negative to positive. - Distinguish between crossing the x-axis and merely touching it. - Check the direction of each sign change.

1. The zeros of \(f'\) are \(x=-2\), \(x=1\), and \(x=3\). 2. At \(x=-2\), \(f'\) changes from negative to positive, so \(f\) changes from decreasing to increasing. Therefore, \(f\) has a local minimum at \(x=-2\). 3. At \(x=1\), \(f'\) touches zero without changing sign, so there is no local extremum. 4. At \(x=3\), \(f'\) changes from positive to negative, so \(f\) has a local maximum, not a local minimum.

The only local minimum is at \(x=-2\).

- A zero of the derivative by itself is not enough to classify a critical number. - Compare the derivative's sign immediately to the left and right of each marked x-value. - Translate each sign pattern into the way the original function moves.

1. At \(x=-1\), \(g'\) changes from positive to negative, so \(g\) has a local maximum. 2. At \(x=1\), \(g'=0\) but \(g'\) is negative on both sides, so \(g\) has neither a local maximum nor a local minimum. 3. At \(x=4\), \(g'\) changes from negative to positive, so \(g\) has a local minimum.

\(x=-1\): local maximum. \(x=1\): neither. \(x=4\): local minimum.

- A critical number can occur where the derivative is zero or where it does not exist. - The first derivative test depends on the signs on the two sides, not on the derivative's value at the critical number itself. - Translate each sign pattern into increasing or decreasing behavior of the original function.

1. At \(x=-2\), \(g'\) is positive on both sides, so \(g\) has neither a local maximum nor a local minimum there. 2. At \(x=0\), \(g'\) changes from positive to negative. Even though \(g'(0)\) does not exist, continuity of \(g\) and the sign change imply a local maximum at \(x=0\). 3. At \(x=3\), \(g'\) changes from negative to positive, so \(g\) has a local minimum.

\(x=-2\): neither. \(x=0\): local maximum. \(x=3\): local minimum.

- Factor the first derivative to find its zeros and their multiplicities. - Check the sign of the first derivative on both sides of each critical number. - A horizontal tangent without a first-derivative sign change is not a local extremum. - After classifying extrema, check concavity to decide whether a remaining horizontal tangent is an inflection point.

1. Factor the first derivative: \(f'(x)=x(x-3)^2\). Thus \(f'(x)=0\) at \(x=0\) and \(x=3\). 2. Since \((x-3)^2>0\) away from \(x=3\), the sign of \(f'\) is negative for \(x<0\) and positive for \(x>0\), except that \(f'(3)=0\). Therefore, \(f'\) changes from negative to positive at \(x=0\), so the first derivative test gives a local minimum there. 3. At \(x=3\), \(f'\) is positive on both sides, so there is no local extremum. 4. Differentiate \(f'\): \(f''(x)=3x^2-12x+9=3(x-1)(x-3)\). The second derivative changes sign at \(x=3\), so the horizontal tangent there is a stationary inflection point.

Horizontal tangents occur at \(x=0\) and \(x=3\). There is a local minimum at \(x=0\) and a stationary inflection point at \(x=3\).

- Local extrema require critical points. - Compare the degree of a polynomial with the degree of its derivative. - A polynomial of degree \(m\) has at most \(m\) real zeros. - Check the multiplicity of each derivative zero to determine whether its sign changes.

1. A local extremum of a differentiable polynomial can occur only at a zero of \(f'\). 2. Since \(f\) is nonconstant and has degree \(n\ge 1\), \(f'\) has degree \(n-1\) and therefore has at most \(n-1\) real zeros. Thus \(f\) has at most \(n-1\) local extrema. 3. In part b), \(f'(x)=(x-3)^2(x+1)\) has degree \(3\), so \(f\) has degree \(4\). 4. The derivative has a double zero at \(x=3\) and a simple zero at \(x=-1\). Its sign does not change at the double zero, but it does change at the simple zero. Therefore, \(f\) has exactly one local extremum, at \(x=-1\).

a) The derivative has degree \(n-1\), so it has at most \(n-1\) real zeros; therefore, \(f\) has at most \(n-1\) local extrema. b) \(f\) has degree \(4\) and exactly one local extremum, at \(x=-1\).

- Compare each new derivative with \(f'\). - Consider how each transformation affects derivative signs, zeros, and intervals where the derivative is zero. - Visualize whether each transformation moves the graph horizontally or only vertically. - For a horizontal shift, track how an extremum input \(x_E\) changes.

1. For part a), \(g'(x)=f'(x)\). The slope behavior, including any intervals where the derivative is zero, is unchanged. Also, subtracting the same constant from all nearby function values preserves every local comparison, so the local extrema occur at exactly the same \(x\)-values. 2. For part b), \(h'(x)=5f'(x)\). Because \(5>0\), the derivative has the same signs, zeros, and zero intervals as \(f'\). Multiplying all nearby function values by a positive constant also preserves local comparisons, so the local extrema occur at exactly the same \(x\)-values and retain their types. 3. For part c), \(k'(x)=f'(x+4)\). If \(x_E\) is an extremum input of \(f\), the corresponding extremum input of \(k\) is \(x_E-4\). Thus the extremum set is shifted left \(4\) units. It is generally different, although it can coincide with the original set for special functions whose extremum inputs are invariant under a shift of \(4\).

a) Yes. The local extrema occur at the same \(x\)-values because \(g'(x)=f'(x)\) and a vertical shift preserves local comparisons. b) Yes. The local extrema occur at the same \(x\)-values because \(h'(x)=5f'(x)\) with \(5>0\), so slope behavior and local comparisons are preserved. c) Not necessarily. The extremum inputs shift from \(x_E\) to \(x_E-4\), so they are generally different but may coincide for special functions.

- Find the critical numbers of the cubic from its first derivative. - Use the sign changes of the first derivative to classify the turning points. - Compare a horizontal line \(y=k\) with the local maximum and local minimum values.

1. Define \(g(x)=x^3-12x+10\). Its derivative is \(g'(x)=3x^2-12=3(x-2)(x+2)\), so the critical numbers are \(x=-2\) and \(x=2\). 2. The derivative is positive on \(( -\infty,-2)\), negative on \((-2,2)\), and positive on \((2,\infty)\). Thus \(g'\) changes from positive to negative at \(x=-2\), giving a local maximum, and from negative to positive at \(x=2\), giving a local minimum by the first derivative test. The values are \(g(-2)=26\) and \(g(2)=-6\). 3. Because \(g\) is a cubic with positive leading coefficient, \(g(x)\to-\infty\) as \(x\to-\infty\) and \(g(x)\to\infty\) as \(x\to\infty\). Therefore a horizontal line \(y=k\) intersects the graph once when \(k<-6\) or \(k>26\), twice when \(k=-6\) or \(k=26\), and three times when \(-6<k<26\).

One real solution: \(k<-6\) or \(k>26\) Two real solutions: \(k=-6\) or \(k=26\) Three real solutions: \(-6<k<26\)

- Use the first derivative for increasing and decreasing intervals. - An isolated zero of the derivative does not necessarily split a monotonic interval. - Use the second derivative for concavity. - Build sign charts using the zeros of each derivative.

1. The first derivative is \(f'(x)=x^3-3x^2=x^2(x-3)\). It is negative for \(x<3\), except that it equals zero at the isolated value \(x=0\), and it is positive for \(x>3\). Therefore, \(f\) is strictly decreasing on \((-\infty,3)\) and strictly increasing on \((3,\infty)\). 2. The second derivative is \(f''(x)=3x^2-6x=3x(x-2)\). It is positive on \((-\infty,0)\) and \((2,\infty)\), so the graph is concave up there. It is negative on \((0,2)\), so the graph is concave down there.

Strictly decreasing: \((-\infty,3)\) Strictly increasing: \((3,\infty)\) Concave up: \((-\infty,0)\) and \((2,\infty)\) Concave down: \((0,2)\)

- The sign of the first derivative determines monotonicity. - The sign of the second derivative determines concavity. - Use the zeros of each derivative to make sign intervals. - Check one test value in each interval.

1. The first derivative is \(g'(x)=\frac{1}{2}x^2-2x+1.5=\frac{1}{2}(x-1)(x-3)\). It is positive for \(x<1\), negative for \(1<x<3\), and positive for \(x>3\). Therefore, \(g\) is strictly increasing on \((-\infty,1)\) and \((3,\infty)\), and strictly decreasing on \((1,3)\). 2. The second derivative is \(g''(x)=x-2\). It is negative for \(x<2\), so the graph is concave down on \((-\infty,2)\). It is positive for \(x>2\), so the graph is concave up on \((2,\infty)\).

Strictly increasing: \((-\infty,1)\) and \((3,\infty)\) Strictly decreasing: \((1,3)\) Concave down: \((-\infty,2)\) Concave up: \((2,\infty)\)

- A point on the y-axis has \(x=0\). - Use the product rule to find the derivative. - The exponential factor is always positive. - Determine when the quadratic factor has no sign-changing zeros.

1. The y-intercept is \(g_k(0)=-k\). Setting \(-k=-5\) gives \(k=5\). 2. Use the product rule: \(g_k'(x)=(x^2+2x-k)e^x\). 3. Since \(e^x>0\), critical numbers come from \(x^2+2x-k=0\). 4. The quadratic has no two distinct real zeros when its discriminant is nonpositive: \(4+4k\leq0\), so \(k\leq-1\). 5. For \(k<-1\), the derivative is always positive. For \(k=-1\), the derivative is \((x+1)^2e^x\), which is nonnegative and does not change sign at \(x=-1\). Thus, there are no local extrema exactly when \(k\leq-1\).

a) \(k=5\) b) \(k\leq-1\)

- Differentiate each product separately, then simplify. - Find every factor of the first derivative that can be zero in the interval. - A horizontal tangent does not always indicate a local extremum. - Use the sign of the first derivative on both sides of each critical number.

1. Differentiate using the product and sum rules: \(f'(x)=2x\sin(x)+(x^2-2)\cos(x)+2\cos(x)-2x\sin(x)=x^2\cos(x)\). 2. Solve \(f'(x)=0\). This gives \(x=0\) or \(\cos(x)=0\), so the critical numbers in the interval are \(x=0\) and \(x=\pm\frac{\pi}{2}\). 3. Since \(f'(x)=x^2\cos(x)\), the derivative changes from positive to negative at \(x=\frac{\pi}{2}\), so there is a local maximum. It changes from negative to positive at \(x=-\frac{\pi}{2}\), so there is a local minimum. At \(x=0\), the derivative is positive on both sides, so there is no extremum. 4. Evaluate the function: \(f\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}-2\), and \(f\left(-\frac{\pi}{2}\right)=2-\frac{\pi^2}{4}\). The critical point at \(x=0\) is a stationary inflection point.

Local maximum: \(\left(\frac{\pi}{2},\frac{\pi^2}{4}-2\right)\). Local minimum: \(\left(-\frac{\pi}{2},2-\frac{\pi^2}{4}\right)\). The point \((0,0)\) is a stationary inflection point, not a local extremum.

- The first derivative gives the sign of the slope. - Find where the derivative is zero. - Make a sign chart for the derivative. - Substitute each critical number into the original function.

1. The derivative is \(f'(x)=-x^2+2x+3=-(x+1)(x-3)\). The critical numbers are \(x=-1\) and \(x=3\). 2. The derivative is negative for \(x<-1\), positive for \(-1<x<3\), and negative for \(x>3\). Therefore, \(f\) is strictly decreasing on \((-\infty,-1)\) and \((3,\infty)\), and strictly increasing on \((-1,3)\). 3. The derivative changes from negative to positive at \(x=-1\), so \((-1,-\frac{8}{3})\) is a local minimum. It changes from positive to negative at \(x=3\), so \((3,8)\) is a local maximum.

Strictly increasing: \((-1,3)\) Strictly decreasing: \((-\infty,-1)\) and \((3,\infty)\) Local minimum: \((-1,-\frac{8}{3})\) Local maximum: \((3,8)\)

- A zero of the derivative is necessary but not sufficient for an extremum. - Factor the derivative and examine its sign on both sides of its zero. - A squared factor cannot make the derivative negative.

1. Differentiate: \(g'(x)=x^2+2x+1=(x+1)^2\). 2. The only critical number is \(x=-1\). 3. Since \(g'(x)\geq0\) for all \(x\) and is positive on both sides of \(-1\), the derivative does not change sign there. Therefore, \(g\) has no local extremum.

The function has no local extrema.

- Find all points where the first derivative is zero. - Use the sign of the first derivative on both sides of each critical number to classify extrema. - A horizontal tangent does not always indicate a local maximum or minimum. - After the first-derivative classification, check concavity if you need to identify a stationary inflection point.

1. Differentiate: \(f'(x)=x^3-3x^2=x^2(x-3)\). 2. The critical numbers are \(x=0\) and \(x=3\). 3. The first derivative is negative on both sides of \(x=0\), so it does not change sign there. Thus, \(x=0\) is not a local extremum. 4. At \(x=3\), \(f'\) changes from negative to positive, so the first derivative test gives a local minimum. Since \(f(3)=-\frac{19}{4}\), the local minimum is \(\left(3,-\frac{19}{4}\right)\). 5. The second derivative is \(f''(x)=3x(x-2)\), which changes sign at \(x=0\). Therefore, \((0,2)\) is a stationary inflection point.

The only local extremum is the local minimum \(\left(3,-\frac{19}{4}\right)\). The horizontal tangent at \((0,2)\) is a stationary inflection point, not an extremum.

- Use a double-angle identity to factor the derivative. - Set each factor of the derivative equal to zero. - Use a sign chart for the first derivative. - Keep only critical numbers in the stated interval.

1. Differentiate: \(g'(x)=2\cos(x)-2\sin(2x)\). 2. Use \(\sin(2x)=2\sin(x)\cos(x)\): \(g'(x)=2\cos(x)(1-2\sin(x))\). 3. The interior critical numbers are \(x=\frac{\pi}{6}\), \(x=\frac{\pi}{2}\), and \(x=\frac{5\pi}{6}\). 4. The derivative is positive on \(\left(0,\frac{\pi}{6}\right)\), negative on \(\left(\frac{\pi}{6},\frac{\pi}{2}\right)\), positive on \(\left(\frac{\pi}{2},\frac{5\pi}{6}\right)\), and negative on \(\left(\frac{5\pi}{6},\pi\right)\). 5. Therefore, the derivative changes from positive to negative at \(x=\frac{\pi}{6}\) and \(x=\frac{5\pi}{6}\), producing local maxima. It changes from negative to positive at \(x=\frac{\pi}{2}\), producing a local minimum. 6. The function values are \(g\left(\frac{\pi}{6}\right)=\frac{3}{2}\), \(g\left(\frac{\pi}{2}\right)=1\), and \(g\left(\frac{5\pi}{6}\right)=\frac{3}{2}\).

Local maxima: \(\left(\frac{\pi}{6},\frac{3}{2}\right)\) and \(\left(\frac{5\pi}{6},\frac{3}{2}\right)\). Local minimum: \(\left(\frac{\pi}{2},1\right)\). Strictly increasing on \(\left(0,\frac{\pi}{6}\right)\) and \(\left(\frac{\pi}{2},\frac{5\pi}{6}\right)\). Strictly decreasing on \(\left(\frac{\pi}{6},\frac{\pi}{2}\right)\) and \(\left(\frac{5\pi}{6},\pi\right)\).

- Apply the quotient rule. - Determine which factors can equal zero. - Check that the derivative changes sign at the solutions.

1. The quotient rule gives \(f_a'(x)=a\frac{1-x^2}{(x^2+x+1)^2}\). 2. The denominator is positive for every real \(x\), and \(a\ne0\). Therefore, \(f_a'(x)=0\) exactly when \(1-x^2=0\). 3. The solutions are \(x=-1\) and \(x=1\). The numerator changes sign at both values, so both are local-extremum x-coordinates for every \(a\).

\(x=-1\) and \(x=1\)

- First determine how the parameter affects the domain. - For \(t>0\), use the sign of the first derivative on the intervals separated by its zeros. - Classify each critical number from the direction of the derivative sign change before using the extremum value in part d).

1. The denominator \(x^2+t\) has no real zero exactly when \(t>0\). Therefore, \(D_t=\mathbb{R}\) for \(t>0\). 2. Since \(f_t(-x)=\frac{-tx}{x^2+t}=-f_t(x)\), every graph is symmetric about the origin. 3. Differentiate: \(f_t'(x)=\frac{t(x^2+t)-tx(2x)}{(x^2+t)^2}=\frac{t(t-x^2)}{(x^2+t)^2}\). 4. For \(t>0\), the derivative is zero at \(x=\pm\sqrt{t}\). It is negative when \(|x|>\sqrt{t}\) and positive when \(|x|<\sqrt{t}\). Thus \(f_t'\) changes from negative to positive at \(x=-\sqrt{t}\), giving a local minimum, and from positive to negative at \(x=\sqrt{t}\), giving a local maximum by the first derivative test. There are no other critical points. 5. The local maximum has y-coordinate \(f_t(\sqrt{t})=\frac{t\sqrt{t}}{2t}=\frac{\sqrt{t}}{2}\). 6. Set \(\frac{\sqrt{t}}{2}=1\). Then \(\sqrt{t}=2\), so \(t=4\).

a) \(t>0\) b) \(f_t(-x)=-f_t(x)\), so every graph is symmetric about the origin. c) For \(t>0\), the extrema occur at \(x=\pm\sqrt{t}\). d) \(t=4\)

- Compare \(g_{k,m}(-x)\) with \(g_{k,m}(x)\). - Find and classify the critical point using the first derivative. - Substitute the given point into the function. - Solve \(x^2+m=0\) when \(m<0\).

1. Since \(g_{k,m}(-x)=\frac{k}{(-x)^2+m}=g_{k,m}(x)\), every graph is symmetric about the y-axis. 2. Differentiate: \(g_{k,m}'(x)=\frac{-2kx}{(x^2+m)^2}\). For \(m>0\), the denominator is always positive, and the only critical point is \(x=0\). 3. The critical point is \(\left(0, \frac{k}{m}\right)\). If \(k>0\), the derivative changes from positive to negative, so it is a local maximum. If \(k<0\), the derivative changes from negative to positive, so it is a local minimum. 4. With \(k=1\), the point condition gives \(\frac{1}{1+m}=0.25=\frac{1}{4}\). Therefore, \(1+m=4\), so \(m=3\). 5. For \(m<0\), vertical asymptotes occur where \(x^2+m=0\). Thus, \(x=\pm\sqrt{-m}\).

a) \(g_{k,m}(-x)=g_{k,m}(x)\), so every graph is symmetric about the y-axis. b) The extremum is \(\left(0, \frac{k}{m}\right)\): a local maximum if \(k>0\), and a local minimum if \(k<0\). c) \(m=3\) d) \(x=-\sqrt{-m}\) and \(x=\sqrt{-m}\)

- The denominator is never zero, so zeros come from the numerator. - Compare polynomial growth with exponential growth. - Rewrite the quotient as a product with \(e^{-x}\) before differentiating. - Use a sign chart for the first derivative.

1. Since the denominator is positive, the zeros come from \(x^2-3x+1=0\). The quadratic formula gives \(x=\frac{3\pm\sqrt{5}}{2}\). 2. As \(x\to\infty\), the exponential denominator grows faster than the quadratic numerator, so \(g(x)\to0\). The horizontal asymptote is \(y=0\). 3. As \(x\to-\infty\), the numerator approaches \(\infty\) and \(e^x\to0^+\), so \(g(x)\to\infty\). 4. Rewrite \(g(x)=(x^2-3x+1)e^{-x}\). Then \(g'(x)=(-x^2+5x-4)e^{-x}\). Since the exponential factor is positive, critical numbers satisfy \(-x^2+5x-4=0\), so \(x=1\) and \(x=4\). 5. The first derivative changes from negative to positive at \(x=1\), giving a local minimum, and from positive to negative at \(x=4\), giving a local maximum. The coordinates are \(\left(1,-\frac{1}{e}\right)\) and \(\left(4,\frac{5}{e^4}\right)\).

a) \(x=\frac{3\pm\sqrt{5}}{2}\). b) \(\lim_{x\to\infty}g(x)=0\); horizontal asymptote \(y=0\). c) \(\lim_{x\to-\infty}g(x)=\infty\). d) Local minimum: \(\left(1,-\frac{1}{e}\right)\). Local maximum: \(\left(4,\frac{5}{e^4}\right)\).

- Exclude zeros of the denominator from the domain. - Compare \(f(-x)\) with \(f(x)\). - Use the quotient rule and simplify the derivative completely. - The squared denominator of the derivative is positive. - A change from negative to positive derivative indicates a local minimum.

1. The denominator is zero at \(x=\pm1\), so the domain is \(\mathbb{R}\setminus\{-1,1\}\). The numerator is zero at \(x=\pm\sqrt{2}\), so these are the zeros. 2. Since \(f(-x)=f(x)\), the function is even. The vertical asymptotes are \(x=-1\) and \(x=1\). The numerator and denominator have the same degree and leading coefficient, so the horizontal asymptote is \(y=1\). 3. The quotient rule gives \(f'(x)=\frac{2x(x^2-1)-2x(x^2-2)}{(x^2-1)^2}=\frac{2x}{(x^2-1)^2}\). 4. The denominator of \(f'\) is positive throughout the domain, so the derivative is negative when \(x<0\) and positive when \(x>0\). Accounting for the vertical asymptotes, \(f\) is strictly decreasing on \((-\infty,-1)\) and \((-1,0)\), and strictly increasing on \((0,1)\) and \((1,\infty)\). 5. At \(x=0\), the derivative changes from negative to positive, so \((0,f(0))=(0,2)\) is a local minimum.

a) Domain: \(\mathbb{R}\setminus\{-1,1\}\). Zeros: \(x=\pm\sqrt{2}\). b) \(f\) is even. Vertical asymptotes: \(x=-1\) and \(x=1\). Horizontal asymptote: \(y=1\). c) Strictly decreasing on \((-\infty,-1)\) and \((-1,0)\); strictly increasing on \((0,1)\) and \((1,\infty)\). d) Local minimum: \((0,2)\).

- Treat domain and asymptotic behavior before differentiating. - Since the denominator of \(h'\) is positive on the domain, its numerator determines the derivative sign. - Classify the critical number from the sign change of the first derivative.

1. The denominator is zero at \(x=\pm3\), so the domain is \(\mathbb{R}\setminus\{-3,3\}\). Since \(h(0)=-\frac{1}{3}\), the y-intercept is \(\left(0,-\frac{1}{3}\right)\). The numerator \(x^2+3\) is always positive, so there are no real zeros. 2. As \(x\to\pm\infty\), the ratio of leading coefficients is \(1\), so \(h(x)\to1\) and the horizontal asymptote is \(y=1\). 3. Near the excluded values, \(\lim_{x\to-3^-}h(x)=\infty\), \(\lim_{x\to-3^+}h(x)=-\infty\), \(\lim_{x\to3^-}h(x)=-\infty\), and \(\lim_{x\to3^+}h(x)=\infty\). Thus \(x=-3\) and \(x=3\) are vertical asymptotes. 4. The quotient rule gives \(h'(x)=\frac{2x(x^2-9)-2x(x^2+3)}{(x^2-9)^2}=\frac{-24x}{(x^2-9)^2}\). 5. The derivative is positive immediately to the left of \(0\) and negative immediately to the right. Thus \(h'\) changes from positive to negative at \(x=0\), so the first derivative test gives a local maximum there. Its coordinates are \(\left(0,-\frac{1}{3}\right)\).

a) Domain: \(\mathbb{R}\setminus\{-3,3\}\). Y-intercept: \(\left(0,-\frac{1}{3}\right)\). No real zeros. b) Horizontal asymptote: \(y=1\). Vertical asymptotes: \(x=-3\) and \(x=3\), with \(\lim_{x\to-3^-}h(x)=\infty\), \(\lim_{x\to-3^+}h(x)=-\infty\), \(\lim_{x\to3^-}h(x)=-\infty\), and \(\lim_{x\to3^+}h(x)=\infty\). c) \(h'(x)=\frac{-24x}{(x^2-9)^2}\). Local maximum: \(\left(0,-\frac{1}{3}\right)\).

- Set \(x=0\) for the y-intercept and set the numerator equal to zero for x-intercepts. - Compare polynomial and exponential growth. - Rewrite the quotient as a product with \(e^{-x}\). - Use the sign of the first derivative to classify critical points.

1. The y-intercept is \((0,-3)\). Since the denominator is positive, x-intercepts satisfy \(x^2-3=0\), so they are \((-\sqrt{3},0)\) and \((\sqrt{3},0)\). 2. As \(x\to\infty\), exponential growth in the denominator dominates the quadratic numerator, so \(g(x)\to0\). As \(x\to-\infty\), \(x^2-3\to\infty\) and \(e^x\to0^+\), so \(g(x)\to\infty\). 3. Rewrite \(g(x)=(x^2-3)e^{-x}\). Then \(g'(x)=(-x^2+2x+3)e^{-x}\). Since the exponential factor is positive, critical numbers satisfy \(-x^2+2x+3=0\), or \((x-3)(x+1)=0\). Thus \(x=-1\) and \(x=3\). 4. The first derivative changes from negative to positive at \(x=-1\), so \((-1,-2e)\) is a local minimum. It changes from positive to negative at \(x=3\), so \(\left(3,\frac{6}{e^3}\right)\) is a local maximum.

a) X-intercepts: \((-\sqrt{3},0)\) and \((\sqrt{3},0)\). Y-intercept: \((0,-3)\). b) \(\lim_{x\to\infty}g(x)=0\), and \(\lim_{x\to-\infty}g(x)=\infty\). c) Local minimum: \((-1,-2e)\). Local maximum: \(\left(3,\frac{6}{e^3}\right)\).

- The argument of a logarithm must be positive. - A logarithm equals zero when its argument equals \(1\). - Use the direction of the first-derivative sign change to classify the critical point. - Relate the logarithm's endpoint behavior to vertical asymptotes.

1. The logarithm requires \(-x^2+6x-8>0\). Factoring gives \(-(x-2)(x-4)>0\), so the domain is \((2,4)\). 2. A zero satisfies \(6x-x^2-8=1\), so \((x-3)^2=0\). Thus the only zero is \(x=3\). 3. The derivative is \(f'(x)=\frac{-2x+6}{-x^2+6x-8}\). It is positive for \(2<x<3\) and negative for \(3<x<4\). Thus \(f'\) changes from positive to negative at \(x=3\), so the first derivative test gives a local maximum at \((3,0)\). 4. As \(x\to2^+\) or \(x\to4^-\), the logarithm argument approaches \(0^+\), so \(f(x)\to-\infty\). Therefore, the vertical asymptotes are \(x=2\) and \(x=4\).

a) Domain: \((2,4)\). Zero: \(x=3\). b) Local maximum: \((3,0)\). c) \(\lim_{x\to2^+}f(x)=-\infty\) and \(\lim_{x\to4^-}f(x)=-\infty\). Vertical asymptotes: \(x=2\) and \(x=4\).

- Analyze the sign of the first derivative. - A global minimum gives the smallest y-value on the graph. - Compare that minimum value with \(0\).

1. Differentiate: \(f'(x)=\frac{1}{x}-\frac{2}{x^2}=\frac{x-2}{x^2}\). 2. Since the denominator is positive, the derivative is negative on \((0,2)\) and positive on \((2,\infty)\). Therefore, \(x=2\) gives a global minimum. 3. The minimum value is \(f(2)=\ln(2)+1>0\). 4. Since the smallest function value is positive, every point on the graph lies above the x-axis, so the graph has no x-intercepts.

Global minimum: \((2,1+\ln(2))\). Since \(1+\ln(2)>0\), the graph has no x-intercepts.

- What equation must hold at an interior local extremum? - Factor the derivative before solving. - How many distinct real roots does the quadratic factor have? - What happens to the sign of a polynomial at a simple zero?

1. Differentiate: \(f'(x) = 8x^3 - 12x^2 - 18x = 2x(4x^2 - 6x - 9)\). 2. One critical number is \(x = 0\). The quadratic factor has discriminant \((-6)^2 - 4(4)(-9) = 180 > 0\), so it has two distinct real zeros: \(x = \frac{3 - 3\sqrt{5}}{4}\) and \(x = \frac{3 + 3\sqrt{5}}{4}\). 3. Thus, \(f'\) has three distinct simple real zeros. A polynomial changes sign at each simple zero, so \(f'\) changes sign at all three critical numbers. 4. Therefore, the first derivative test shows that \(f\) has exactly three local extrema.

The three critical numbers are \(x = \frac{3 - 3\sqrt{5}}{4}\), \(x = 0\), and \(x = \frac{3 + 3\sqrt{5}}{4}\). Because \(f'\) changes sign at each one, the graph has exactly three local extrema.

- Set the first derivative equal to zero. - Use the discriminant to show whether the quadratic factor can be zero. - Examine the sign of the derivative on either side of the critical number. - Evaluate the original function to obtain the coordinates.

1. Differentiate: \(f'(x) = 4x^3 + 4x^2 + 4x = 4x(x^2 + x + 1)\). 2. The quadratic factor \(x^2 + x + 1\) has discriminant \(1 - 4 = -3\), so it is always positive and has no real zeros. 3. Therefore, the only critical number is \(x = 0\). Because the quadratic factor is positive, the sign of \(f'(x)\) is the sign of \(x\). Thus, \(f'\) changes from negative to positive at \(x = 0\). 4. By the first derivative test, \(f\) has a local minimum at \(x = 0\). Since \(f(0) = 7\), the point is \((0, 7)\).

The graph has exactly one local extremum, a local minimum at \((0, 7)\).

- A horizontal tangent occurs where the derivative is \(0\). - Factor the derivative. - Use the first derivative test to check for an extremum. - A double zero of the derivative often produces no sign change.

1. Differentiate: \(g'(x)=x^2+4x+4=(x+2)^2\). 2. The derivative is \(0\) only at \(x=-2\), so the graph has a horizontal tangent there. 3. Since \(g'(x)\geq0\) and is positive on both sides of \(-2\), the derivative does not change sign. Therefore, \(g\) has no local extremum at \(x=-2\).

The graph has a horizontal tangent at \(x=-2\), but no local extremum because \(g'(x)=(x+2)^2\) does not change sign.

- Find where the first derivative is zero. - Factor the cubic derivative. - Use sign changes of the derivative to classify the critical points. - Use the leading term to determine the end behavior.

1. Differentiate: \(f'(x) = 2x^3 - 8x = 2x(x - 2)(x + 2)\). The critical numbers are \(x = -2\), \(x = 0\), and \(x = 2\). 2. Evaluate the function: \(f(-2) = -2\), \(f(0) = 6\), and \(f(2) = -2\). 3. The sign of \(f'\) changes from negative to positive at \(x = -2\), from positive to negative at \(x = 0\), and from negative to positive at \(x = 2\). Therefore, \((-2, -2)\) and \((2, -2)\) are local minima, and \((0, 6)\) is a local maximum. 4. Since \(f(x) \to \infty\) as \(x \to \pm\infty\), the two local minima are also absolute minima. There is no absolute maximum.

1. \(x = -2\), \(x = 0\), and \(x = 2\) 2. \(f(-2) = -2\), \(f(0) = 6\), and \(f(2) = -2\) 3. Local maximum: \((0, 6)\); local and absolute minima: \((-2, -2)\) and \((2, -2)\); there is no absolute maximum.

- Determine which powers can appear in an even polynomial. - Differentiate and use the symmetry of the resulting odd function. - Focus on the sign of the first derivative immediately to the left and right of \(0\). - Apply the first derivative test to that sign change.

1. Y-axis symmetry means \(f\) is even, so its polynomial expression contains only even powers of \(x\). 2. Differentiating produces only odd powers of \(x\). Every term of \(f'\) therefore contains a positive power of \(x\), so \(f'(0)=0\). 3. The derivative of an even function is odd, so \(f'(-x)=-f'(x)\). 4. Because \(f\) is nonconstant, the first nonzero term of \(f'\) near \(0\) has odd degree. Thus \(f'\) has opposite signs on the two sides of \(0\): it changes either from negative to positive or from positive to negative. By the first derivative test, \(f\) therefore has a local extremum at \(0\), not a stationary inflection point.

a) An even polynomial contains only even powers, so its derivative contains only odd powers and \(f'(0)=0\). b) The derivative is odd and changes sign at \(0\). Therefore, \(f\) has a local extremum there, not a stationary inflection point.

- Evaluate the first derivative at \(x=0\). - Factor the derivative into expressions whose signs are easy to identify. - Even powers are never negative. - A local extremum requires a sign change in the derivative.

1. Differentiate: \(f'(x)=5x^4+6x^2=x^2(5x^2+6)\). 2. Since \(f'(0)=0\), the necessary critical-point condition is satisfied. 3. For every \(x\neq0\), both factors are positive, so \(f'(x)>0\) on both sides of \(0\). 4. The derivative does not change sign, so there is no local extremum at \(x=0\).

a) \(f'(0)=0\) b) No local extremum occurs because \(f'\) is positive on both sides of \(0\).

- Factor the first derivative and find its zeros. - Use a sign chart to determine where the function increases or decreases. - Evaluate the original function at each critical number.

1. Differentiate and factor: \(f'(x)=x^2-2x-3=(x+1)(x-3)\). The critical numbers are \(x=-1\) and \(x=3\). 2. The derivative is positive on \((-\infty,-1)\), negative on \((-1,3)\), and positive on \((3,\infty)\). 3. Thus, \(f\) is strictly increasing on \((-\infty,-1)\) and \((3,\infty)\), and strictly decreasing on \((-1,3)\). 4. At \(x=-1\), the derivative changes from positive to negative, so there is a local maximum. At \(x=3\), it changes from negative to positive, so there is a local minimum. 5. Evaluate the function: \(f(-1)=\frac{20}{3}\) and \(f(3)=-4\). The local maximum is \(\left(-1,\frac{20}{3}\right)\), and the local minimum is \((3,-4)\).

Strictly increasing on \((-\infty,-1)\) and \((3,\infty)\); strictly decreasing on \((-1,3)\). Local maximum: \(\left(-1,\frac{20}{3}\right)\). Local minimum: \((3,-4)\).

- Factor the first derivative completely. - Use a sign chart to determine the monotonic intervals. - A positive-to-negative sign change gives a local maximum; a negative-to-positive change gives a local minimum.

1. Differentiate and factor: \(f'(x)=-4x^3+16x=-4x(x-2)(x+2)\). The critical numbers are \(x=-2\), \(x=0\), and \(x=2\). 2. The derivative is positive on \((-\infty,-2)\), negative on \((-2,0)\), positive on \((0,2)\), and negative on \((2,\infty)\). 3. Therefore, \(f\) is strictly increasing on \((-\infty,-2)\) and \((0,2)\), and strictly decreasing on \((-2,0)\) and \((2,\infty)\). 4. The derivative changes from positive to negative at \(x=-2\) and \(x=2\), giving local maxima. It changes from negative to positive at \(x=0\), giving a local minimum. 5. Since \(f(-2)=16\), \(f(0)=0\), and \(f(2)=16\), the local maxima are \((-2,16)\) and \((2,16)\), and the local minimum is \((0,0)\).

Strictly increasing on \((-\infty,-2)\) and \((0,2)\); strictly decreasing on \((-2,0)\) and \((2,\infty)\). Local maxima: \((-2,16)\) and \((2,16)\). Local minimum: \((0,0)\).

- Distinguish necessary and sufficient conditions. - Test the first two claims with \(x^3\) and \(x^4\). - Interpret a derivative sign change as a change in monotonicity.

1. Statement 1 is false. For \(f(x)=x^3\), \(f'(0)=0\), but no local extremum occurs. 2. Statement 2 is false. For \(f(x)=x^4\), there is a local minimum at \(0\), but \(f''(0)=0\). 3. Statement 3 is true. A change from positive to negative derivative gives a local maximum; a change from negative to positive gives a local minimum.

1. False 2. False 3. True

- Find all critical numbers of the quartic. - Make a complete sign chart for the first derivative. - Evaluate the function at every critical number. - Combine the extrema and end behavior to count x-axis crossings.

1. Let \(f(x)=\frac{1}{4}x^4-\frac{1}{3}x^3-x^2+2\). Its derivative is \(f'(x)=x(x+1)(x-2)\), so the critical numbers are \(x=-1,0,2\). 2. The derivative is negative on \((-\infty,-1)\), positive on \((-1,0)\), negative on \((0,2)\), and positive on \((2,\infty)\). Thus, the first derivative test gives local minima at \(x=-1\) and \(x=2\), and a local maximum at \(x=0\). 3. The corresponding values are \(f(-1)=\frac{19}{12}>0\), \(f(0)=2>0\), and \(f(2)=-\frac{2}{3}<0\). 4. Since \(f(x)\to\infty\) as \(x\to\pm\infty\), the graph remains positive through the two left monotonic pieces, crosses once while decreasing from \(2\) to \(-\frac{2}{3}\), and crosses once while increasing from \(-\frac{2}{3}\) to \(\infty\). Therefore, there are exactly two real zeros.

The equation has exactly \(2\) real solutions.

- Factor each derivative and find its zeros. - A simple zero causes a sign change. - Use the derivative's sign before and after each zero. - Positive-to-negative change means a local maximum.

1. Since \(f'(x)=(x-1)(x-3)\), both \(f'\) and \(g'\) have simple zeros at \(x=1\) and \(x=3\). Each derivative changes sign at both values, so \(f\) and \(g\) have local extrema there. 2. The derivative \(h'\) has simple zeros at \(x=1\), \(x=2\), and \(x=3\), so it changes sign at each. Thus \(h\) also has local extrema at \(x=1\) and \(x=3\). 3. At \(x=2\), \(h'\) changes from positive to negative, so \(h\) has a local maximum there.

1. \(f\) and \(g\) have local extrema at \(x=1\) and \(x=3\). 2. \(h\) also has local extrema at \(x=1\) and \(x=3\). 3. The additional extremum is a local maximum at \(x=2\).

- Horizontal tangents occur at zeros of the derivative. - Multiplicity affects whether a factor changes sign. - Apply the first derivative test at each zero. - Check concavity to classify a non-extremum horizontal tangent.

1. The derivative is \(0\) at \(x=4\) and \(x=-1\). 2. At \(x=4\), \(k'\) changes from negative to positive, so \(k\) has a local minimum. 3. At \(x=-1\), the squared factor does not change sign, so \(k'\) is negative on both sides and there is no extremum. Since \(k''(x)=(3x-7)(x+1)\) changes sign at \(-1\), the point is a stationary inflection point.

1. Horizontal tangents occur at \(x=-1\) and \(x=4\). 2. There is a local minimum at \(x=4\). 3. At \(x=-1\), there is a stationary inflection point, not an extremum.

- What does a horizontal tangent tell you about the first derivative? - What signs must a strictly decreasing function have immediately before and after a zero? - Which local extremum occurs when the derivative changes from positive to negative? - How does the monotonic behavior of \(f'\) determine concavity?

1. A horizontal tangent gives \(f'(3)=0\). 2. Because \(f'\) is strictly decreasing, \(f'(x)>0\) for \(x<3\) near \(3\) and \(f'(x)<0\) for \(x>3\) near \(3\). 3. The derivative changes from positive to negative, so by the first derivative test, \(f\) has a local maximum at \(x=3\). 4. A decreasing first derivative means the tangent slopes decrease as \(x\) increases, so the graph is concave down on the interval.

The point at \(x=3\) is a local maximum, and the graph is concave down on the interval.

- Determine how an odd power changes sign. - Use the first derivative test. - Compare even and odd exponents. - For even \(k\), examine the second derivative to classify the point.

1. The derivative is \(0\) at \(x=-3\). Because the exponent \(5\) is odd, \(f'(x)<0\) for \(x<-3\) and \(f'(x)>0\) for \(x>-3\). 2. The derivative changes from negative to positive, so \(f\) has a local minimum at \(x=-3\). 3. If \(k\) is odd, the derivative changes from negative to positive, so \(g\) has a local minimum. If \(k\) is even, the derivative is nonnegative on both sides and no extremum occurs. In that case \(g''(x)=k(x+3)^{k-1}\) changes sign, so the point is a stationary inflection point.

a) The zero is \(x=-3\); the derivative is negative to the left and positive to the right. b) There is a local minimum at \(x=-3\). c) Odd \(k\): local minimum. Even \(k\): stationary inflection point.

- Local extrema require zeros of the derivative. - Factor out the greatest power of \(x\). - Complete the square in the remaining quadratic. - Check whether the derivative changes sign at its zero.

1. Differentiate: \(f'(x)=x^4-2x^3+\frac{3}{2}x^2=x^2\left(x^2-2x+\frac{3}{2}\right)\). 2. Rewrite the quadratic factor as \((x-1)^2+\frac{1}{2}\), which is always positive. 3. Thus \(f'(x)\geq0\) for all \(x\), and the only zero is the double zero \(x=0\). 4. The derivative is positive on both sides of \(0\), so it does not change sign. Therefore, \(f\) has no local extrema.

The function has no local extrema because \(f'(x)=x^2\left((x-1)^2+\frac{1}{2}\right)\geq0\) and does not change sign at its only zero, \(x=0\).

- How does the multiplicity of a zero of the derivative affect sign changes? - Check the sign of the derivative on both sides of each critical number. - A critical number is not automatically a local extremum.

1. Differentiate: \(f'(x)=x^5-2x^4+x^3=x^3(x-1)^2\). 2. The critical numbers are \(x=0\) and \(x=1\). 3. At \(x=0\), the factor \(x^3\) has odd multiplicity, while \((x-1)^2\) is nonnegative. Thus, \(f'\) changes from negative to positive, so \(f\) has a local minimum at \(x=0\). Since \(f(0)=1\), the point is \((0,1)\). 4. At \(x=1\), the factor \((x-1)^2\) has even multiplicity, so \(f'\) does not change sign. Therefore, \(x=1\) is not a local extremum.

The only local extremum is the local minimum \((0,1)\).

- Identify the numerator and denominator before applying the quotient rule. - The denominator of the derivative is always positive. - Use a sign change to determine which critical point is the local maximum. - Eliminate \(k\) from the maximum coordinates.

1. The quotient rule gives \(g_k'(x)=\frac{(x^2+k)-x(2x)}{(x^2+k)^2}=\frac{k-x^2}{(x^2+k)^2}\). 2. The denominator is positive, so critical points satisfy \(k-x^2=0\), giving \(x=\pm\sqrt{k}\). The derivative changes from positive to negative at \(x=\sqrt{k}\), so the local maximum is \(\left(\sqrt{k}, \frac{1}{2\sqrt{k}}\right)\). 3. At the maximum, \(x=\sqrt{k}>0\), so \(k=x^2\). Substituting into the y-coordinate gives \(y=\frac{1}{2x}\). Thus the locus is \(y=\frac{1}{2x}\) for \(x>0\).

1. \(g_k'(x)=\frac{k-x^2}{(x^2+k)^2}\) 2. \(\left(\sqrt{k}, \frac{1}{2\sqrt{k}}\right)\) 3. \(y=\frac{1}{2x}\) for \(x>0\)

- Evaluate the function at \(x=0\). - Use the product rule and the chain rule. - The exponential factor is always positive. - Compare the x- and y-coordinates of the maximum.

1. For every \(t>0\), \(f_t(0)=0\), so all graphs pass through \((0, 0)\). If \(x\neq0\) were common to graphs with distinct parameters \(t\) and \(s\), then \(e^{1-x/t}=e^{1-x/s}\), which would imply \(t=s\). Thus the origin is the only common point. 2. By the product and chain rules, \(f_t'(x)=e^{1-x/t}\left(1-\frac{x}{t}\right)\). Since the exponential factor is positive, the only critical point is \(x=t\). 3. The derivative is positive for \(x<t\) and negative for \(x>t\), so this point is the unique local maximum. Its y-coordinate is \(f_t(t)=t\), giving the local maximum \((t, t)\). 4. Because \(x=t\) and \(y=t\), the locus is \(y=x\). The condition \(t>0\) gives \(x>0\).

a) \((0, 0)\) b) \((t, t)\) c) \(y=x\) for \(x>0\)

- Use the derivative rule for \(b^x\). - Rewrite \(4\) as \(2^2\) and use \(\ln 4=2\ln 2\). - Exponential factors are always positive. - Use the sign change of the first derivative to classify the critical point.

1. Use \(\frac{d}{dx}b^x=(\ln b)b^x\): \(f'(x)=(\ln 4)4^x-8(\ln 2)2^x\). Since \(\ln 4=2\ln 2\) and \(4^x=(2^x)^2\), \(f'(x)=2(\ln 2)2^x(2^x-4)\). 2. The factors \(2\ln 2\) and \(2^x\) are positive, so \(f'(x)=0\) exactly when \(2^x-4=0\). Thus, \(2^x=4\), so \(x=2\). The derivative changes from negative to positive at \(x=2\), so this critical point is a local minimum. 3. Evaluate the function: \(f(2)=4^2-8\cdot 2^2=16-32=-16\). Therefore, the local minimum is \((2, -16)\).

a) \(f'(x)=(\ln 4)4^x-8(\ln 2)2^x=2(\ln 2)2^x(2^x-4)\) b) Local minimum: \((2, -16)\)

- Apply the quotient rule and factor the numerator. - Check that each critical point belongs to the domain. - Use a sign chart to classify the extrema. - Substitute each point into the line equation.

1. By the quotient rule, \(f_k'(x)=\frac{2x(x-k)-x^2}{(x-k)^2}=\frac{x(x-2k)}{(x-k)^2}\). 2. The critical points are \(x=0\) and \(x=2k\), both of which are in the domain because \(k\neq0\). Their coordinates are \((0, 0)\) and \((2k, 4k)\). 3. The derivative changes sign at each critical point, so both are local extrema. Their classifications depend on the sign of \(k\): for \(k>0\), \((0, 0)\) is a local maximum and \((2k, 4k)\) is a local minimum; for \(k<0\), the classifications are reversed. 4. Both points satisfy \(y=2x\): \(0=2(0)\) and \(4k=2(2k)\).

a) \((0, 0)\) and \((2k, 4k)\). For \(k>0\), they are a local maximum and local minimum, respectively; for \(k<0\), the classifications are reversed. b) \(y=2x\)

- Apply the quotient rule and factor the derivative. - Check that the critical points are in the domain. - Evaluate the function at each critical point. - Substitute each point into \(y=x^2\).

1. By the quotient rule, \(f_t'(x)=\frac{2tx(x-t)-tx^2}{(x-t)^2}=\frac{tx(x-2t)}{(x-t)^2}\). 2. The critical points are \(x=0\) and \(x=2t\). Both are in the domain because \(t\neq0\). 3. Their coordinates are \((0, 0)\) and \((2t, 4t^2)\). A sign chart for the derivative confirms that both are local extrema. 4. The first point satisfies \(0=0^2\), and the second satisfies \(4t^2=(2t)^2\). Therefore, every local extremum lies on \(y=x^2\).

The local extrema are \((0, 0)\) and \((2t, 4t^2)\), and both satisfy \(y=x^2\).

- Use the denominator to impose the vertical-asymptote condition. - Set the first derivative equal to zero at the required extremum x-value. - Factor the derivative to find any other critical point. - Use polynomial division for the slant asymptote.

1. A vertical asymptote at \(x=1\) requires \(1+b=0\), so \(b=-1\). Then \(f(x)=\frac{x^2+a}{x-1}\). 2. Differentiate: \(f'(x)=\frac{2x(x-1)-(x^2+a)}{(x-1)^2}=\frac{x^2-2x-a}{(x-1)^2}\). 3. The condition \(f'(3)=0\) gives \(9-6-a=0\), so \(a=3\). The numerator at \(x=1\) is then \(4\), confirming a vertical asymptote rather than a hole. 4. Now \(f'(x)=\frac{(x-3)(x+1)}{(x-1)^2}\). The derivative changes from positive to negative at \(x=-1\), so \((-1, -2)\) is a local maximum. It changes from negative to positive at \(x=3\), so \((3, 6)\) is a local minimum. 5. The vertical asymptote is \(x=1\). Polynomial division gives \(f(x)=x+1+\frac{4}{x-1}\), so the slant asymptote is \(y=x+1\).

a) \(a=3\), \(b=-1\); local minimum at \((3, 6)\) b) Local maximum at \((-1, -2)\); asymptotes: \(x=1\) and \(y=x+1\)

- Use the sign of \(f'\) to determine where \(f\) increases or decreases. - A zero of \(f'\) gives a local extremum only when the derivative changes sign. - The direction of the sign change distinguishes a local maximum from a local minimum. - Use monotonicity to compare \(f(2)\) and \(f(3)\).

1. The derivative is zero and changes sign at \(x=-2\), \(x=1\), and \(x=3\), so \(f\) has exactly three local extrema in \((-3,4)\). Statement a) is true. 2. On \((1,3)\), \(f'(x)<0\), so \(f\) is strictly decreasing there. Statement b) is true. 3. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum, not a local minimum. Statement c) is false. 4. Because \(f\) is decreasing from \(x=2\) to \(x=3\), \(f(2)>f(3)\). Statement d) is false.

a) True b) True c) False; \(f\) has a local maximum at \(x = 1\). d) False; \(f(2) > f(3)\).

- Find the zeros of the derivative graph. - Negative derivative means decreasing; positive derivative means increasing. - Negative-to-positive gives a local minimum. - Positive-to-negative gives a local maximum.

1. The derivative graph crosses the x-axis at \(x=2\) and \(x=6\), so these are the critical-value candidates. 2. At \(x=2\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 3. At \(x=6\), \(f'\) changes from positive to negative, so \(f\) has a local maximum.

a) \(x=2\) and \(x=6\) b) Local minimum at \(x=2\); local maximum at \(x=6\)

- Horizontal tangents occur at zeros of the first derivative. - The second derivative is the slope of the first derivative graph. - Use sign changes in \(f'\) to identify extrema. - For a non-extremum horizontal tangent, check whether the slope of \(f'\) changes sign.

1. The derivative is \(0\) at \(x=-2\) and \(x=2\). 2. At \(x=2\), the graph of \(f'\) has a local maximum with horizontal tangent, so \(f''(2)=0\). Therefore, the second derivative test is inconclusive there. 3. At \(x=-2\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. At \(x=2\), \(f'\) is negative on both sides, so there is no local extremum. 4. Near \(x=2\), the graph of \(f'\) changes from increasing to decreasing. Thus \(f''\) changes from positive to negative, so \(f\) changes concavity at \(x=2\). Because \(f'(2)=0\), the point is a stationary inflection point.

a) \(x=-2\) and \(x=2\) b) \(f''(2)=0\), so the second derivative test is inconclusive at \(x=2\). c) Local maximum at \(x=-2\); stationary inflection point at \(x=2\)

- Use the sign of the derivative for monotonicity. - Use sign changes of the derivative to classify local extrema. - Turning points of the derivative correspond to possible inflection points of the original function. - The derivative does not determine the original function's vertical shift.

1. On \((1,4)\), \(f'(x)<0\), so \(f\) is strictly decreasing. Statement 1 is true. 2. At \(x=-3\), the derivative changes from negative to positive, so \(f\) has a local minimum, not a local maximum. Statement 2 is false. 3. On \((-3,1)\), \(f'(x)>0\), so \(f\) is strictly increasing. Statement 3 is true. 4. At \(x=1\), the derivative changes from positive to negative, so \(f\) has a local maximum. Statement 4 is true. 5. Inflection points of \(f\) correspond to local extrema of \(f'\). The displayed derivative graph has two local extrema, so statement 5 is false. 6. A derivative graph determines changes in \(f\), but not its vertical position. Therefore, the sign of \(f(x)\) cannot be determined. Statement 6 is false.

1. True 2. False 3. True 4. True 5. False 6. False

- Find the zeros of \(f'\). - Use sign changes in \(f'\) to classify local extrema of \(f\). - A zero of \(f'\) without a sign change does not produce a local extremum of \(f\). - For a stationary inflection point, also check whether \(f'\) has a local extremum at that zero.

1. The zeros of \(f'\) are \(x=-3\), \(x=2\), and \(x=5\). 2. At \(x=-3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 3. At \(x=5\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 4. At \(x=2\), \(f'(2)=0\), but \(f'\) remains positive on both sides. Therefore, \(f\) has no local extremum there. The graph of \(f'\) has a local minimum at \(x=2\), so \(f''\) changes sign there. Thus, \(f\) has a stationary inflection point at \(x=2\).

a) Local minimum at \(x=-3\); local maximum at \(x=5\) b) Stationary inflection point at \(x=2\)

- Identify the points where the graph has a horizontal tangent. - A horizontal tangent has slope zero. - Use whether the graph rises or falls on each side to determine the sign of the derivative. - A stationary inflection point has slope zero and a change in concavity, but the derivative does not change sign.

1. From the graph, the local minimum is \((-2, -3)\), and the stationary inflection point is \((0, -1)\). 2. Both points have horizontal tangents, so \(f'(-2)=0\) and \(f'(0)=0\). 3. Near \(x=-2\), the graph decreases to the left and increases to the right. Thus, \(f'(x)\) changes from negative to positive. 4. Near \(x=0\), the graph increases on both sides. Thus, \(f'(x)\) is positive on both sides and does not change sign. The graph also changes concavity at \(x=0\), which confirms that the point is a stationary inflection point.

a) Local minimum: \((-2, -3)\); stationary inflection point: \((0, -1)\) b) \(f'(-2)=0\) and \(f'(0)=0\) c) At \(x=-2\), \(f'\) changes from negative to positive. At \(x=0\), \(f'\) is positive on both sides and does not change sign.

- Apply the chain rule. - The exponential factor is always positive. - Use sign changes of the derivative to classify interior extrema. - The function increases where its derivative is positive.

1. By the chain rule, \(f'(x)=g'(x)e^{g(x)}\). Since \(e^{g(x)}>0\), \(f'(x)\) has the same zeros and signs as \(g'(x)\). 2. The graph of \(g'\) has zeros at \(x=-2\), \(x=1\), and \(x=3\). At \(x=-2\), the sign changes from negative to positive, so \(f\) has an interior local minimum. At \(x=1\), the sign changes from positive to negative, so \(f\) has an interior local maximum. At \(x=3\), the sign changes from negative to positive, so \(f\) has an interior local minimum. 3. In the displayed domain \([-3,4]\), the graph of \(g'\) is positive on \((-2,1)\) and \((3,4)\). Therefore, \(f\) is strictly increasing on those maximal open intervals.

a) \(f'(x)=g'(x)e^{g(x)}\), and the exponential factor is always positive. b) Interior local minima at \(x=-2\) and \(x=3\); interior local maximum at \(x=1\). c) \((-2,1)\) and \((3,4)\)

- Apply the quotient rule and simplify the numerator carefully. - Evaluate the derivative at the x-coordinate of \(P\) to find the tangent slope. - Horizontal tangents occur where the derivative is zero. - Classify each critical number from the direction of the first-derivative sign change.

1. Let \(u(x)=4x-4\) and \(v(x)=x^2+3\). Then \(u'(x)=4\) and \(v'(x)=2x\). By the quotient rule, \(f'(x)=\frac{4(x^2+3)-(4x-4)(2x)}{(x^2+3)^2}\) \(=\frac{4x^2+12-8x^2+8x}{(x^2+3)^2}\) \(=\frac{-4x^2+8x+12}{(x^2+3)^2}\). 2. At \(x=1\), \(f'(1)=\frac{-4+8+12}{(1+3)^2}=1\). Thus, the slope at \(P\) is \(1\). 3. The denominator of \(f'\) is always positive, so horizontal tangents occur where \(-4x^2+8x+12=0\). Dividing by \(-4\) gives \(x^2-2x-3=0\), so \(x=-1\) or \(x=3\). The derivative is negative on \((-\infty, -1)\), positive on \((-1, 3)\), and negative on \((3, \infty)\). Therefore, \(x=-1\) gives a local minimum and \(x=3\) gives a local maximum. \(f(-1)=-2\) and \(f(3)=\frac{2}{3}\). Hence, \(T=(-1, -2)\) and \(H=\left(3, \frac{2}{3}\right)\).

a) \(f'(x)=\frac{-4x^2+8x+12}{(x^2+3)^2}\) b) The slope at \(P\) is \(1\). c) \(T=(-1, -2)\) and \(H=\left(3, \frac{2}{3}\right)\)

- Differentiate the product before using the point \(P\) to write the tangent line. - The acute inclination angle comes from the positive tangent slope. - The exponential factor in \(f'\) is always positive, so a simpler factor determines the derivative sign chart. - For part c, the sign change of \(f'\), not the appearance of a graph, must justify the extremum classification.

1. Differentiate using the product and chain rules: \(f'(x)=(0.5-0.5x)e^{-0.5x}\). 2. At \(x=0\), \(f'(0)=0.5\). Since the tangent passes through \((0,1)\), its equation is \(y=0.5x+1\). 3. The angle of inclination satisfies \(\tan\alpha=0.5\). Therefore, \(\alpha=\arctan(0.5)\approx26.6^\circ\). 4. Because \(e^{-0.5x}>0\), the sign of \(f'(x)\) is the sign of \(1-x\). Hence \(f'(x)>0\) for \(x<1\), \(f'(1)=0\), and \(f'(x)<0\) for \(x>1\). 5. By the first derivative test, \(x=1\) is a local maximum. The function is increasing on \(( -\infty,1)\) and decreasing on \((1,\infty)\).

a) \(y=0.5x+1\) b) \(\alpha\approx26.6^\circ\) c) Critical number: \(x=1\). Since \(f'\) changes from positive to negative there, the first derivative test gives a local maximum. Increasing on \(( -\infty,1)\); decreasing on \((1,\infty)\).

- Factor the first derivative so its sign is easy to analyze. - A local maximum is confirmed by a positive-to-negative sign change of the first derivative. - Eliminate the parameter from the maximum coordinates for the locus. - Use the maximum's x-coordinate to distinguish the displayed family members.

1. By the product and chain rules, \(f_k'(x)=x(2-kx)e^{-kx}\). 2. For \(x>0\), both \(x\) and \(e^{-kx}\) are positive. Therefore, \(f_k'(x)>0\) when \(x<\frac{2}{k}\) and \(f_k'(x)<0\) when \(x>\frac{2}{k}\). The nonzero critical point \(x=\frac{2}{k}\) is therefore a local maximum. 3. If the maximum has x-coordinate \(1\), then \(\frac{2}{k}=1\), so \(k=2\). Its y-coordinate is \(f_2(1)=e^{-2}\). 4. At any local maximum, \(k=\frac{2}{x}\). Substitution into \(y=x^2e^{-kx}\) gives \(y=e^{-2}x^2\), with \(x>0\). 5. For \(k=0.5\), the maximum occurs at \(x=4\), matching p. For \(k=1.0\), it occurs at \(x=2\), matching q.

a) \(k=2\); maximum \((1,e^{-2})\) b) \(y=e^{-2}x^2\) for \(x>0\) c) p: \(k=0.5\); q: \(k=1.0\)

- Use zeros and sign changes of \(f'\) to classify local extrema. - The function increases where its derivative is positive. - The maximum value of the derivative is the greatest slope of the function.

1. The derivative is zero at \(x=-2\) and \(x=4\). At \(x=-2\), it changes from negative to positive, so \(f\) has a local minimum. At \(x=4\), it changes from positive to negative, so \(f\) has a local maximum. 2. The derivative is positive for \(-2<x<4\), so \(f\) is strictly increasing on \((-2,4)\). 3. The greatest slope of \(f\) is the maximum value of \(f'\). The vertex of the derivative graph is \((1,4)\), so the greatest slope is \(4\) at \(x=1\).

a) Local minimum at \(x=-2\); local maximum at \(x=4\). b) \((-2,4)\). c) Greatest slope \(4\) at \(x=1\).

- Use the sign of \(f'\) to determine increasing and decreasing behavior. - At each zero, check whether \(f'\) changes sign. - A derivative value of \(0\) means the original function has a horizontal tangent.

1. On \((-2,0)\), \(f'(x)>0\), so \(f\) is strictly increasing. Statement a) is false. 2. At \(x=0\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. Statement b) is true. 3. At \(x=3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum, not a stationary inflection point. Statement c) is false. 4. Since \(f'(3)=0\), the slope of \(f\) at \(x=3\) is zero. Therefore, the tangent is horizontal. Statement d) is true.

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

- Use a clearly visible intercept or extremum from the graph. - Set the first derivative equal to zero. - Use the discriminant to determine when the derivative has two distinct real roots.

1. The graph shows a zero at \(x=3\) that is also a local minimum. Using the zero, \(f_k(3)=18-9k=0\), so \(k=2\). 2. The first derivative is \(f_k'(x)=x^2-2kx+3\). It has two distinct real roots exactly when its discriminant is positive: \(4k^2-12>0\). 3. Therefore, \(k^2>3\), so \(k<-\sqrt{3}\) or \(k>\sqrt{3}\). At the two simple critical points, the derivative changes sign, giving two local extrema.

a) \(k=2\) b) \(k<-\sqrt{3}\) or \(k>\sqrt{3}\)

- The value of \(f''(2)\) is the slope of the graph of \(f'\) at \(x=2\). - Check whether \(f'\) changes sign at its zero. - To classify a stationary inflection point, also check whether \(f'\) has a local extremum there.

1. Because \(f'\) has a local minimum at \(x=2\), its slope there is zero. Therefore, \(f''(2)=0\), so the second derivative test is inconclusive. 2. The graph of \(f'\) touches the x-axis at \(x=2\) but remains nonnegative on both sides. Thus, \(f'\) does not change sign, and \(f\) has no local extremum at \(x=2\). 3. Since \(f'(2)=0\) and \(f'\) has a local minimum there, \(f''\) changes from negative to positive. Therefore, the graph of \(f\) changes from concave down to concave up and has a stationary inflection point at \(x=2\).

a) The second derivative test is inconclusive because \(f''(2)=0\). b) There is no local extremum because \(f'\) does not change sign. The point is a stationary inflection point at \(x=2\).

- A corner in the graph of \(f'\) means its derivative is undefined there. - Determine the sign of \(f'\) on each side of \(x=2\). - A change from positive to negative indicates a local maximum of \(f\).

1. The graph of \(f'\) has a corner at \(x=2\), so \(f'\) is not differentiable there. Therefore, \(f''(2)\) does not exist, and the second derivative test cannot be applied. 2. To the left of \(x=2\), \(f'(x)>0\), so \(f\) is increasing. To the right, \(f'(x)<0\), so \(f\) is decreasing. 3. Because \(f'\) changes from positive to negative, the first derivative test shows that \(f\) has a local maximum at \(x=2\).

a) \(f''(2)\) does not exist because the graph of \(f'\) has a corner at \(x=2\). b) Local maximum at \(x=2\)

- Use the sign of \(f^{\prime}(x)\) to determine where \(f\) increases or decreases. - At a zero of \(f^{\prime}\), check the direction of the sign change to classify a local extremum. - The slope of the graph of \(f^{\prime}\) gives the sign of \(f^{\prime\prime}\). - If a function is increasing, compare its values at the two x-values in their left-to-right order.

1. On \((-2,2)\), \(f'(x)>0\). Therefore, \(f\) is strictly increasing on \((-2,2)\), so a) is true. 2. At \(x=2\), \(f'\) changes from positive to negative. Therefore, \(f\) has a local maximum, not a local minimum, so b) is false. 3. For \(0<x<4\), the graph of \(f'\) is decreasing, so \(f''(x)<0\). Therefore, \(f\) is concave down, so c) is true. 4. Since \(f\) is strictly increasing from \(x=-2\) to \(x=0\), \(f(0)>f(-2)=0\). Therefore, d) is false.

a) True b) False; \(f\) has a local maximum at \(x=2\). c) True d) False; \(f(0)>0\).

- Use the derivative's sign for monotonicity and local extrema. - A local extremum of the derivative can indicate a concavity change in the original function. - Use monotonicity to compare \(f(0)\) and \(f(1)\).

1. On \((-2,2)\), the derivative is negative, so \(f\) is strictly decreasing there. Statement a) is true. 2. At \(x=-2\), the derivative changes from positive to negative, so \(f\) has a local maximum. Statement b) is false. 3. The derivative graph has a local minimum at \(x=0\), so its slope changes from negative to positive. Thus, the concavity of \(f\) changes and statement c) is true. 4. Since \(f\) is strictly decreasing from \(x=0\) to \(x=1\), \(f(1)<f(0)=4\). Statement d) is false.

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

- Identify each horizontal tangent and decide whether the graph changes direction there. - The sign of \(f'\) is determined by whether the function is increasing or decreasing. - A stationary inflection point has a horizontal tangent and a change in concavity without a change from increasing to decreasing or vice versa.

1. From the graph, the local minimum is approximately \((-5,-2.0)\), the local maximum is approximately \((-1,2.0)\), and the stationary inflection point is approximately \((3,0.8)\). 2. Near \(x=-5\), the graph changes from decreasing to increasing, so \(f'\) changes from negative to positive. 3. Near \(x=-1\), the graph changes from increasing to decreasing, so \(f'\) changes from positive to negative. 4. Near \(x=3\), the graph is decreasing on both sides of the horizontal tangent, so \(f'\) stays negative on both sides. The concavity changes there, making the point a stationary inflection point rather than a local extremum.

a) Local minimum \(\approx(-5,-2.0)\); local maximum \(\approx(-1,2.0)\); stationary inflection point \(\approx(3,0.8)\) b) At \(x=-5\), \(f'\) changes from negative to positive. At \(x=-1\), \(f'\) changes from positive to negative. At \(x=3\), \(f'\) stays negative on both sides.

- The function decreases where its derivative is negative. - Find the x-intercepts of the derivative graph. - At each intercept, determine whether the derivative changes from negative to positive or from positive to negative.

1. The function \(f\) is decreasing where \(f^{\prime}(x)<0\). The derivative is negative on \((-4, -2)\) and \((1, 3)\). 2. The zeros of \(f^{\prime}\) are \(x=-2\), \(x=1\), and \(x=3\). 3. At \(x=-2\), \(f^{\prime}\) changes from negative to positive, so \(f\) has a local minimum. 4. At \(x=1\), \(f^{\prime}\) changes from positive to negative, so \(f\) has a local maximum. 5. At \(x=3\), \(f^{\prime}\) changes from negative to positive, so \(f\) has a local minimum.

a) \((-4, -2)\) and \((1, 3)\) b) Local minima at \(x=-2\) and \(x=3\); local maximum at \(x=1\).

- The y-values of the derivative graph give the slope of the original function. - Use zeros and sign changes of \(f^{\prime}\) to classify local extrema. - The slope of the derivative graph gives the sign of \(f^{\prime\prime}\). - Read \(f^{\prime}(-1)\) directly from the derivative graph.

1. For \(-2<x<1\), \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \((-2,1)\), so a) is true. 2. At \(x=-2\), \(f'\) changes from positive to negative. Therefore, \(f\) has a local maximum, not a local minimum, so b) is false. 3. For \(0<x<3\), the graph of \(f'\) is increasing, so \(f''(x)>0\). Therefore, \(f\) is concave up, so c) is true. 4. The tangent slope at \(x=-1\) is \(f'(-1)\). From the graph, \(f'(-1)=-1\), so d) is true.

a) True b) False; \(f\) has a local maximum at \(x=-2\). c) True d) True

- Read where the derivative graph lies above the x-axis. - Use maximal open intervals within the displayed domain. - At each zero, compare the sign of the derivative on the two sides.

1. The derivative is positive on \((-3,1)\) and \((4,6)\). Therefore, \(f\) is strictly increasing on those maximal open intervals. 2. At \(x=-3\), \(f'\) changes from negative to positive, so \(f\) has a local minimum. 3. At \(x=1\), \(f'\) changes from positive to negative, so \(f\) has a local maximum. 4. At \(x=4\), \(f'\) changes from negative to positive, so \(f\) has a local minimum.

a) \((-3,1)\) and \((4,6)\) b) Local minima at \(x=-3\) and \(x=4\); local maximum at \(x=1\).

- Use the position of the derivative graph relative to the x-axis to determine monotonicity. - Classify a critical point from the direction of the derivative's sign change. - The greatest downward slope corresponds to the lowest value of the derivative.

1. The derivative is negative on \((-2,2)\), so \(g\) is strictly decreasing on \((-2,2)\). 2. At \(x=-2\), \(g'\) changes from positive to negative, so \(g\) has a local maximum. At \(x=2\), \(g'\) changes from negative to positive, so \(g\) has a local minimum. 3. The greatest downward slope is the minimum value of \(g'\). The vertex of the derivative graph is \((0,-4)\), so it occurs at \(x=0\) and the slope is \(-4\).

1. \((-2,2)\) 2. Local maximum at \(x=-2\); local minimum at \(x=2\). 3. At \(x=0\), the slope is \(-4\).

- Critical points occur where the first derivative is zero or undefined. - Use the chain rule to find \(f'(x)\). - The exponential factor is always positive. - Use the sign change of the derivative to classify the critical point.

1. By the chain rule, \(f'(x)=h'(x)e^{h(x)}\). Since \(e^{h(x)}>0\) for every \(x\), \(f'(x)\) and \(h'(x)\) have the same zeros and the same sign. 2. The graph shows that \(h'(x)=0\) at \(x=1\). Therefore, \(f'(1)=0\). For \(x<1\), \(h'(x)<0\), so \(f'(x)<0\). For \(x>1\), \(h'(x)>0\), so \(f'(x)>0\). Because \(f'\) changes from negative to positive at \(x=1\), \(f\) has a local minimum there.

1) \(x=1\) 2) The point is a local minimum because \(f'\) changes from negative to positive at \(x=1\).

- Use the necessary first-derivative condition. - Factor the derivative. - Check the derivative's sign on both sides of the critical point.

1. The derivative is \(g_a'(x)=xe^{ax}(2+ax)\). 2. Requiring \(g_a'(2)=0\) gives \(2+2a=0\), so \(a=-1\). 3. For \(a=-1\), \(g_{-1}'(x)=xe^{-x}(2-x)\), which changes from positive to negative at \(x=2\). Therefore, the point is a local maximum.

\(a=-1\); local maximum

- Find the maximum of the given function. - Compare its exact maximum height with the marked graph maxima. - A multiplier outside the function creates a vertical scale.

1. The derivative of \(h\) is \(h'(x)=(2-x)e^{-0.5x}\). It changes from positive to negative at \(x=2\), so the maximum occurs there. Its value is \(h(2)=\frac{4}{e}\), which matches graph q. 2. All three graphs have the same zero and maximum x-coordinate, so they differ by a vertical scale factor. Their marked maximum heights are \(\frac{2}{e}\), \(\frac{4}{e}\), and \(\frac{6}{e}\). 3. Graph p has half the height of q, so its equation is \(xe^{-0.5x}\). Graph r has \(1.5\) times the height of q, so its equation is \(3xe^{-0.5x}\). 4. The family is \(f_k(x)=kxe^{-0.5x}\), where \(k>0\).

a) Graph q b) p: \(xe^{-0.5x}\); r: \(3xe^{-0.5x}\) c) \(f_k(x)=kxe^{-0.5x}\), \(k>0\)

- Zeros of the derivative are candidates for local extrema of the original function. - Relate the tangent angle to slope using the tangent function. - Use the sign change of the derivative to distinguish a local maximum from a local minimum. - Write a quadratic in factored form when its zeros are known.

1. The zeros of \(f^{\prime}\) are \(x=1\) and \(x=5\). 2. At \(x=1\), \(f^{\prime}\) changes from negative to positive, so \(f\) has a local minimum. At \(x=5\), \(f^{\prime}\) changes from positive to negative, so \(f\) has a local maximum. 3. A tangent angle of \(45^\circ\) corresponds to slope \(\tan(45^\circ)=1\). The graph shows \(f^{\prime}(x)=1\) at \(x=2\) and \(x=4\). 4. Using the zeros, write \(f^{\prime}(x)=a(x-1)(x-5)\). Since \(P=(2, 1)\) lies on the graph, \(1=a(2-1)(2-5)=-3a\), so \(a=-\frac{1}{3}\). 5. Expanding gives \(f^{\prime}(x)=-\frac{1}{3}x^2+2x-\frac{5}{3}\). Therefore, \(a=-\frac{1}{3}\), \(b=2\), and \(c=-\frac{5}{3}\).

a) Local minimum at \(x=1\); local maximum at \(x=5\). b) \(x=2\) and \(x=4\) c) \(a=-\frac{1}{3}\), \(b=2\), and \(c=-\frac{5}{3}\)

- Determine which factor of the derivative controls its sign. - Use maximal open intervals separated by the critical numbers. - Classify each critical point from the direction of the derivative's sign change.

1. The quotient rule gives \(f'(x)=\frac{16-4x^2}{(x^2+4)^2}\). 2. The denominator is always positive. The numerator is zero at \(x=-2\) and \(x=2\), negative for \(x<-2\) and \(x>2\), and positive for \(-2<x<2\). 3. Therefore, \(f\) is strictly decreasing on \((-\infty,-2)\) and \((2,\infty)\), and strictly increasing on \((-2,2)\). 4. The derivative changes from negative to positive at \(x=-2\), so \((-2,-1)\) is a local minimum. It changes from positive to negative at \(x=2\), so \((2,1)\) is a local maximum.

Strictly increasing on \((-2,2)\); strictly decreasing on \((-\infty,-2)\) and \((2,\infty)\). Local minimum: \((-2,-1)\) Local maximum: \((2,1)\)

- Use the sign of the displayed derivative for monotonicity. - Classify each derivative zero from the direction of its sign change. - Differentiate once more only for the concavity statement. - Remember that a concavity change at a point outside the domain is not an inflection point.

1. On \((0,2)\), \(f'(x)<0\), so \(f\) is strictly decreasing. Statement (1) is true. 2. At \(x=2\), \(f'\) changes from negative to positive, so \(f\) has a local minimum, not a local maximum. Statement (2) is false. 3. Differentiate the given derivative: \(f''(x)=\frac{8}{x^3}\). It is negative on \((-\infty,0)\) and positive on \((0,\infty)\), but \(x=0\) is not in the domain. There is no point in the domain where concavity changes, so statement (3) is true. 4. At \(x=-2\), \(f'\) changes from positive to negative, so the first derivative test gives a local maximum. Statement (4) is true.

(1) True. (2) False. (3) True. (4) True.

- Use the nonnegative square in the numerator to understand the point \(x=1\) before differentiating. - Factor the derivative into pieces whose signs can be read on intervals. - Classify each critical number from the direction of the first-derivative sign change.

1. For \(x>0\), \((\ln x)^2\ge0\) and \(x>0\), so \(f(x)\ge0\). Since \(f(1)=0\), \(x=1\) is a local minimum. 2. Differentiate: \(f'(x)=\frac{\ln x(2-\ln x)}{x^2}\). The critical numbers are \(x=1\) and \(x=e^2\). 3. Because \(x^2>0\), the derivative is negative on \((0,1)\), positive on \((1,e^2)\), and negative on \((e^2,\infty)\). Thus \(f'\) changes from negative to positive at \(x=1\), confirming a local minimum, and from positive to negative at \(x=e^2\), giving a local maximum by the first derivative test. 4. The local maximum value is \(f(e^2)=\frac{4}{e^2}\). Therefore the local extrema are \((1,0)\) and \(\left(e^2,\frac{4}{e^2}\right)\).

a) \(x=1\) is a local minimum because \(f(x)\ge0\) and \(f(1)=0\). b) Local minimum: \((1,0)\). Local maximum: \(\left(e^2,\frac{4}{e^2}\right)\). c) Decreasing on \((0,1)\) and \((e^2,\infty)\); increasing on \((1,e^2)\).

- First use the sign of the factors in the original function to understand the zero at \(x=2\). - Factor the first derivative completely. - Use the derivative sign on each interval to classify both critical numbers.

1. Since \((x-2)^2\ge0\) and \(e^x>0\), \(f(x)\ge0\) for all \(x\). Because \(f(2)=0\), \(x=2\) is a local minimum. 2. Differentiate: \(f'(x)=2(x-2)e^x+(x-2)^2e^x=xe^x(x-2)\). The critical numbers are \(x=0\) and \(x=2\). 3. Since \(e^x>0\), \(f'\) is positive on \(( -\infty,0)\), negative on \((0,2)\), and positive on \((2,\infty)\). Thus \(f'\) changes from positive to negative at \(0\), giving a local maximum, and from negative to positive at \(2\), giving a local minimum. 4. \(f(0)=4\) and \(f(2)=0\), so the extrema are \((0,4)\) and \((2,0)\).

a) \(x=2\) is a local minimum because \(f(x)\ge0\) and \(f(2)=0\). b) Local maximum: \((0,4)\). Local minimum: \((2,0)\). c) Increasing on \(( -\infty,0)\) and \((2,\infty)\); decreasing on \((0,2)\).

- Use the two displayed graphs to form a conjecture about the x-coordinate of the extremum, then prove it from the derivative. - Factor the derivative so its sign can be determined without a second derivative. - Compare the sign of \(1-\ln x\) on the two sides of \(e\), then account for the sign of \(a\).

1. The two displayed positive-\(a\) graphs suggest that their extrema occur at the same x-coordinate. 2. a) Substituting \(P(e^2,2)\) gives \(\frac{2a}{e^2}=2\), so \(a=e^2\). 3. b) The quotient rule gives \(f_a'(x)=a\frac{1-\ln x}{x^2}\). Since \(a\ne0\) and \(x^2>0\), the only critical point occurs when \(\ln x=1\), so \(x=e\). 4. The factor \(1-\ln x\) is positive for \(0<x<e\) and negative for \(x>e\). Therefore, if \(a>0\), \(f_a'\) changes from positive to negative and \(x=e\) is a local maximum. If \(a<0\), the signs reverse and \(x=e\) is a local minimum. 5. Every extremum has x-coordinate \(e\), so all extrema lie on the vertical line \(x=e\). 6. c) At the zero \(x=1\), \(f_a'(1)=a\).

a) \(a=e^2\) b) The only critical point is at \(x=e\); it is a local maximum for \(a>0\) and a local minimum for \(a<0\). All extrema lie on \(x=e\). c) \(m=a\)

- The exponential factor is never zero. - Use the product rule and chain rule. - Compare the two x-coordinate expressions. - An exponential equals \(1\) when its exponent is zero.

1. Since the exponential factor is positive, the zero is \(x=k\). 2. The derivative is \(f_k'(x)=(1-0.5x+0.5k)e^{1-0.5x}\). Setting it equal to zero gives \(x=k+2\). The derivative changes from positive to negative, so this is a local maximum. 3. The distance between the zero and the local maximum is \((k+2)-k=2\). 4. The slope at the zero is \(f_k'(k)=e^{1-0.5k}\). Setting this equal to \(1\) gives \(1-0.5k=0\), so \(k=2\).

a) \(x=k\) b) Local maximum x-coordinate: \(x=k+2\); the distance from the zero is \(2\). c) \(k=2\)

- Use the first derivative to locate the critical number. - Determine the derivative sign on each side of that value and classify it from the sign change. - Use the tangent slope at \(x=1\) for part b). - Combine end behavior with the maximum when determining the range.

1. Differentiate using the product rule: \(f'(x)=2-2\ln x\). Set the derivative equal to zero: \(2-2\ln x=0\), so \(x=e\). The derivative is positive on \((0,e)\) and negative on \((e,\infty)\). Thus \(f'\) changes from positive to negative at \(x=e\), so the first derivative test gives a local maximum there. The function is strictly increasing on \((0,e)\) and strictly decreasing on \((e,\infty)\). The y-coordinate is \(f(e)=2e\). 2. At \(x=1\), \(f(1)=4\) and \(f'(1)=2\). Thus, the tangent line is \(y=2(x-1)+4=2x+2\). Setting \(y=0\) gives \(x=-1\), so the x-intercept is \((-1, 0)\). 3. Rewrite \(f(x)=4x-2x\ln x\). The negative term dominates as \(x\to\infty\), so \(f(x)\to-\infty\). Since the function increases to its global maximum \(2e\) and then decreases without bound, its range is \((-\infty, 2e]\).

a) Local maximum: \((e, 2e)\); increasing on \((0,e)\); decreasing on \((e,\infty)\) b) Tangent line: \(y=2x+2\); x-intercept: \((-1, 0)\) c) \(\lim_{x\to\infty}f(x)=-\infty\); range: \((-\infty, 2e]\)

- The exponential factor is always positive, so the zero comes from the remaining factor. - Compare linear growth with exponential decay for the long-run behavior. - Use the sign of \(g'(x)\) on either side of its critical point to classify the local extremum. - A positive volume-change rate means the reservoir's volume is increasing.

1. Since \(e^{-0.5x}>0\), \(g(x)=0\) only when \(x+1=0\). Thus, the zero is \(x=-1\). As \(x\to\infty\), exponential decay dominates linear growth, so \(g(x)\to0\). 2. Differentiate: \(g'(x)=e^{-0.5x}(0.5-0.5x)\). The derivative is zero at \(x=1\), and it changes from positive to negative. Therefore, the graph has a local maximum at \(\left(1,2e^{-0.5}\right)\approx(1,1.21)\). 3. The graph gives \(g(0)=1\,\text{m}^3/\text{day}\). Because this rate is positive, the water volume is increasing on the reference date.

a) Zero at \(x=-1\); \(\lim_{x\to\infty}g(x)=0\) b) Local maximum at \(\left(1,2e^{-0.5}\right)\approx(1,1.21)\) c) \(1\,\text{m}^3/\text{day}\); the water volume is increasing

- Use the product rule and chain rule together. - Find the intercepts of the derivative algebraically. - Remember that an exponential factor is always positive. - Use the sign of the first derivative to classify the critical point.

1. Differentiate using the product and chain rules: \(f'(x)=5e^{-0.5x}+5x\cdot(-0.5)e^{-0.5x}\) \(=(5-2.5x)e^{-0.5x}\). 2. Differentiate again: \(f''(x)=-2.5e^{-0.5x}-0.5(5-2.5x)e^{-0.5x}\) \(=(1.25x-5)e^{-0.5x}\). 3. Since the exponential factor is always positive, \(f'(x)=0\) when \(5-2.5x=0\), so the x-intercept is \((2, 0)\). Also, \(f'(0)=5\), so the y-intercept is \((0, 5)\). For large positive \(x\), \(f'(x)\) approaches \(0\) from below. 4. Since the exponential factor is positive, \(f'(x)>0\) for \(x<2\) and \(f'(x)<0\) for \(x>2\). Therefore, \(f\) changes from increasing to decreasing at \(x=2\), so it has a local maximum there.

a) \(f'(x)=(5-2.5x)e^{-0.5x}\); \(f''(x)=(1.25x-5)e^{-0.5x}\) b) The derivative graph has intercepts \((0, 5)\) and \((2, 0)\), and approaches the x-axis from below as \(x\to\infty\). c) Since \(f'(x)>0\) for \(x<2\) and \(f'(x)<0\) for \(x>2\), \(f\) has a local maximum at \(x=2\).

- Locate every x-intercept of the derivative graph. - Compare the sign of the derivative immediately to the left and right of each zero. - A zero of the derivative alone does not guarantee an extremum.

1. The zeros of \(g'\) are \(x=-3\), \(x=0\), and \(x=3\). 2. At \(x=-3\), \(g'\) changes from positive to negative, so \(g\) has a local maximum. 3. At \(x=0\), \(g'\) is negative on both sides, so \(g\) has neither a local maximum nor a local minimum there. 4. At \(x=3\), \(g'\) changes from negative to positive, so \(g\) has a local minimum.

\(x=-3\): local maximum \(x=0\): neither \(x=3\): local minimum

- Find the degree of the derivative. - An extremum requires a sign change in the derivative. - Consider simple, double, and triple zeros of a cubic. - Compare the derivative's signs for very large positive and negative x-values.

1. The derivative of a fourth-degree polynomial is a cubic polynomial. 2. A local extremum occurs at a derivative zero where the derivative changes sign. 3. A cubic can have three simple real zeros, producing three sign changes and three local extrema. It can also have one simple real zero and two nonreal zeros, one simple and one double real zero, or one triple real zero. Each of those cases produces exactly one sign change and one local extremum. 4. Because a cubic has opposite end behavior as \(x\to-\infty\) and \(x\to\infty\), the total number of sign-changing real zeros is odd. Therefore, a fourth-degree polynomial has exactly one or exactly three local extrema, never exactly two.

The claim is true. The cubic derivative changes sign at either one or three real zeros, so the fourth-degree polynomial has either one or three local extrema.

- Use the chain rule for each composition. - Use the given sign of \(g'\). - A horizontal tangent occurs where the first derivative equals zero. - Use the sign change of the first derivative to classify the critical point. - Can a product be zero if neither factor is ever zero?

1. By the chain rule, \(f'(x)=2x\,g'(x^2-4)\). 2. Since \(g'(x^2-4)<0\) for every \(x\), the derivative \(f'(x)\) can equal zero only when \(2x=0\). Thus \(x=0\) is the only critical point. 3. For \(x<0\), both \(2x\) and \(g'(x^2-4)\) are negative, so \(f'(x)>0\). For \(x>0\), \(2x>0\) while \(g'(x^2-4)<0\), so \(f'(x)<0\). The derivative changes from positive to negative at \(x=0\), so \(f\) has a local maximum there. 4. For \(h\), the chain rule gives \(h'(x)=g'(x^3+x)(3x^2+1)\). The first factor is always negative and \(3x^2+1>0\) for every real \(x\). Therefore \(h'(x)<0\) everywhere. Since a differentiable function can have a local extremum only at a point where its derivative is zero, \(h\) has no local extrema.

a) \(f'(x)=2x\,g'(x^2-4)\) b) \(f'(x)>0\) for \(x<0\) and \(f'(x)<0\) for \(x>0\), so \(f\) has a local maximum at \(x=0\). c) \(h'(x)=g'(x^3+x)(3x^2+1)<0\) for every real \(x\), so \(h\) has no local extrema.

- Determine the domain before reasoning about extrema. - Once the derivative is factored into sign-known pieces, identify its sign on each side of \(0\). - Use the direction of the derivative sign change to classify the critical point.

1. The denominator requires \(1-ax^2>0\). 2. If \(a<0\), then \(1-ax^2\geq1\), so the domain is \(\mathbb{R}\). If \(a>0\), then \(|x|<\frac{1}{\sqrt{a}}\), so the domain is \(\left(-\frac{1}{\sqrt{a}},\frac{1}{\sqrt{a}}\right)\). 3. Since \(g_a(-x)=g_a(x)\), every graph is symmetric about the y-axis. 4. For \(a>0\), as \(x\) approaches either endpoint from inside the domain, the positive denominator approaches \(0\), so \(g_a(x)\to\infty\). 5. Write \(g_a(x)=a(1-ax^2)^{-1/2}\). The chain rule gives \(g_a'(x)=\frac{a^2x}{(1-ax^2)^{3/2}}\). 6. On the domain, the denominator and \(a^2\) are positive. Thus, the derivative is negative for \(x<0\) and positive for \(x>0\). The derivative changes from negative to positive at \(x=0\), so the first derivative test gives a local minimum there.

a) If \(a<0\), the domain is \(\mathbb{R}\). If \(a>0\), the domain is \(\left(-\frac{1}{\sqrt{a}},\frac{1}{\sqrt{a}}\right)\). b) \(g_a(-x)=g_a(x)\). c) From within the domain, \(g_a(x)\to\infty\) at both endpoints. d) \(g_a'(x)=\frac{a^2x}{(1-ax^2)^{3/2}}\); the derivative changes from negative to positive at \(x=0\), so there is a local minimum there.

- Rewrite the variable exponent in a form that can be differentiated. - Determine the sign of the derivative from the factor that can change sign. - Classify the critical number from the derivative sign change.

1. Rewrite the function: \(f(x)=e^{\frac{\ln x}{x}}\). Let \(h(x)=\frac{\ln x}{x}\). Then \(h'(x)=\frac{1-\ln x}{x^2}\). 2. Apply the chain rule: \(f'(x)=x^{1/x}\frac{1-\ln x}{x^2}\). 3. Since \(x^{1/x}>0\) and \(x^2>0\), critical numbers satisfy \(1-\ln x=0\), so \(x=e\). 4. The derivative is positive for \(0<x<e\) and negative for \(x>e\). Thus \(f'\) changes from positive to negative at \(x=e\), so the first derivative test gives a local maximum there. 5. The corresponding y-value is \(f(e)=e^{1/e}\).

a) \(f'(x)=x^{1/x}\frac{1-\ln x}{x^2}\) b) Local maximum at \(\left(e, e^{1/e}\right)\)

- Compare the growth rates of logarithmic and linear functions. - Apply the quotient rule. - Eliminate the parameter from the maximum-point coordinates. - Include the attainable x-values in the locus domain.

1. As \(x\to0^+\), \(\ln(x)+t\to-\infty\) and \(x\to0^+\), so \(h_t(x)\to-\infty\). Thus \(x=0\) is a vertical asymptote. As \(x\to\infty\), \(\frac{\ln(x)+t}{x}\to0\), so \(y=0\) is a horizontal asymptote. 2. The quotient rule gives \(h_t'(x)=\frac{1-\ln(x)-t}{x^2}\). Setting the derivative equal to zero gives \(x=e^{1-t}\). 3. The derivative changes from positive to negative at this value, so it is a local maximum. Its y-coordinate is \(h_t(e^{1-t})=e^{t-1}\). Thus the local maximum is \((e^{1-t}, e^{t-1})\). 4. The coordinates satisfy \(xy=1\), so the locus is \(y=\frac{1}{x}\). Because \(x=e^{1-t}>0\), the locus domain is \(x>0\).

a) Vertical asymptote: \(x=0\); horizontal asymptote: \(y=0\) b) \((e^{1-t}, e^{t-1})\) c) \(y=\frac{1}{x}\) for \(x>0\)

- Factor the first derivative and use its zeros to divide the real line into sign intervals. - Classify the local extrema from the direction of each first-derivative sign change. - A critical number where the first derivative keeps the same sign is not a local extremum. - Use concavity only for the separate stationary-inflection classification.

1. Differentiate: \(g'(x)=-\frac14x^4+x^2=x^2\left(1-\frac14x^2\right)\). 2. The critical numbers are \(x=-2,0,2\). The sign of \(g'\) is negative on \(( -\infty,-2)\), positive on \((-2,0)\), positive on \((0,2)\), and negative on \((2,\infty)\). 3. At \(x=-2\), \(g'\) changes from negative to positive, so the first derivative test gives a local minimum. Since \(g(-2)=-\frac{16}{15}\), the point is \((-2,-\frac{16}{15})\). 4. At \(x=2\), \(g'\) changes from positive to negative, so the first derivative test gives a local maximum. Since \(g(2)=\frac{16}{15}\), the point is \((2,\frac{16}{15})\). 5. At \(x=0\), \(g'\) is positive on both sides, so there is no local extremum. Also \(g''(x)=-x^3+2x\) changes from negative to positive at \(0\), so \((0,0)\) is a stationary inflection point.

a) \(g'(x) = -\frac{1}{4}x^4 + x^2\); critical numbers: \(x = -2\), \(x = 0\), and \(x = 2\) b) Local minimum: \((-2, -\frac{16}{15})\); stationary inflection point: \((0, 0)\); local maximum: \((2, \frac{16}{15})\)

- Each local extremum requires a sign-changing zero of the first derivative. - Compare the degree of a polynomial with the degree of its derivative. - To show the lower bound is attainable, test the derivative of an explicit fifth-degree polynomial. - Relate the parity of the degree to derivative end signs.

1. Four local extrema require at least four distinct zeros of \(g'\) at which the derivative changes sign. Therefore, \(g'\) must have degree at least \(4\), so \(g\) must have degree at least \(5\). 2. Degree \(5\) is attainable. For example, \(g(x)=\frac15x^5-\frac53x^3+4x\) has derivative \(g'(x)=x^4-5x^2+4=(x+2)(x+1)(x-1)(x-2)\). The derivative changes sign at all four simple zeros, so this fifth-degree polynomial has exactly four local extrema. 3. If \(g\) had even degree, then \(g'\) would have odd degree and opposite signs at the two ends. A continuous derivative with four sign-changing zeros would have the same sign after four changes, which is incompatible with opposite end signs. 4. If \(g\) has odd degree, then \(g'\) has even degree and can have the same sign at both ends, allowing four sign changes. Thus the degree must be odd.

a) The least possible degree is \(5\). b) The degree must be odd.

- Use the absence of any other horizontal tangents to control whether the first derivative can change sign. - Classify a horizontal tangent by what the derivative does immediately on its two sides. - Track the monotonic behavior from the left end behavior through \(P\) and \(Q\). - Count x-axis crossings only after the possible sign patterns are settled.

1. Because there are no horizontal tangents for \(x<0\), the polynomial \(h'\) has a constant sign on \((-\infty,0)\). Since \(h(x)\to-\infty\) as \(x\to-\infty\) while \(h(0)=2\), \(h\) must be increasing there, so \(h'>0\) for \(x<0\). 2. If \(h'\) changed from positive to negative at \(P\), then \(P\) would be a local maximum. To reach the higher point \(Q=(2,5)\), the derivative would later have to change back to positive, creating another zero of \(h'\) between \(P\) and \(Q\), contrary to the hypothesis. Thus \(h'\) stays positive on both sides of \(0\). Because \(h'\) is a polynomial and \(h'(0)=0\), this zero has even multiplicity. Therefore \(h''\) has an odd-multiplicity zero at \(0\) and changes sign there, so \(P\) is a stationary inflection point. 3. The function is increasing from \(P\) to \(Q\), so \(h'>0\) immediately to the left of \(Q\). At \(Q\), either \(h'\) changes from positive to negative, giving a local maximum; then \(h'<0\) for all \(x>2\), so \(h(x)\to-\infty\) as \(x\to\infty\). Or \(h'\) remains positive on both sides of \(2\). In that case the zero of the polynomial \(h'\) at \(2\) has even multiplicity, so \(h''\) changes sign there; \(Q\) is a stationary inflection point, and \(h(x)\to\infty\) as \(x\to\infty\). 4. The function rises from \(-\infty\) to \(P=(0,2)\), so it has exactly one zero to the left of \(P\). If \(Q\) is a stationary inflection point there are no more zeros; if \(Q\) is a local maximum, the later decrease to \(-\infty\) creates exactly one additional zero. Thus the possible totals are one or two.

a) \(P\) must be a stationary inflection point. b) \(Q\) can be a local maximum, giving \(h(x)\to-\infty\), or a stationary inflection point, giving \(h(x)\to\infty\), as \(x\to\infty\). c) Minimum: \(1\) real zero; maximum: \(2\) real zeros.

- Find all critical numbers of the fifth-degree polynomial. - Make a complete sign chart for the first derivative. - Evaluate the function at each critical number. - Combine the monotonicity with the end behavior to count sign changes of the function.

1. Let \(f(x)=3x^5-25x^3+60x-20\). Then \(f'(x)=15(x^4-5x^2+4)=15(x^2-1)(x^2-4)\), so the critical numbers are \(x=-2,-1,1,2\). 2. The derivative is positive on \((-\infty,-2)\), negative on \((-2,-1)\), positive on \((-1,1)\), negative on \((1,2)\), and positive on \((2,\infty)\). Thus the first derivative test gives local maxima at \(x=-2\) and \(x=1\), and local minima at \(x=-1\) and \(x=2\). 3. The corresponding values are \(f(-2)=-36\), \(f(-1)=-58\), \(f(1)=18\), and \(f(2)=-4\). 4. Since the polynomial has odd degree and positive leading coefficient, \(f(x)\to-\infty\) as \(x\to-\infty\), and \(f(x)\to\infty\) as \(x\to\infty\). 5. The function remains negative through the first two monotonic pieces, crosses once while rising from \(-58\) to \(18\), crosses once while falling from \(18\) to \(-4\), and crosses once while rising from \(-4\) to \(\infty\). Therefore, there are exactly three real zeros.

The equation has exactly \(3\) real solutions.

- Compare \(f_k(-x)\) with \(f_k(x)\). - Factor the first derivative and make a sign chart at all three critical numbers. - Eliminate \(k\) from a maximum's coordinates. - Use the horizontal distance between the maxima as the triangle's base.

1. Since \(f_k(-x)=f_k(x)\), the graph is symmetric about the y-axis. The exponential factor never equals zero, so the only x- and y-intercept is the origin. 2. By the product and chain rules, \(f_k'(x)=2x(1-kx^2)e^{-kx^2}\). The critical numbers are \(x=0\) and \(x=\pm\frac{1}{\sqrt{k}}\). 3. Because the exponential factor is positive, a sign chart for \(2x(1-kx^2)\) gives \(f_k'>0\) on \((-\infty,-1/\sqrt{k})\), \(f_k'<0\) on \((-1/\sqrt{k},0)\), \(f_k'>0\) on \((0,1/\sqrt{k})\), and \(f_k'<0\) on \((1/\sqrt{k},\infty)\). Thus, the first derivative test gives a local minimum at \(x=0\) and local maxima at \(x=\pm\frac{1}{\sqrt{k}}\). 4. The extremum values are \(f_k(0)=0\) and \(f_k\left(\pm\frac{1}{\sqrt{k}}\right)=\frac{1}{ek}\), so the local maxima are \(\left(\pm\frac{1}{\sqrt{k}},\frac{1}{ek}\right)\). 5. At a maximum, \(k=\frac{1}{x^2}\). Substitution into \(y=\frac{1}{ek}\) gives \(y=\frac{x^2}{e}\), with \(x\ne0\). 6. The triangle has base \(\frac{2}{\sqrt{k}}\) and height \(\frac{1}{ek}\). Therefore, \(A(k)=\frac{1}{ek^{3/2}}\). Setting \(A(k)=\frac{1}{e}\) gives \(k=1\).

a) The graph is symmetric about the y-axis; the only intercept is \((0, 0)\). b) Local minimum: \((0, 0)\); local maxima: \(\left(\pm\frac{1}{\sqrt{k}}, \frac{1}{ek}\right)\) c) \(y=\frac{x^2}{e}\) for \(x\neq0\) d) \(A(k)=\frac{1}{ek^{3/2}}\); the area is \(\frac{1}{e}\) when \(k=1\).

- Focus on the numerator of the first derivative. - Determine when that numerator is linear rather than quadratic. - Check the constant-function case separately. - For \(a\ne0\), use the discriminant to count the derivative's real zeros.

1. Apply the quotient rule: \(g_{a,b}'(x)=\frac{-ax^2+2(1-b)x+a}{(x^2+1)^2}\). 2. If \(a=0\), the derivative numerator is \(2(1-b)x\). For \(b\ne1\), it has one simple zero at \(x=0\), so the derivative changes sign and the function has exactly one local extremum. 3. If \(a=0\) and \(b=1\), then \(g_{0,1}(x)=1\) for every real \(x\). Under the non-strict local-extremum definition, every point is both a local maximum and a local minimum, so the function has infinitely many local extrema, not exactly one. 4. If \(a\ne0\), the derivative numerator is quadratic with discriminant \(D=4(1-b)^2+4a^2>0\). Thus, it has two distinct real zeros. Because the denominator is always positive and the quadratic numerator changes sign at each simple zero, the function has exactly two local extrema.

a) Yes. For \(a=0\) and \(b\ne1\), the function has exactly one local extremum at \(x=0\). For \(a=0\), \(b=1\), the function is constant and has infinitely many non-strict local extrema. b) For \(a\ne0\), \(D=4(1-b)^2+4a^2>0\), so the function has exactly two local extrema.

- Differentiate both functions. - Determine whether each derivative has real zeros and whether its sign changes. - A cubic's derivative is quadratic. - A repeated zero of the derivative does not produce a sign change.

1. Since \(f'(x)=3x^2+1>0\) for every real \(x\), \(f\) has no local extrema. 2. Since \(g'(x)=3x^2-3=3(x-1)(x+1)\), its derivative changes sign at \(x=-1\) and \(x=1\). Thus, \(g\) has two local extrema. 3. The derivative of any cubic polynomial is quadratic. If it has no real zeros, the cubic has no local extrema. If it has two distinct real zeros, its sign changes at both and the cubic has two local extrema. If it has one repeated real zero, the derivative does not change sign there, so the cubic has a stationary inflection point rather than a local extremum. Therefore, exactly one local extremum is impossible.

a) \(f\) has \(0\) local extrema; \(g\) has \(2\) local extrema. b) A quadratic derivative has either no real zeros, two distinct real zeros, or one repeated zero. These cases produce \(0\), \(2\), or \(0\) local extrema, respectively.

- Calculate the first two derivatives and substitute \(x=2\). - The equations \(g'(2)=0\) and \(g''(2)=0\) are not sufficient by themselves. - Check the sign of \(g'\) on each side of \(x=2\). - A very flat graph can still have a local minimum.

1. Differentiate: \(g'(x)=0.4x^3-2.4x^2+4.8x-3.2=0.4(x-2)^3\) and \(g''(x)=1.2x^2-4.8x+4.8=1.2(x-2)^2\). 2. Therefore, \(g'(2)=0\) and \(g''(2)=0\). 3. For \(x<2\), \(g'(x)=0.4(x-2)^3<0\). For \(x>2\), \(g'(x)>0\). Thus, \(g\) changes from decreasing to increasing at \(x=2\). 4. By the first derivative test, \(g\) has a local minimum at \(x=2\), not a stationary inflection point. Equivalently, \(g'''(2)=0\) and \(g^{(4)}(2)=2.4>0\), so the first nonzero higher derivative has even order and is positive.

a) \(g'(2)=0\) and \(g''(2)=0\) b) The claim is false. Since \(g'\) changes from negative to positive, \(g\) has a local minimum at \(x=2\), not a stationary inflection point.

- Apply the quotient rule carefully. - Set the numerator of the derivative equal to zero. - Compare both coordinates of the maximum as \(k\) changes. - Compare the degrees of the numerator and denominator.

1. The quotient rule gives \(f_k'(x)=\frac{4(x^2+k)-8x^2}{(x^2+k)^2}=\frac{4k-4x^2}{(x^2+k)^2}\). 2. The derivative is zero at \(x=\sqrt{k}\). It changes from positive to negative, so this is a local maximum. Its value is \(f_k(\sqrt{k})=\frac{2}{\sqrt{k}}\), giving the local maximum \(\left(\sqrt{k}, \frac{2}{\sqrt{k}}\right)\). 3. As \(k\) increases, the maximum moves right and downward. Thus p corresponds to \(k=1\), q to \(k=4\), and r to \(k=9\). 4. The denominator has higher degree than the numerator, so \(\lim_{x\to\infty}f_k(x)=0\), independent of \(k\).

a) \(f_k'(x)=\frac{4k-4x^2}{(x^2+k)^2}\) b) \(\left(\sqrt{k}, \frac{2}{\sqrt{k}}\right)\) c) p: \(k=1\); q: \(k=4\); r: \(k=9\) d) \(0\), independent of \(k\)