- The sign of \(f'\) gives the direction of travel. - A negative-to-positive change in \(f'\) gives a local minimum. - The sign of \(f''\) gives concavity. - Use \(f(0)=1\) to place the landmark vertically.

1. Since \(f'(x)<0\) for \(x<0\), the graph decreases as it approaches \(x=0\) from the left. 2. Since \(f'(x)>0\) for \(x>0\), the graph increases to the right of \(x=0\). Thus \(f'\) changes from negative to positive, so \(x=0\) is a local minimum. 3. The anchor \(f(0)=1\) makes the minimum point \((0,1)\). 4. Because \(f''(x)>0\) everywhere, both the left and right branches are concave up. Blueprint: decreasing and concave up to \((0,1)\), then increasing and concave up.

Decreasing and concave up for \(x<0\); local minimum at \((0,1)\); increasing and concave up for \(x>0\).

- Above the x-axis, the derivative has what sign? - Translate a derivative sign into the direction of the original graph. - When the sign changes at a zero, what shape does the original graph have nearby?

1. The derivative \(f'(x)=2-x\) is positive for \(x<2\) and negative for \(x>2\). 2. Therefore, a graph of \(f\) rises up to \(x=2\) and falls after \(x=2\). 3. The graph has a local maximum at \(x=2\). Its vertical position is not determined by \(f'\).

A graph of \(f\) rises for \(x<2\), has a local maximum at \(x=2\), and falls for \(x>2\). The height of the graph cannot be determined from \(f'\) alone.

- What does a horizontal derivative graph say about the slope of the original graph? - Does a constant slope make a curve or a straight line? - Could two vertically shifted straight lines share the same derivative?

1. The derivative is the constant \(f'(x)=-2\), so the slope of \(f\) is always \(-2\). 2. A function with this derivative has the form \(f(x)=-2x+C\), where \(C\) is any constant. 3. Thus its graph is a straight line sloping downward. The derivative graph does not determine \(C\) or the line's height.

The graph of \(f\) is a decreasing straight line with slope \(-2\). Its height is not determined.

- Combine the sign of \(f'\) and the sign of \(f''\) on each interval. - Split the blueprint at \(-2\), \(0\), and \(1\). - Use the supplied function values only to place the named landmarks.

1. On \(( -\infty,-2)\), \(f'>0\) and \(f''<0\): increasing, concave down. 2. At \(x=-2\), \(f'\) changes positive to negative, so \((-2,4)\) is a local maximum. 3. On \((-2,0)\), \(f'<0\) and \(f''<0\): decreasing, concave down. 4. At \(x=0\), concavity changes from down to up, so \((0,1)\) is an inflection point. 5. On \((0,1)\), \(f'<0\) and \(f''>0\): decreasing, concave up. 6. At \(x=1\), \(f'\) changes negative to positive, so \((1,0)\) is a local minimum. 7. On \((1,\infty)\), \(f'>0\) and \(f''>0\): increasing, concave up.

Increasing/concave down on \(( -\infty,-2)\); local max \((-2,4)\); decreasing/concave down on \((-2,0)\); inflection \((0,1)\); decreasing/concave up on \((0,1)\); local min \((1,0)\); increasing/concave up on \((1,\infty)\).

- Read the two derivative signs in each row as two different graph features. - A turning point depends on a change in the first derivative's sign. - An inflection point depends on a change in concavity. - Use the supplied points only after deciding which kind of feature each one has.

1. Positive to negative \(f'\) at \(-2\) gives a local maximum at \((-2,4)\). 2. Negative to positive \(f'\) at \(1\) gives a local minimum at \((1,0)\). 3. The sign of \(f''\) changes from negative to positive at \(x=0\), so \((0,2)\) is an inflection point. 4. The first interval is increasing and concave down; the second is decreasing and concave down; the third is decreasing and concave up; the fourth is increasing and concave up.

Local maximum \((-2,4)\); local minimum \((1,0)\); inflection point \((0,2)\). On \((-3,-2)\): increasing, concave down. On \((-2,0)\): decreasing, concave down. On \((0,1)\): decreasing, concave up. On \((1,2)\): increasing, concave up.

- Zeros of \(f^{\prime}\) give locations of horizontal tangents on \(f\). - Where \(f^{\prime}>0\), \(f\) increases. - Where \(f^{\prime}<0\), \(f\) decreases. - Where \(f^{\prime}\) changes from increasing to decreasing, \(f\) changes concavity.

1. The derivative is \(0\) at \(x=-2\) and \(x=4\). It changes from negative to positive at \(x=-2\), so a possible graph of \(f\) has a local minimum there. It changes from positive to negative at \(x=4\), so the graph has a local maximum there. The graph of \(f^{\prime}\) increases up to \(x=1\) and decreases after \(x=1\), so \(f\) changes from concave up to concave down and has an inflection point at \(x=1\). Any vertical shift with these slope and concavity properties is acceptable. 2. Since \(f^{\prime}<0\) on \((-3, -2)\), \(f\) is decreasing there. Since \(f^{\prime}>0\) on \((-2, 4)\), \(f\) is increasing there. Since \(f^{\prime}<0\) on \((4, 5)\), \(f\) is decreasing there.

1. One possible graph has a local minimum at \(x=-2\), an inflection point at \(x=1\), and a local maximum at \(x=4\), with slopes matching the displayed derivative. 2. Decreasing on \((-3, -2)\) and \((4, 5)\); increasing on \((-2, 4)\)

- A zero of the derivative without a sign change does not produce a local extremum. - Use whether \(f^{\prime}\) is increasing or decreasing to determine the concavity of \(f\). - Determine whether \(f^{\prime}\) is increasing or decreasing on each side of \(x=3\). - A horizontal tangent together with a change in concavity indicates a stationary inflection point.

1. The derivative changes from negative to positive at \(x=0\), so a possible graph of \(f\) has a local minimum at \(x=0\). The graph of \(f^{\prime}\) increases up to \(x=1\), decreases from \(x=1\) to \(x=3\), and increases after \(x=3\). Therefore, \(f\) is concave up, then concave down, then concave up, with inflection points at \(x=1\) and \(x=3\). At \(x=3\), \(f^{\prime}=0\), but \(f^{\prime}\) is positive on both sides, so \(f\) remains increasing and has no local extremum there. 2. The graph of \(f^{\prime}\) has a local minimum at \(x=3\). Therefore, \(f^{\prime}\) changes from decreasing to increasing, so the concavity of \(f\) changes. Because \(f^{\prime}(3)=0\), \(f\) has a stationary inflection point at \(x=3\).

1. One possible graph has a local minimum at \(x=0\), an inflection point at \(x=1\), and a stationary inflection point at \(x=3\); it remains increasing through \(x=3\). 2. At \(x=3\), \(f\) has a stationary inflection point: its tangent is horizontal, but the function remains increasing through the point.

- Find the zeros of \(f^{\prime}\). - Use the sign of \(f^{\prime}\) on each interval to determine whether \(f\) increases or decreases. - A sign change at a zero indicates a local extremum. - Turning points of \(f^{\prime}\) indicate changes in the concavity of \(f\).

1. The zeros of \(f^{\prime}\) are \(x=-3\), \(x=-1\), and \(x=2\). These are the x-values where \(f\) has horizontal tangents. The graph of \(f^{\prime}\) has a local maximum at approximately \(x=-2.1\) and a local minimum at approximately \(x=0.8\), so a corresponding graph of \(f\) changes concavity at those x-values. Any vertical shift with these slope and concavity properties is possible. 2. At \(x=-3\), \(f^{\prime}\) changes from negative to positive, so \(f\) has a local minimum. At \(x=-1\), it changes from positive to negative, so \(f\) has a local maximum. At \(x=2\), it changes from negative to positive, so \(f\) has a local minimum. There are no stationary inflection points because the derivative changes sign at every zero.

1. One possible graph has local minima at \(x=-3\) and \(x=2\), a local maximum at \(x=-1\), and inflection points at approximately \(x=-2.1\) and \(x=0.8\). 2. Horizontal tangents occur at \(x=-3\) (local minimum), \(x=-1\) (local maximum), and \(x=2\) (local minimum). There are no stationary inflection points.

- First use the zeros and sign of \(f'\) to organize the monotonicity and extrema. - The original function is concave up where its derivative graph is increasing and concave down where the derivative graph is decreasing. - The vertex of the derivative graph supplies the inflection input of the cubic. - Put the three significant x-values in order before writing the final blueprint.

1. The zeros of \(f'\) are \(x=-1\) and \(x=3\). The derivative is positive for \(x<-1\), negative for \(-1<x<3\), and positive for \(x>3\). Hence \(f\) has a local maximum at \(x=-1\) and a local minimum at \(x=3\). 2. The same sign chart gives increasing intervals \(( -\infty,-1)\) and \((3,\infty)\), with decreasing interval \((-1,3)\). 3. The graph of \(f'\) decreases until its vertex at \(x=1\) and increases afterward. Thus \(f''<0\) for \(x<1\) and \(f''>0\) for \(x>1\). Therefore, \(f\) is concave down on \(( -\infty,1)\), concave up on \((1,\infty)\), and has an inflection point at \(x=1\). 4. Left to right, the graph increases concave down to a local maximum at \(x=-1\), decreases concave down to the inflection input \(x=1\), continues decreasing concave up to a local minimum at \(x=3\), then increases concave up.

a) Local maximum at \(x=-1\); local minimum at \(x=3\). b) Increasing on \(( -\infty,-1)\) and \((3,\infty)\); decreasing on \((-1,3)\). c) Concave down on \(( -\infty,1)\); concave up on \((1,\infty)\); inflection-point x-coordinate \(1\). d) Left-to-right features: maximum at \(-1\), inflection at \(1\), minimum at \(3\), with the monotonicity/concavity sequence stated above.

- A squared factor in \(f'\) gives a zero where the sign of \(f'\) may stay the same. - Differentiate the supplied expression for \(f'\) to get \(f''\); do not integrate. - Split the blueprint at every x-value where either monotonicity or concavity changes.

1. Since \(f'(x)=-(x+1)^2(x-2)\), the squared factor does not change sign at \(x=-1\). Thus \(f'>0\) on both sides of \(-1\). At \(x=2\), \(f'\) changes from positive to negative. 2. Therefore \((-1,0)\) is not a local extremum, while \((2,\tfrac{27}{4})\) is a local maximum. 3. Differentiate the given derivative: \(f''(x)=-3x^2+3=3(1-x^2)\). Hence \(f''=0\) at \(x=-1\) and \(x=1\), and its sign changes at both values. 4. At \(x=-1\), the horizontal tangent together with the concavity change makes \((-1,0)\) a stationary inflection point. At \(x=1\), \((1,4)\) is a nonstationary inflection point. 5. The graph is increasing/concave down on \(( -\infty,-1)\), increasing/concave up on \((-1,1)\), increasing/concave down on \((1,2)\), and decreasing/concave down on \((2,\infty)\).

Stationary inflection \((-1,0)\); nonstationary inflection \((1,4)\); local maximum \((2,\tfrac{27}{4})\). Blueprint: ↑/concave down on \(( -\infty,-1)\), ↑/concave up on \((-1,1)\), ↑/concave down on \((1,2)\), and ↓/concave down on \((2,\infty)\).

- Use all breakpoints where either derivative sign changes. - Endpoint values matter for absolute extrema even though endpoints are not interior critical numbers. - Keep local classification and absolute comparison as separate steps.

1. \((-4,-1)\): increasing and concave down. 2. At \((-1,5)\), \(f'\) changes positive to negative, so there is a local maximum. 3. \((-1,0)\): decreasing and concave down. 4. At \((0,3)\), concavity changes down to up, so this is an inflection point. 5. \((0,2)\): decreasing and concave up. 6. At \((2,1)\), \(f'\) changes negative to positive, so there is a local minimum. 7. \((2,4)\): increasing and concave up. 8. Compare the supplied candidate values at the endpoints and local extrema: \(0,5,1,4\). The absolute maximum is \(5\) at \(x=-1\); the absolute minimum is \(0\) at the endpoint \(x=-4\).

Blueprint: ↑/concave down on \((-4,-1)\); local max \((-1,5)\); ↓/concave down to inflection \((0,3)\); ↓/concave up to local min \((2,1)\); ↑/concave up on \((2,4)\). Absolute max \(5\) at \(x=-1\); absolute min \(0\) at \(x=-4\).

- Compare the signs of the factored first derivative on the intervals separated by its zeros. - Differentiate the given derivative to check where the curve changes concavity. - The supplied function values place each turning or inflection point vertically; no formula for the original function is needed.

1. The zeros of \(f'\) are \(x=-2\) and \(x=1\). The factored derivative is negative on \((-\infty,-2)\), positive on \((-2,1)\), and negative on \((1,\infty)\). 2. Therefore, \((-2,0)\) is a local minimum and \((1,\frac{9}{2})\) is a local maximum. 3. Differentiate: \(f''(x)=-2x-1\). This is positive for \(x<-\frac{1}{2}\) and negative for \(x>-\frac{1}{2}\), so \((-\frac{1}{2},\frac{9}{4})\) is an inflection point. 4. From left to right, the graph decreases and is concave up before \(x=-2\); increases and is concave up between \(-2\) and \(-\frac{1}{2}\); increases and is concave down between \(-\frac{1}{2}\) and \(1\); then decreases and is concave down after \(x=1\).

Local minimum: \((-2,0)\). Inflection point: \((-\frac{1}{2},\frac{9}{4})\). Local maximum: \((1,\frac{9}{2})\). Decreasing and concave up on \((-\infty,-2)\); increasing and concave up on \((-2,-\frac{1}{2})\); increasing and concave down on \((-\frac{1}{2},1)\); decreasing and concave down on \((1,\infty)\).

- Use the ordered simple zeros of \(f'\) to build its sign chart. - Differentiate \(f'\) before building the concavity chart. - Merge the two ordered breakpoint lists only after both charts are correct. - The anchor and stated end behavior place the qualitative graph; no integration is needed.

1. The critical numbers are \(-2,-1,1,2\). Because all four zeros of \(f'\) are simple, its sign alternates. Since \(f'>0\) for large \(|x|\), the sign pattern is \(+,-,+,-,+\). 2. Therefore \(x=-2\) and \(x=1\) are local maxima, while \(x=-1\) and \(x=2\) are local minima. The maximal increasing intervals are \(( -\infty,-2)\), \((-1,1)\), and \((2,\infty)\); the maximal decreasing intervals are \((-2,-1)\) and \((1,2)\). 3. Differentiate: \(f''(x)=4x^3-10x=2x(2x^2-5)\). Thus the inflection inputs are \(-\sqrt{5/2},0,\sqrt{5/2}\). Each zero is simple, so concavity changes at each one. 4. Testing signs gives concave down on \(( -\infty,-\sqrt{5/2})\), concave up on \((-\sqrt{5/2},0)\), concave down on \((0,\sqrt{5/2})\), and concave up on \((\sqrt{5/2},\infty)\). 5. Left-to-right significant inputs are \(-2,-\sqrt{5/2},-1,0,1,\sqrt{5/2},2\). At \(x=0\), the graph passes through \((0,0)\) and changes concavity. Combine this order with the stated end behavior and interval directions.

Local maxima at \(x=-2,1\); local minima at \(x=-1,2\). Increasing on \(( -\infty,-2)\), \((-1,1)\), \((2,\infty)\); decreasing on \((-2,-1)\), \((1,2)\). Inflection inputs \(-\sqrt{5/2},0,\sqrt{5/2}\). Concave down on \(( -\infty,-\sqrt{5/2})\) and \((0,\sqrt{5/2})\); concave up on \((-\sqrt{5/2},0)\) and \((\sqrt{5/2},\infty)\). Significant order: \(-2,-\sqrt{5/2},-1,(0,0),1,\sqrt{5/2},2\), with left end down and right end up.