AP Calculus AB Quiz: Sketching Graphs Of Functions And Derivatives

What this quiz covers

This quiz focuses on Sketching Graphs Of Functions And Derivatives, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $f''(x)>0$ for $x<1$, $f''(1)=0$, and $f''(x)<0$ for $x>1$. (correct answer)
2. $f''(x)<0$ for $x<1$, $f''(1)=0$, and $f''(x)>0$ for $x>1$.
3. $f''(x)=0$ for all $x$.
4. $f''(x)>0$ for all $x$ and never equals $0$.
5. $f''$ is undefined at $x=1$ and negative on both sides.

Explanation: This question tests graph-derivative reasoning, linking the concavity of f to the signs and zeros of f''. Concave up intervals correspond to f'' > 0, and concave down to f'' < 0. An inflection point at x=1 means f'' changes sign there, from positive to negative for a shift from concave up to down. Thus, f'' > 0 for x<1, =0 at x=1, and <0 for x>1. A tempting distractor is choice B, which flips the signs and would reverse the concavity change. A transferable sketch-checking strategy is to plot the sign changes in f'' and verify they match the desired concavity shifts on f's graph.

Question 2

1. A curve above the $x$-axis that is increasing on $(0,5)$.
2. A curve below the $x$-axis that is decreasing on $(0,5)$.
3. A curve above the $x$-axis that is decreasing on $(0,5)$. (correct answer)
4. A curve that crosses the $x$-axis within $(0,5)$ from negative to positive.
5. A horizontal line at $0$ on $(0,5)$.

Explanation: This question tests graph-derivative reasoning by connecting the function's monotonicity and concavity to its derivative's behavior. On (0,5), f increasing means f' > 0, so above the x-axis, and concave down means f'' < 0, so f' is decreasing. Thus, f' is positive and decreasing on (0,5), matching choice C. This ensures the slope starts higher and reduces while staying positive, consistent with slowing increase and downward curvature. A tempting distractor is A, positive and increasing, but that would imply f'' > 0, making f concave up instead of down. For checking sketches, always link concavity to whether f' is increasing or decreasing, and confirm signs match monotonicity.

Question 3

1. $f'$ is negative for $x<-1$, positive on $(-1,2)$, negative for $x>2$, with zeros at $x=-1,2$.
2. $f'$ is positive for $x<-1$, negative on $(-1,2)$, positive for $x>2$, with zeros at $x=-1,2$. (correct answer)
3. $f'$ is positive for all $x$ and never crosses the $x$-axis.
4. $f'$ has zeros at $x=-1,2$ but is negative on both sides of each zero.
5. $f'$ has a single zero at $x=2$ and is positive for $x<2$, negative for $x>2$.

Explanation: This question tests graph-derivative reasoning, focusing on how the monotonicity of a function $f$ relates to the sign of its derivative $f'$. When $f$ is increasing on an interval, $f'$ is positive there, and when $f$ is decreasing, $f'$ is negative. The points where $f$ changes from increasing to decreasing or vice versa are critical points where $f'$ equals zero, such as at $x=-1$ and $x=2$ in this case. Thus, $f'$ should be positive for $x<-1$, negative on $(-1,2)$, and positive for $x>2$, matching the given behavior of $f$. A tempting distractor is choice A, which flips the signs and would incorrectly suggest $f$ is decreasing where it should be increasing. A transferable sketch-checking strategy is to select test points in each interval and verify if the sign of $f'$ aligns with the expected increasing or decreasing trend of $f$.

Question 4

1. $f'$ crosses from negative to positive at $x=-3$ and from positive to negative at $x=1$.
2. $f'$ is zero at $x=-3,1$ and positive everywhere else.
3. $f'$ crosses from positive to negative at $x=-3$ and from negative to positive at $x=1$. (correct answer)
4. $f'$ has no zeros and stays negative for all $x$.
5. $f'$ is zero only at $x=-3$ and changes sign there.

Explanation: This question tests graph-derivative reasoning, specifically identifying sign changes in f' at local extrema of f. A local maximum occurs where f' changes from positive to negative, as the function shifts from increasing to decreasing. A local minimum occurs where f' changes from negative to positive, marking a shift from decreasing to increasing. Here, f' should cross from positive to negative at x=-3 (for the maximum) and from negative to positive at x=1 (for the minimum). A tempting distractor is choice A, which reverses the sign changes and would swap the maximum and minimum locations. A transferable sketch-checking strategy is to use the first derivative test by checking sign changes around critical points to confirm the type of extremum.

Question 5

1. A curve always above the $x$-axis except touching it once at $x=-1$ and once at $x=2$.
2. A curve crossing the $x$-axis at $x=-1$ and $x=2$, positive on $(-\infty,-1)$ and $(2,\infty)$, negative on $(-1,2)$. (correct answer)
3. A curve crossing the $x$-axis at $x=-1$ and $x=2$, negative on $(-\infty,-1)$ and $(2,\infty)$, positive on $(-1,2)$.
4. A line with negative slope crossing the $x$-axis at $x=2$ only.
5. A curve that is negative for all $x$.

Explanation: This question tests graph-derivative reasoning by matching the derivative's graph to the function's monotonicity intervals. The function f increasing on (-∞, -1), decreasing on (-1, 2), and increasing on (2, ∞) implies f' > 0 on (-∞, -1) and (2, ∞), f' < 0 on (-1, 2), and f' = 0 at x = -1 and x = 2 assuming critical points there. Thus, f' crosses the x-axis at these points, with positive signs outside and negative between, matching choice B. This sign pattern ensures the correct changes in f's direction at the specified points. A tempting distractor is C, which has negative outside and positive between, but that would make f decreasing outside and increasing between, reversing the given intervals. A transferable strategy for checking sketches is to plot sign charts for f' based on f's behavior and verify that zeros of f' correspond to transitions in monotonicity.

Question 6

1. Local minima at $x=-2$ and $x=2$, local maximum at $x=0$.
2. Local maxima at $x=-2$ and $x=2$, local minimum at $x=0$.
3. Local maximum at $x=-2$, local minimum at $x=0$, local maximum at $x=2$. (correct answer)
4. Local minimum at $x=-2$, local maximum at $x=0$, local minimum at $x=2$.
5. No local extrema because $f'$ is zero only at isolated points.

Explanation: This problem requires interpreting f' sign changes to identify extrema of f. The sign pattern +,-,+,- means f' is positive before -2, negative between -2 and 0, positive between 0 and 2, and negative after 2. When f' changes from + to -, f has a local maximum; when f' changes from - to +, f has a local minimum. At x = -2: f' goes from + to -, so f has a local max. At x = 0: f' goes from - to +, so f has a local min. At x = 2: f' goes from + to -, so f has a local max. Choice D incorrectly identifies the extrema types. Remember: + to - crossing means local max; - to + crossing means local min.

Question 7

1. $f$ decreases on $(-2,4)$ and increases on $(-\infty,-2)$ and $(4,\infty)$. (correct answer)
2. $f$ increases on $(-2,4)$ and decreases on $(-\infty,-2)$ and $(4,\infty)$.
3. $f$ increases for all $x$.
4. $f$ decreases for all $x$.
5. $f$ has a local maximum at $x=1$ only.

Explanation: This problem tests your understanding of how the sign of the derivative determines where a function increases or decreases. A parabola opening upward with zeros at x = -2 and x = 4 has its vertex (minimum point) at x = 1, and is negative between its roots and positive outside them. Since f'(x) < 0 on (-2,4), the function f decreases on this interval. Since f'(x) > 0 on (-∞,-2) and (4,∞), the function f increases on these intervals. Option B reverses all the monotonicity, claiming f increases where f' < 0, which violates the fundamental relationship between a function and its derivative. The correct answer properly identifies that f decreases on (-2,4) and increases on (-∞,-2) and (4,∞).

Question 8

1. $f$ is concave down everywhere and has a local maximum at $x=0$.
2. $f$ is concave up everywhere and has a local minimum at $x=0$. (correct answer)
3. $f$ is concave up everywhere and has a local maximum at $x=0$.
4. $f$ is concave down everywhere and has a local minimum at $x=0$.
5. $f$ changes concavity at $x=0$ and has no critical points.

Explanation: This question tests graph-derivative reasoning, connecting the behavior of f' to the concavity and extrema of f. If f' is increasing everywhere, then f'' > 0, meaning f is concave up for all x. With f'(0)=0 and f' increasing, f' is negative before x=0 and positive after, indicating a local minimum at x=0. This combination results in f being concave up with a local minimum at x=0. A tempting distractor is choice C, which suggests a local maximum, but that would require f' to be decreasing, not increasing. A transferable sketch-checking strategy is to analyze the monotonicity of f' to determine concavity and use sign changes in f' to identify minima or maxima.

Question 9

1. $f'(x)>0$ and $f''(x)>0$.
2. $f'(x)<0$ and $f''(x)>0$. (correct answer)
3. $f'(x)>0$ and $f''(x)<0$.
4. $f'(x)<0$ and $f''(x)<0$.
5. $f'(x)=0$ and $f''(x)>0$ for all $x$.

Explanation: This question tests graph-derivative reasoning, connecting the signs of f' and f'' to the monotonicity and concavity of f. For f decreasing on (0,4), f' < 0 there, indicating a negative slope. For concave up, f'' > 0, meaning the graph curves upward like a U-shape while overall descending. This combination shows f decreasing but with slopes becoming less negative over the interval. A tempting distractor is choice D, which has f'' < 0, but that would make f concave down, not up. A transferable sketch-checking strategy is to combine sign charts for f' and f'' to predict both direction and curvature on f's graph.

Question 10

1. $f$ has a local maximum at $x=-2$ and a local minimum at $x=3$.
2. $f$ has local minima at both $x=-2$ and $x=3$.
3. $f$ is increasing everywhere and has no local extrema at $x=-2$ or $x=3$. (correct answer)
4. $f$ is decreasing on $(-2,3)$ and increasing elsewhere.
5. $f$ has local maxima at both $x=-2$ and $x=3$.

Explanation: This question tests graph-derivative reasoning, examining how zeros of f' without sign changes affect f's monotonicity. With f' = 0 at x=-2 and x=3 but > 0 elsewhere, f' never changes sign and remains non-negative. This means f is increasing everywhere, and the zeros do not create local extrema since there's no shift from increasing to decreasing or vice versa. Such points might be horizontal inflections but do not halt the overall increasing trend. A tempting distractor is choice A, which assumes sign changes for extrema, but here f' stays positive around the zeros. A transferable sketch-checking strategy is to check for sign changes around f''s zeros; no change means no extremum, even if f'=0.

Question 11

1. $f$ has local minima at $x=-1$ and $x=2$.
2. $f$ has a local maximum at $x=-1$ and a local minimum at $x=2$. (correct answer)
3. $f$ has local maxima at $x=-1$ and $x=2$.
4. $f$ has a local minimum at $x=-1$ and a local maximum at $x=2$.
5. $f$ has no local extrema because $f'$ changes sign only once.

Explanation: This question tests graph-derivative reasoning, using the sign chart of f' to determine the extrema of f. Where f' > 0 on (-5,-1) and (2,5), f is increasing, and where < 0 on (-1,2), f is decreasing. This creates a local maximum at x=-1 (shift from increasing to decreasing) and a local minimum at x=2 (shift from decreasing to increasing). The signs indicate f rises to a peak at -1, falls to a valley at 2, then rises again. A tempting distractor is choice D, which swaps the extrema types but mismatches the sign changes. A transferable sketch-checking strategy is to mark critical points from f''s zeros and use adjacent signs to classify maxima or minima.

Question 12

1. $f'$ is positive and increasing on $(0,5)$.
2. $f'$ is negative and decreasing on $(0,5)$.
3. $f'$ is positive and decreasing on $(0,5)$. (correct answer)
4. $f'$ is negative and increasing on $(0,5)$.
5. $f'$ is zero on $(0,5)$.

Explanation: This question combines two conditions about f to determine properties of f'. If f is increasing on (0,5), then f' > 0 throughout this interval. If f is concave down on (0,5), then f' must be decreasing on this interval. Combining these: f' is both positive and decreasing on (0,5). This could be a graph starting at some positive value and decreasing toward (but not reaching) zero. Choice A incorrectly claims f' is increasing, which would make f concave up, not down. To verify: increasing f means positive f'; concave down f means decreasing f'.

Question 13

1. $f$ has exactly one critical point, which is a point of inflection.
2. $f$ has two critical points; it increases, then decreases, then increases. (correct answer)
3. $f$ has two critical points; it decreases, then increases, then decreases.
4. $f$ is increasing for all $x$.
5. $f$ is decreasing for all $x$.

Explanation: This question tests graph-derivative reasoning by deducing the function's critical points and monotonicity from the derivative's graph. An upward-opening parabola for f' with vertex below the x-axis and two distinct zeros means f' > 0 outside the zeros, < 0 between, crossing from positive to negative at the left zero and negative to positive at the right. This implies f increases left, decreases middle (local max at left zero), and increases right (local min at right zero), giving two critical points with the pattern increases-decreases-increases. The vertex below ensures the negative region exists for the middle decrease. A tempting distractor is C, suggesting decreases-increases-decreases, but that would require f' negative-positive-negative, matching a downward-opening parabola instead. To check sketches effectively, trace the sign of f' across its zeros to map out f's increasing and decreasing intervals.

Question 14

1. $f$ increases on $(-2,0)$, decreases on $(0,4)$, with a local maximum at $x=0$.
2. $f$ decreases on $(-2,0)$, increases on $(0,4)$, with a local minimum at $x=0$. (correct answer)
3. $f$ increases on both intervals and has no extremum at $x=0$.
4. $f$ decreases on both intervals and has no extremum at $x=0$.
5. $f$ has a horizontal tangent at $x=0$ but continues decreasing for $x>0$.

Explanation: This question tests graph-derivative reasoning, relating the signs and zeros of f' to the monotonicity and extrema of f. Where f' is negative on (-2,0), f is decreasing, and where positive on (0,4), f is increasing. The zero of f' at x=0 indicates a critical point, with the sign change from negative to positive signaling a local minimum. Thus, f decreases to a minimum at x=0 and then increases. A tempting distractor is choice A, which reverses the intervals and would imply a local maximum instead of a minimum. A transferable sketch-checking strategy is to trace the sign chart of f' and ensure it produces the correct increasing/decreasing patterns on f's graph.

Question 15

1. A function with a local max at $x=-2$ and a local min at $x=1$.
2. A function with local minima at both $x=-2$ and $x=1$.
3. A function with a horizontal tangent at $x=-2$ but no extremum there, and a local extremum at $x=1$. (correct answer)
4. A function with no critical points.
5. A function with a local extremum at $x=-2$ but no extremum at $x=1$.

Explanation: This question tests understanding of how f' crossing versus touching the x-axis affects f. When f' crosses the x-axis at x = 1, it changes sign, so f has a local extremum there. When f' only touches (is tangent to) the x-axis at x = -2, it doesn't change sign, so f has a horizontal tangent but no extremum at that point. This could occur when f'(x) = (x + 2)²(x - 1), making f' ≥ 0 near x = -2 with equality only at x = -2. Choice A incorrectly assumes both points must be extrema. The key insight is that sign changes in f' create extrema in f, while tangencies without sign change create horizontal tangents without extrema.

Question 16

1. $f'$ is positive and increasing on $(0,5)$.
2. $f'$ is negative and decreasing on $(0,5)$.
3. $f'$ is positive and decreasing on $(0,5)$. (correct answer)
4. $f'$ is negative and increasing on $(0,5)$.
5. $f'$ is zero everywhere on $(0,5)$.

Explanation: This problem requires understanding how both monotonicity and concavity of f constrain f'. Since f is increasing on (0, 5), we must have f' > 0 throughout this interval. Since f is concave down on (0, 5), we have f'' < 0, which means f' is decreasing on this interval. Therefore, f' must be positive and decreasing on (0, 5). Choice A incorrectly claims f' is increasing, which would make f concave up rather than down. To verify such problems, remember that f' > 0 means f increases, while f'' < 0 (equivalently, f' decreasing) means f is concave down.

Question 17

1. $f$ has local minima at both $x=-1$ and $x=2$.
2. $f$ has a local maximum at $x=-1$ and a local minimum at $x=2$. (correct answer)
3. $f$ has a local minimum at $x=-1$ and a local maximum at $x=2$.
4. $f$ is increasing on $(-1,2)$.
5. $f$ has no local extrema.

Explanation: This problem tests your ability to translate the derivative's sign changes into the original function's behavior. When f' crosses the x-axis, f has a critical point. Since f' crosses at x = -1 and x = 2, these are critical points of f. The sign of f' determines whether each critical point is a maximum or minimum: if f' changes from positive to negative, f has a local maximum; if f' changes from negative to positive, f has a local minimum. Since f' is negative between -1 and 2, it must be positive before -1 and positive after 2, giving the pattern (+,-,+). Option C reverses the extrema types. The correct answer identifies that f has a local maximum at x = -1 (where f' goes from + to -) and a local minimum at x = 2 (where f' goes from - to +).

Question 18

1. $f$ increases on $(-\infty,1)$, decreases on $(1,4)$, then increases on $(4,\infty)$.
2. $f$ decreases on $(-\infty,1)$, increases on $(1,4)$, then decreases on $(4,\infty)$, with a local min at $1$ and local max at $4$. (correct answer)
3. $f$ is constant on $(1,4)$ and decreasing elsewhere.
4. $f$ increases for all $x$ and has no critical points.
5. $f$ decreases for all $x$ and has no critical points.

Explanation: This question tests graph-derivative reasoning by translating the derivative's sign and zeros to the function's behavior. f' negative for x<1, positive on (1,4), negative for x>4, crossing at 1 and 4, means f decreases to 1, increases to 4, then decreases after. This gives a local min at 1 (f' negative to positive) and local max at 4 (positive to negative). The pattern matches choice B's description of monotonicity and extrema. A tempting distractor is A, which swaps signs to positive-negative-positive, leading to increases-decreases-increases and max at 1, min at 4 instead. To check sketches, construct interval sign charts for f' and ensure transitions at zeros correspond to correct min/max types.

Question 19

1. A local maximum at $x=3$.
2. A local minimum at $x=3$.
3. A horizontal tangent at $x=3$ but no local extremum there. (correct answer)
4. A discontinuity at $x=3$.
5. A vertical tangent at $x=3$.

Explanation: This problem explores a special case where the derivative touches but doesn't cross the x-axis. When f' touches the x-axis at x = 3 and stays positive on both sides, it means f'(3) = 0 but f' doesn't change sign. Since f' remains positive near x = 3, the function f continues to increase through this point. The horizontal tangent at x = 3 (due to f'(3) = 0) combined with f increasing on both sides means there's neither a local maximum nor minimum—just a horizontal tangent. Choice B incorrectly identifies this as a local minimum, which would require f' to change from negative to positive. To identify extrema, always check whether f' actually changes sign at critical points; a mere touch of the x-axis without sign change indicates a horizontal tangent with no extremum.

Question 20

1. Crosses the $x$-axis at $x=-2$ and $x=1$, positive between them.
2. Touches the $x$-axis at $x=-2$ only, negative elsewhere.
3. Crosses the $x$-axis at $x=-2$ and $x=1$, negative between them. (correct answer)
4. A horizontal line at $y=0$.
5. Crosses the $x$-axis once at $x=0$ from negative to positive.

Explanation: This problem combines information about critical points and monotonicity to determine the derivative's behavior. Since f has horizontal tangents at x = -2 and x = 1, we know f'(-2) = 0 and f'(1) = 0, so f' must touch or cross the x-axis at these points. The fact that f decreases on (-2, 1) tells us that f' < 0 throughout this interval. For f' to be zero at the endpoints but negative between them, it must cross the x-axis at both points, being positive outside the interval (-2, 1) and negative inside. Choice A incorrectly shows f' as positive between the zeros, which would make f increasing there. When sketching f' from information about f, ensure the sign of f' between critical points matches the given monotonicity of f.