AP Calculus BC 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 BC.

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. A curve increasing to a local maximum at $x=0$, then decreasing afterward (correct answer)
2. A curve decreasing to a local minimum at $x=0$, then increasing afterward
3. A curve increasing for all $x$ with an inflection point at $x=0$
4. A curve decreasing for all $x$ with a horizontal tangent at $x=0$
5. A curve with no horizontal tangents on $[-3,3]$

Explanation: This question tests graph-derivative reasoning, particularly interpreting the sign of the derivative to determine the function's monotonicity and extrema. Given f'>0 for x<0, f'(0)=0, and f'<0 for x>0, f is increasing left of 0 up to a local maximum at x=0, then decreasing right of 0. This sign pattern around x=0 confirms a change from positive to negative derivative, hallmark of a local maximum. The behavior holds on the closed interval [-3,3] without additional critical points implied. A tempting distractor like choice B fails because it describes a local minimum, which would require f' to change from negative to positive at x=0, opposite the given signs. To check sketches transferably, use a sign chart for f' to map out increasing and decreasing intervals and identify extremum types at zeros.

Question 2

1. A parabola opening up with zeros at $x=-2,2$ only.
2. A cubic crossing the $x$-axis at $x=-2,0,2$ with sign pattern $+,-,+,-$.
3. A cubic crossing the $x$-axis at $x=-2,0,2$ with sign pattern $-,+,-,+$. (correct answer)
4. A line with a single zero at $x=0$ and positive slope.
5. A quartic touching (not crossing) the $x$-axis at $x=-2,0,2$ and nonnegative elsewhere.

Explanation: This question tests your ability to translate between a function's critical points and its derivative's zeros. Since f has local maxima at x = -2 and x = 2, f' must equal zero at these points, and since f has a local minimum at x = 0, f' must also equal zero there. At a local maximum, f' changes from positive to negative (function goes from increasing to decreasing), so f' must cross from positive to negative at x = -2 and x = 2. At a local minimum, f' changes from negative to positive, so f' must cross from negative to positive at x = 0. This gives us the sign pattern: negative for x < -2, positive for -2 < x < 0, negative for 0 < x < 2, and positive for x > 2, which matches choice C. Choice B incorrectly has f' positive for x < -2, which would make f increasing before the first maximum. When sketching derivatives from extrema, always check that the derivative's sign changes match the type of extremum: positive-to-negative for maxima, negative-to-positive for minima.

Question 3

1. $f$ decreases on $(-1,3)$ and increases on $(-\infty,-1)\cup(3,\infty)$, with a local max at $x=-1$ and local min at $x=3$ (correct answer)
2. $f$ increases on $(-1,3)$ and decreases on $(-\infty,-1)\cup(3,\infty)$, with a local min at $x=-1$ and local max at $x=3$
3. $f$ increases for all $x$, with a horizontal tangent at $x=-1$ and $x=3$
4. $f$ decreases for all $x$, with a horizontal tangent only at $x=1$
5. $f$ has local minima at both $x=-1$ and $x=3$

Explanation: This question tests graph-derivative reasoning, focusing on deducing the function's behavior from the shape and zeros of its derivative graph. An upward-opening parabola for f' with zeros at x=-1 and x=3 means f' is negative between these roots and positive outside, indicating f decreases on (-1,3) and increases elsewhere. This creates a local maximum at x=-1 where f switches from increasing to decreasing, and a local minimum at x=3 where it switches back to increasing. The parabolic shape ensures smooth transitions without additional zeros or discontinuities. A tempting distractor like choice B fails because it reverses the intervals, suggesting f increases between -1 and 3 while decreasing outside, which would require f' to be positive inside and negative outside, contradicting an upward-opening parabola. To check sketches transferably, integrate the derivative's sign and shape mentally to predict the function's monotonicity and extrema.

Question 4

1. A cubic crossing the $x$-axis at $x=-2,0,2$ with sign pattern $(-,+,-,+)$ from left to right
2. A cubic crossing the $x$-axis at $x=-2,0,2$ with sign pattern $(+,-,+,-)$ from left to right (correct answer)
3. An upward-opening parabola tangent to the $x$-axis at $x=0$ and positive elsewhere
4. A line with positive slope crossing the $x$-axis only at $x=0$
5. A cubic that is negative for all $x<0$ and positive for all $x>0$ with a single zero at $x=0$

Explanation: This question tests graph-derivative reasoning, specifically identifying the graph of the derivative from the function's critical points and their nature. Local maxima on f occur where f' changes from positive to negative, while local minima occur where f' changes from negative to positive. Here, for local maxima at x=-2 and x=2, f' must switch from positive to negative at those points, and for the local minimum at x=0, it must switch from negative to positive. A cubic polynomial for f' with roots at x=-2,0,2 and the sign pattern (+,-,+,-) aligns with these changes, as it starts positive left of -2, becomes negative between -2 and 0, positive between 0 and 2, and negative right of 2. A tempting distractor like choice A fails because its sign pattern (-,+,-,+) would instead produce local minima at x=-2 and x=2 with a maximum at x=0, reversing the given extremum types. To check sketches transferably, always apply the first derivative test by plotting sign changes around critical points to confirm monotonicity intervals.

Question 5

1. An increasing line crossing the $x$-axis at $x=0$. (correct answer)
2. A decreasing line crossing the $x$-axis at $x=0$.
3. A constant negative horizontal line.
4. A constant positive horizontal line.
5. A parabola opening down with vertex at $x=0$ and value $0$ there.

Explanation: This problem combines monotonicity and concavity information to determine the derivative's graph. Since f decreases on (-2, 0) and increases on (0, 5), we know f'(x) < 0 for -2 < x < 0 and f'(x) > 0 for 0 < x < 5, with f'(0) = 0. Additionally, f is concave up for x > 0, meaning f'' > 0 for x > 0, which tells us f' is increasing for x > 0. An increasing line crossing the x-axis at x = 0 satisfies all conditions: it's negative before x = 0, positive after x = 0, and always increasing. Choice B (decreasing line) would make f concave down everywhere, contradicting the given concavity. To find f' from combined information, first mark where f' = 0 from extrema, then use monotonicity for f's sign pattern and concavity for whether f' increases or decreases.

Question 6

1. $f$ has a local maximum at $x=0$.
2. $f$ has a local minimum at $x=0$.
3. $f$ is increasing for all $x$ and has a horizontal tangent at $x=0$ but no extremum there. (correct answer)
4. $f$ is decreasing for all $x$ and has a cusp at $x=0$.
5. $f$ changes from increasing to decreasing at $x=0$.

Explanation: This question involves a subtle case where f' touches but doesn't cross the x-axis. Since f'(x) > 0 for all x ≠ 0 and f'(0) = 0, the function f is always increasing (never decreasing). The fact that f' only touches the x-axis at x=0 means f' doesn't change sign there—it remains non-negative. This creates a horizontal tangent at x=0 without an extremum, similar to y = x³ at the origin. Choice A incorrectly identifies this as a local maximum, missing that f continues to increase through x=0. When f' touches the x-axis without crossing, check whether f' changes sign: if not, there's no extremum despite the horizontal tangent.

Question 7

1. $f'(x)$ is negative for $x<2$, zero at $x=2$, positive for $x>2$.
2. $f'(x)$ is positive for all $x$ and never equals $0$.
3. $f'(x)$ is positive for $x<2$, zero at $x=2$, negative for $x>2$. (correct answer)
4. $f'(x)$ is zero for all $x<2$ and positive for $x>2$.
5. $f'(x)$ is negative for all $x$ but equals $0$ at $x=2$.

Explanation: This problem asks you to deduce the derivative's behavior from the original function's monotonicity. Since f is increasing on (-∞,2), we know f'(x) ≥ 0 for x < 2, with f'(x) > 0 except possibly at isolated points. Since f is decreasing on (2,∞), we know f'(x) ≤ 0 for x > 2, with f'(x) < 0 except possibly at isolated points. At x=2, where f changes from increasing to decreasing, f must have a local maximum, so f'(2) = 0. Therefore, f'(x) is positive for x < 2, zero at x = 2, and negative for x > 2. Choice A incorrectly has f'(x) negative for x < 2, which would make f decreasing there instead of increasing. Always verify that your derivative's sign matches the function's increasing/decreasing behavior.

Question 8

1. $f'(x)$ changes from positive to negative at $x=0$.
2. $f'(x)$ changes from negative to positive at $x=0$.
3. $f'(x)$ is positive for $x\ne0$ near $0$ and $f'(0)=0$. (correct answer)
4. $f'(x)$ is negative for $x\ne0$ near $0$ and $f'(0)=0$.
5. $f'(x)$ is undefined at $x=0$.

Explanation: This question tests graph-derivative reasoning by exploring how f' behaves around a horizontal tangent while maintaining monotonicity in f. A horizontal tangent at x = 0 means f'(0) = 0, and f increasing on both sides requires f' ≥ 0 nearby with no sign change. Thus, f' can be positive for x ≠ 0 near 0 and zero at 0, allowing f to increase through the point without an extremum. This is possible even if f' approaches 0 from positive values on both sides. A tempting distractor like choice A fails because changing from positive to negative would create a local maximum in f, contradicting the increasing on both sides. To check sketches, confirm that no sign change in f' around a zero maintains monotonicity in f, while verifying tangent conditions.

Question 9

1. A graph increasing for $x<1$ and decreasing for $x>1$, with a local maximum at $x=1$ (correct answer)
2. A graph decreasing for $x<1$ and increasing for $x>1$, with a local minimum at $x=1$
3. A graph increasing for all $x$, with no critical points
4. A graph decreasing for all $x$, with no critical points
5. A constant positive horizontal line

Explanation: This question tests graph-derivative reasoning, specifically recognizing how the concavity of the function relates to the monotonicity of its derivative. Concave up on (-∞,1) means f''>0 there, so f' is increasing; concave down on (1,∞) means f''<0, so f' is decreasing, implying f' has a local maximum at x=1. An inflection point at x=1 occurs where f'' changes sign, consistent with f' reaching a peak. This setup allows f' to be positive or negative but must reflect the monotonicity shift. A tempting distractor like choice B fails because it would require f' to have a local minimum at x=1, which would correspond to f'' changing from positive to negative in reverse, not matching the concavity. To check sketches transferably, analyze the sign of the second derivative to determine where the first derivative increases or decreases.

Question 10

1. $f$ has local minima at $x=-1$ and $x=2$ and a local maximum between them.
2. $f$ has a local maximum at $x=-1$ and a local minimum at $x=2$; increasing on $(-1,2)$. (correct answer)
3. $f$ is increasing on $(-\infty,-1)$, decreasing on $(-1,2)$, and increasing on $(2,\infty)$.
4. $f$ is decreasing on $(-\infty,-1)$, increasing on $(-1,2)$, and decreasing on $(2,\infty)$.
5. $f$ has no relative extrema because $f'(x)$ is zero only twice.

Explanation: This question tests your ability to translate between a derivative graph and the original function's behavior. Since f'(x) crosses the x-axis at x=-1 and x=2, these are critical points where f has horizontal tangents. When f'(x) > 0 (between -1 and 2), the function f is increasing; when f'(x) < 0 (outside this interval), f is decreasing. This means f decreases until x=-1, increases from x=-1 to x=2, then decreases again, creating a local minimum at x=-1 and a local maximum at x=2. Choice C incorrectly reverses the increasing/decreasing intervals by confusing the derivative's sign with the function's behavior. To verify your sketch, check that f' changes from negative to positive at local minima and from positive to negative at local maxima.

Question 11

1. A concave-up parabola with vertex at $x=2$
2. A concave-down parabola with vertex at $x=2$ (correct answer)
3. A cubic with inflection point at $x=2$ and increasing for all $x$
4. A constant function
5. A V-shaped absolute value graph with corner at $x=2$

Explanation: This question tests graph-derivative reasoning, linking the derivative's linear form to the function's quadratic shape and concavity. A line for f' with negative slope crossing at x=2 means f' changes from positive to negative at x=2, so f has a maximum there, and constant negative f'' implies concave down everywhere. Integrating yields a downward-opening parabola with vertex at x=2, matching the maximum and concavity. The linear f' ensures no other critical points or inflection points. A tempting distractor like choice A fails because a concave-up parabola would require positive f'', but the negative slope of f' gives negative f''. To check sketches transferably, compute the second derivative from the slope of the first to verify concavity direction.

Question 12

1. $f'(x)$ is zero at $x=-3$ and has a local extremum at $x=1$ (correct answer)
2. $f'(x)$ is zero at $x=1$ and has a local extremum at $x=-3$
3. $f'(x)$ is never zero and is constant on both sides of $x=1$
4. $f'(x)$ has a jump discontinuity at $x=-3$
5. $f'(x)$ is zero at $x=-3$ and changes sign from positive to negative there

Explanation: This question tests graph-derivative reasoning, connecting the function's extrema and inflections to features in the derivative graph. A local minimum at x=-3 requires f'(-3)=0 with f' changing from negative to positive there. An inflection at x=1 means f''(1)=0 with f'' changing sign, which translates to f' having a local extremum at x=1, as local extrema of f' occur where its derivative f'' changes sign at zero. This pairs the zero and the extremum appropriately in f'. A tempting distractor like choice B fails because it reverses the features, placing the zero at the inflection point instead of the extremum. To check sketches transferably, remember that inflections in f correspond to extrema in f', while extrema in f correspond to zeros in f'.

Question 13

1. $f'(x)$ is positive on $(-1,3)$ and negative elsewhere, with zeros at $x=-1,3$.
2. $f'(x)$ is negative on $(-1,3)$ and positive elsewhere, with zeros at $x=-1,3$. (correct answer)
3. $f'(x)$ is negative for all $x$, with zeros at $x=-1,3$.
4. $f'(x)$ is positive for all $x$, with zeros at $x=-1,3$.
5. $f'(x)$ changes sign only at $x=-1$ and never at $x=3$.

Explanation: This question tests graph-derivative reasoning by relating critical points and monotonicity intervals of f to the signs and zeros of f'. Critical points occur where f' = 0, and decreasing intervals of f correspond to f' < 0, while increasing intervals mean f' > 0. The decrease on (-1,3) requires f' < 0 there, increases elsewhere require f' > 0 outside, and critical points at -1 and 3 imply f' zeros there with sign changes. This results in f' positive for x < -1, negative on (-1,3), and positive for x > 3, matching the monotonicity. A tempting distractor like choice A fails because it has f' positive on (-1,3), which would make f increasing there instead of decreasing. To check sketches, ensure zeros of f' align with critical points of f and that the signs of f' match the increasing or decreasing behavior in the specified intervals.

Question 14

1. $f'(x)$ decreases on $(0,2)$ and increases on $(2,5)$, with $f'(2)=3$.
2. $f'(x)$ increases on $(0,2)$ and decreases on $(2,5)$, with $f'(2)=3$. (correct answer)
3. $f'(x)$ is constant on $(0,5)$ and equals $3$.
4. $f'(x)$ increases on $(0,2)$ and increases on $(2,5)$, with $f'(2)=3$.
5. $f'(x)$ decreases on $(0,2)$ and decreases on $(2,5)$, with $f'(2)=3$.

Explanation: This question tests graph-derivative reasoning by linking concavity changes in f to monotonicity shifts in f'. Concave up on (0,2) means f'' > 0, so f' is increasing there, and concave down on (2,5) means f'' < 0, so f' is decreasing there. The change at x = 2 implies a local maximum in f' at 2, consistent with increasing before and decreasing after. The condition f'(2) = 3 specifies a positive value at that point, which fits without requiring a zero. A tempting distractor like choice A fails because decreasing on (0,2) would imply f'' < 0 and concave down, opposite to the given concave up. To check sketches, align increasing segments of f' with concave up regions of f and decreasing segments with concave down, noting points of change for potential extrema in f'.

Question 15

1. A line through the origin with nonzero slope
2. A parabola tangent to the $x$-axis at $x=0$ and opening upward (correct answer)
3. A cubic that crosses the $x$-axis at $x=0$ with nonzero slope
4. A curve that passes through $(0,0)$ with a vertical tangent
5. A curve with a hole at $x=0$

Explanation: This question tests graph-derivative reasoning, particularly identifying function graphs with specific tangent and sign properties at a point. A horizontal tangent at x=0 means f'(0)=0, and crossing the x-axis without changing sign implies f touches zero at x=0 but remains non-negative or non-positive around it. An upward-opening parabola tangent at x=0 fits, as its vertex provides the horizontal tangent and it stays positive elsewhere without sign change. This setup ensures f(0)=0 and f'(0)=0 without crossing to negative values. A tempting distractor like choice A fails because a line with nonzero slope crosses the x-axis and changes sign, violating the no-sign-change condition. To check sketches transferably, evaluate the function's values around the point to confirm tangency and sign consistency.

Question 16

1. $f'(x)$ is negative on $(-\infty,-2)$, positive on $(-2,1)$, negative on $(1,\infty)$ with zeros at $-2,1$.
2. $f'(x)$ is positive on $(-\infty,-2)$, negative on $(-2,1)$, positive on $(1,\infty)$ with zeros at $-2,1$. (correct answer)
3. $f'(x)$ is positive for all $x$ and equals $0$ only at $x=1$.
4. $f'(x)$ is negative for all $x$ and equals $0$ at $x=-2$ and $x=1$.
5. $f'(x)$ is zero for all $x$ between $-2$ and $1$.

Explanation: This problem requires translating the function's behavior into the derivative's sign pattern. Horizontal tangents at x=-2 and x=1 mean f'(-2) = 0 and f'(1) = 0. Since f increases on (-∞,-2), we have f'(x) > 0 for x < -2. Since f decreases on (-2,1), we have f'(x) < 0 for -2 < x < 1. Since f increases on (1,∞), we have f'(x) > 0 for x > 1. This creates the pattern: positive, zero at -2, negative, zero at 1, positive. Choice A incorrectly has f' negative for x < -2 and x > 1, which would make f decrease in those regions instead of increase. To verify derivative sketches, ensure the sign of f' matches whether f is increasing (positive) or decreasing (negative) in each interval.

Question 17

1. $f'(x)$ decreases on $(-1,0)$ and increases on $(0,3)$, with $f'(1)=0$.
2. $f'(x)$ increases on $(-1,0)$ and decreases on $(0,3)$, with $f'(1)=0$. (correct answer)
3. $f'(x)$ increases on $(-1,0)$ and increases on $(0,3)$, with $f'(1)=0$.
4. $f'(x)$ decreases on $(-1,0)$ and decreases on $(0,3)$, with $f'(1)=0$.
5. $f'(x)$ has a jump discontinuity at $x=0$ and $f'(1)=0$.

Explanation: This question tests graph-derivative reasoning by connecting concavity of f to the monotonicity of f'. Concave up regions in f correspond to f'' > 0, meaning f' is increasing there, while concave down regions mean f'' < 0, so f' is decreasing. For f concave up on (-1,0), f' increases on that interval, and concave down on (0,3), f' decreases there, with the change at x = 0 implying a local extremum in f' at 0. The condition f'(1) = 0 adds a zero in f' within the concave down interval, which is consistent as long as the overall monotonicity of f' aligns with the concavity. A tempting distractor like choice A fails because it has f' decreasing on (-1,0), which would imply f'' < 0 and concave down, opposite to the given concave up. To check sketches, match increasing intervals of f' with concave up regions of f and decreasing intervals of f' with concave down regions, verifying any specified values or points.

Question 18

1. $f$ has a local minimum at $x=0$ and a local maximum at $x=4$.
2. $f$ has local maxima at both $x=0$ and $x=4$.
3. $f$ has a local maximum at $x=0$ and a local minimum at $x=4$. (correct answer)
4. $f$ is increasing on all real numbers.
5. $f$ is decreasing on all real numbers.

Explanation: This question tests interpreting a derivative's sign to determine a function's behavior. When f'(x) > 0, the function f is increasing, and when f'(x) < 0, f is decreasing. Given that f' is positive for x < 0, f increases on (-∞, 0); since f' is negative for 0 < x < 4, f decreases on (0, 4); and since f' is positive for x > 4, f increases on (4, ∞). At x = 0, f changes from increasing to decreasing, creating a local maximum. At x = 4, f changes from decreasing to increasing, creating a local minimum. Choice A incorrectly reverses these extrema, suggesting a minimum at x = 0. When reading derivative signs, remember: positive derivative means increasing function, negative derivative means decreasing function, and sign changes indicate extrema.

Question 19

1. A graph of $f'$ that is increasing for $x<0$ and decreasing for $x>0$, with a local maximum at $x=0$. (correct answer)
2. A graph of $f'$ that is decreasing for $x<0$ and increasing for $x>0$, with a local minimum at $x=0$.
3. A graph of $f'$ that is increasing for all $x$, with an inflection at $x=0$.
4. A graph of $f'$ that is constant for $x<0$ and increasing for $x>0$.
5. A graph of $f'$ that is decreasing for all $x$, with a horizontal tangent at $x=0$.

Explanation: This question requires understanding how the second derivative affects the first derivative's graph. Since f''(x) represents the derivative of f'(x), when f''(x) > 0 for x < 0, the function f' is increasing on that interval. Similarly, when f''(x) < 0 for x > 0, the function f' is decreasing on that interval. At x = 0, where f''(0) = 0 and f'' changes from positive to negative, f' has a local maximum because it transitions from increasing to decreasing. Choice B might seem plausible if you confuse the roles of f' and f'', but remember that f'' tells us about the monotonicity of f', not f itself. When sketching derivatives, always remember that the second derivative controls whether the first derivative is increasing or decreasing.

Question 20

1. Positive and decreasing on $(0,2)$, then positive and increasing on $(2,5)$, with a minimum at $x=2$. (correct answer)
2. Negative and decreasing on $(0,2)$, then negative and increasing on $(2,5)$, with a minimum at $x=2$.
3. Positive and increasing on $(0,2)$, then positive and decreasing on $(2,5)$, with a maximum at $x=2$.
4. Zero on $(0,2)$, then positive and increasing on $(2,5)$.
5. Positive and constant on $(0,5)$.

Explanation: This problem connects concavity of f with the behavior of f'. When f is concave down, its derivative f' is decreasing; when f is concave up, f' is increasing. Since f is increasing throughout (0, 5), we know f' must be positive on this entire interval. On (0, 2), f is increasing and concave down, so f' is positive and decreasing. On (2, 5), f is increasing and concave up, so f' is positive and increasing. At x = 2, where concavity changes, f' has a local minimum (switching from decreasing to increasing). Choice C might seem reasonable since it shows f' positive throughout with a maximum at x = 2, but this would make f concave down on (2, 5), contradicting the given information. To sketch f' from f, remember: increasing f means positive f', and concavity determines whether f' is increasing or decreasing.