Sketching Graphs of Functions and Derivatives - AP Calculus AB

A function $f$ has a local maximum at $x=-2$ and a local minimum at $x=1$ ; which could be $f'$ ?
Answer: $f'$ crosses the $x$ -axis at $x=-2$ and $x=1$ , positive for $x<-2$ , negative for $-2<x<1$ , positive for $x>1$ . This problem requires translating local extrema of f into sign changes of f'. At a local maximum, f' changes from positive to negative; at a local minimum, f' changes from negative to positive. Since f has a local max at x = -2, f' must go from positive to negative there, crossing zero. Since f has a local min at x = 1, f' must go from negative to positive there, also crossing zero. Between these points (-2 < x < 1), f' must be negative for f to decrease from the max to the min. Choice A incorrectly omits the zero crossings at the extrema. Remember: local extrema of f correspond to sign-changing zeros of f'.
The graph of $f$ is a smooth cubic with a local maximum left of the origin and a local minimum right of the origin. Which graph could be $f'$ ?
Answer: An upward-opening parabola crossing the $x$ -axis twice, negative between the zeros and positive outside them. This question tests graph-derivative reasoning by requiring identification of the derivative's graph based on the function's behavior. A smooth cubic function with a local maximum left of the origin and a local minimum right of the origin is increasing to the max, decreasing to the min, and increasing afterward. Thus, its derivative f' must be positive left of the max, zero at the max, negative between extrema, zero at the min, and positive afterward. Since f is cubic, f' is quadratic, and the described sign pattern matches an upward-opening parabola crossing the x-axis twice, negative between the zeros and positive outside. A tempting distractor is B, a downward-opening parabola, but it would show positive between zeros and negative outside, corresponding to a local min first and max second, reversing the given extrema order. To check such sketches reliably, always map the function's increasing/decreasing intervals to the derivative's sign and confirm zero crossings align with extrema locations.
The graph of $f'$ is above the $x$ -axis on $( -5,-1)$ and $(2,5)$ , but below on $(-1,2)$ ; which could be $f$ ?
Answer: $f$ has a local maximum at $x=-1$ and a local minimum at $x=2$ . 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.
A graph of $f'$ crosses the $x$ -axis at $x=-1$ and $x=2$ and is negative between them; which could be $f$ ?
Answer: Local maximum at $x=-1$ and local minimum at $x=2$ . This problem tests your understanding of how sign changes in the derivative create extrema in the original function. Since f' crosses the x-axis at x = -1 and x = 2, these are critical points of f. Before x = -1, we can infer f' is positive (though not explicitly stated), making f increasing. Between x = -1 and x = 2, f' is negative, so f is decreasing. After x = 2, f' must be positive again for a crossing to occur, making f increasing. This pattern creates a local maximum at x = -1 (where f changes from increasing to decreasing) and a local minimum at x = 2 (where f changes from decreasing to increasing). Choice B incorrectly reverses these extrema types. Remember that f' crossing from positive to negative creates a local maximum, while crossing from negative to positive creates a local minimum.
The graph of $f'$ is decreasing on $(-\infty,1)$ and increasing on $(1,\infty)$ ; which statement about $f$ must be true?
Answer: $f$ is concave down for $x<1$ and concave up for $x>1$ . This question tests the relationship between the monotonicity of f' and the concavity of f. When f' is decreasing on (-∞, 1), we have f'' < 0 there, making f concave down. When f' is increasing on (1, ∞), we have f'' > 0 there, making f concave up. Therefore, f changes from concave down to concave up at x = 1, which is an inflection point. Choice A reverses the concavity directions by misunderstanding that decreasing f' means negative f''. The key principle is that the derivative of f' is f'', so increasing/decreasing behavior of f' directly determines the sign of f'' and thus the concavity of f.
A function $f$ is increasing for all $x$ and has exactly one inflection point at $x=0$ . Which could be $f'(x)$ ?
Answer: Always positive with a local extremum at $x=0$ . This question asks what the derivative looks like when a function is always increasing but has an inflection point. Since f is increasing for all x, we know $f'(x) > 0$ everywhere. At the inflection point $x = 0$ , the concavity changes, so $f''(0) = 0$ and f' has a local extremum at $x = 0$ . Since f' must stay positive (to keep f increasing), it cannot cross the x-axis but can have a local minimum or maximum while remaining above the x-axis. Option B (crosses the x-axis) would create a critical point of f, contradicting that f is always increasing. The correct answer shows f' always positive with a local extremum at $x = 0$ , which keeps f increasing while allowing the concavity to change at the inflection point.
A differentiable $f$ is concave up for $x<0$ and concave down for $x>0$ ; which could be $f'(x)$ ?
Answer: A function $f'$ that is increasing for $x<0$ and decreasing for $x>0$ (a single peak at $x=0$ ). This question connects concavity of f to the behavior of f'. When f is concave up for x < 0, its derivative f' must be increasing on that interval. When f is concave down for x > 0, its derivative f' must be decreasing there. This means f' increases until x = 0 and then decreases afterward, creating a peak (local maximum) at x = 0. Choice B incorrectly suggests f' has a valley at x = 0, which would make f concave down then up. The key principle is that f'' > 0 means f' is increasing (concave up), while f'' < 0 means f' is decreasing (concave down).
If $f'$ is increasing for all $x$ and $f'(0)=0$ , which statement best matches the graph of $f$ ?
Answer: $f$ is concave up everywhere and has a local minimum at $x=0$ . 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.
The graph of $f$ has horizontal tangents at $x=-2$ and $x=1$ , and $f$ decreases on $(-2,1)$ ; which could be $f'$ ?
Answer: Crosses the $x$ -axis at $x=-2$ and $x=1$ , negative between them. 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.
A function $f$ has $f''(x)>0$ for $x<0$ and $f''(x)<0$ for $x>0$ ; which could be $f'$ ?
Answer: $f'$ is increasing for $x<0$ and decreasing for $x>0$ . This question tests the relationship between f'' and the monotonicity of f'. When f'' > 0, the derivative f' is increasing; when f'' < 0, the derivative f' is decreasing. Since f'' > 0 for x < 0, we know f' increases on (-∞,0). Since f'' < 0 for x > 0, we know f' decreases on (0,∞). This means f' has a maximum at x = 0. Choice B incorrectly reverses this, showing f' with a minimum at x = 0. To determine f' behavior from f'': positive second derivative means f' increases; negative second derivative means f' decreases.
A function $f$ is increasing and concave down on $(0,5)$ ; which could be the graph of $f'$ on $(0,5)$ ?
Answer: $f'$ is positive and decreasing on $(0,5)$ . 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'.
A function $f$ is increasing and concave down on $(0,5)$ ; which could be the graph of $f'(x)$ on $(0,5)$ ?
Answer: $f'$ is positive and decreasing on $(0,5)$ . 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.
A function $f$ is concave up on $(-\infty,0)$ and concave down on $(0,\infty)$ . Which could be $f'(x)$ ?
Answer: An increasing function for $x<0$ and decreasing function for $x>0$ (a peak at $x=0$ ). This question asks you to determine what the derivative looks like given concavity information about the original function. When f is concave up, f'' > 0, which means f' is increasing. When f is concave down, f'' < 0, which means f' is decreasing. Since concavity changes at x = 0, this is an inflection point where f'' = 0, so f' has a local extremum at x = 0. The derivative f' must be increasing for x < 0 and decreasing for x > 0, which means f' has a local maximum at x = 0. Option B (valley at x = 0) would represent a local minimum of f', not a maximum. The correct answer shows f' increasing then decreasing with a peak at x = 0, matching our requirement.
The graph of $f$ has local maxima at $x=-2,3$ and a local minimum at $x=1$ ; which could be $f'(x)$ ?
Answer: A cubic crossing the $x$ -axis at $x=-2,1,3$ with sign pattern $(+,-,+,-)$ . This question tests your ability to connect features of f with the graph of f'. When f has a local maximum at x = -2, f' must equal zero there and change from positive to negative (f increases then decreases). Similarly, at the local minimum x = 1, f' changes from negative to positive, and at the local maximum x = 3, f' changes from positive to negative. This gives f' a sign pattern of (+, -, +, -) on the intervals (-∞, -2), (-2, 1), (1, 3), and (3, ∞). Choice A incorrectly reverses this pattern, starting negative instead of positive. To verify derivative graphs, check that f' crosses the x-axis at each critical point with the correct sign change matching the extremum type.
The graph of $f'$ is positive and decreasing for all $x$ , approaching $0$ as $x\to\infty$ . Which could be $f$ ?
Answer: An increasing concave down function that levels off (approaches a horizontal asymptote). This problem requires translating properties of the derivative to properties of the original function. Since f'(x) > 0 for all x, the function f is always increasing. Since f'(x) is decreasing and approaches 0, the rate of increase is slowing down, which means f is concave down. As f'(x) → 0, the graph of f becomes nearly horizontal, approaching a horizontal asymptote. Option A (concave up) contradicts our concave down requirement. The correct answer describes an increasing, concave down function that levels off, which perfectly matches a function whose derivative is positive, decreasing, and approaching zero.
$f$ is concave up on $(-2,1)$ and concave down on $(1,4)$ . Which statement could describe the sign of $f''$ ?
Answer: $f''(x)>0$ on $(-2,1)$ and $f''(x)<0$ on $(1,4)$ . This question tests graph-derivative reasoning by relating concavity of the function to the sign of its second derivative. Concave up on (-2, 1) means f'' > 0 there, as the second derivative positive indicates the graph curves upward. Concave down on (1, 4) means f'' < 0 there, curving downward. This direct correspondence matches choice A, with the change at x = 1 potentially indicating an inflection point. A tempting distractor is B, which swaps the signs, but that would reverse the concavity intervals, making it up on (1, 4) and down on (-2, 1). A useful strategy for checking sketches is to remember that f'' > 0 where f' is increasing and f'' < 0 where f' is decreasing, linking concavity to the derivative's monotonicity.
The graph of $f'$ is a parabola opening upward with zeros at $x=-2$ and $x=4$ . Which describes $f$ ?
Answer: $f$ decreases on $(-2,4)$ and increases on $(-\infty,-2)$ and $(4,\infty)$ . 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,∞).
If $f'(x)>0$ for $x<2$ , $f'(2)=0$ , and $f'(x)<0$ for $x>2$ , which graph could be $f$ ?
Answer: A graph increasing to a maximum at $x=2$ and decreasing afterward. This question requires translating derivative sign information into the shape of f. When f' > 0, the function f is increasing; when f' < 0, f is decreasing; when f' = 0, f has a horizontal tangent. Since f' > 0 for x < 2, f must increase up to x = 2. Since f' < 0 for x > 2, f must decrease after x = 2. With f'(2) = 0, there's a horizontal tangent at x = 2. This describes a local maximum at x = 2. Choice A incorrectly shows a minimum, reversing the increasing/decreasing behavior. To sketch f from f' signs: positive derivative means upward slope, negative means downward slope.
The graph of $f$ is decreasing everywhere and has a point of inflection at $x=2$ . Which could be $f'(x)$ ?
Answer: Always negative, with a local extremum at $x=2$ . This problem combines information about monotonicity and concavity to determine the derivative's properties. Since f is decreasing everywhere, we know f'(x) < 0 for all x. At the inflection point x = 2, the concavity of f changes, which means f''(2) = 0 and f' has a local extremum at x = 2. Since f' is always negative, it cannot cross the x-axis, but it can have a local maximum or minimum while remaining below the x-axis. Option C (crossing from negative to positive) would make f increase after x = 2, contradicting that f decreases everywhere. The correct answer shows f' always negative with a local extremum at x = 2, which maintains f'(x) < 0 while allowing f'' to change sign at the inflection point.
The graph of $f'$ is an upward-opening parabola with vertex below the $x$ -axis and two distinct zeros. What must be true about $f$ ?
Answer: $f$ has two critical points; it increases, then decreases, then increases. 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.

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