Sketching Graphs of Functions and Derivatives - AP Calculus BC

The graph of $f$ is increasing on $(-\infty,2)$ and decreasing on $(2,\infty)$ . Which could represent $f'(x)$ ?
Answer: $f'(x)$ is positive for $x<2$ , zero at $x=2$ , negative for $x>2$ . 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.
A function $f$ is increasing on $(-\infty,-1)$ , decreasing on $(-1,3)$ , then increasing on $(3,\infty)$ ; which could be $f'(x)$ ?
Answer: A downward-opening parabola with zeros at $x=-1$ and $x=3$ . This problem requires translating intervals of increase/decrease into the derivative's sign. When f is increasing, f' > 0, and when f is decreasing, f' < 0. Since f increases on (-∞, -1), decreases on (-1, 3), then increases on (3, ∞), we need f' to be positive before x = -1, negative between x = -1 and x = 3, and positive after x = 3. This means f' must have zeros at x = -1 and x = 3, crossing from positive to negative at x = -1 and from negative to positive at x = 3. A downward-opening parabola with zeros at these points satisfies this pattern: it's positive between its roots and negative outside them. Choice A (upward-opening parabola) would give the opposite sign pattern, being negative between its roots. To verify derivative graphs from monotonicity, check that the derivative is positive where the function increases and negative where it decreases.
The graph of $f$ is concave up on $(-\infty,1)$ and concave down on $(1,\infty)$ ; which could be $f'(x)$ ?
Answer: A graph increasing for $x<1$ and decreasing for $x>1$ , with a local maximum at $x=1$ 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.
A graph of $f'(x)$ crosses the $x$ -axis at $x=-1$ and $x=2$ , is positive between, and negative outside. Which could be $f$ ?
Answer: $f$ has a local maximum at $x=-1$ and a local minimum at $x=2$ ; increasing on $(-1,2)$ . 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.
A function $f'$ has a local maximum at $x=1$ and crosses the $x$ -axis there from positive to negative; which could be $f$ near $x=1$ ?
Answer: $f$ has a local maximum at $x=1$ and is concave down there. This problem tests understanding of how features of f' translate to features of f. Since f' has a local maximum at x = 1, the second derivative f''(1) < 0, which means f is concave down at x = 1. Since f' crosses the x-axis from positive to negative at x = 1, this means f changes from increasing to decreasing, giving f a local maximum at x = 1. Therefore, f has a local maximum at x = 1 and is concave down there. Choice D might seem plausible as it correctly identifies the local maximum, but it incorrectly states f is concave up when f''(1) < 0 tells us f must be concave down. When analyzing critical points, remember that f' = 0 gives potential extrema, the sign change of f' determines the type of extremum, and the sign of f'' determines concavity.
A graph of $f$ has an inflection point at $x=-1$ and $f'(-1)=5$ ; which could be the graph of $f'$ ?
Answer: $f'$ has a local extremum at $x=-1$ with value $5$ . This question connects inflection points to the derivative's behavior. An inflection point at x = -1 means f changes concavity there, so f'' changes sign at x = -1. Since f'' is the derivative of f', this means f' has a local extremum at x = -1. We're also told f'(-1) = 5, so this extremum has value 5, not zero. The graph of f' should show a local maximum or minimum at x = -1 with y-value 5. Choice B incorrectly suggests f' crosses the x-axis at x = -1, which would mean f'(-1) = 0, contradicting the given information. At inflection points of f, look for extrema in f', and the value of f' at that point gives the height of the extremum, not necessarily zero.
A graph of $f'(x)$ touches the $x$ -axis at $x=0$ without crossing, and $f'(x)>0$ for $x\ne 0$ . Which could be $f$ ?
Answer: $f$ is increasing for all $x$ and has a horizontal tangent at $x=0$ but no extremum there. 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.
A graph of $f$ crosses the $x$ -axis at $x=0$ and has a horizontal tangent there without changing sign; which could be $f$ ?
Answer: A parabola tangent to the $x$ -axis at $x=0$ and opening upward 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.
Function $f$ has a local maximum at $x=-2$ , an inflection point at $x=1$ , and $f'(0)<0$ ; which could be $f'(x)$ ?
Answer: $f'(x)$ crosses from positive to negative at $x=-2$ and has a local extremum at $x=1$ ; also $f'(0)<0$ . This question tests graph-derivative reasoning by linking features like local extrema and inflection points of f to the behavior of f'. A local maximum in f occurs where f' changes from positive to negative, indicating a sign change at that point. An inflection point in f occurs where concavity changes, meaning f'' changes sign, which corresponds to f' having a local extremum since f'' is the derivative of f'. Here, the local maximum at x = -2 requires f' to cross from positive to negative there, the inflection at x = 1 requires f' to have a local extremum at 1, and f'(0) < 0 specifies the sign at that point. A tempting distractor like choice B fails because it has f' crossing from negative to positive at x = -2, which would indicate a local minimum for f instead of the given maximum. To check sketches, identify extrema in f with sign changes in f' and inflections in f with extrema in f', ensuring additional conditions like specific sign values are satisfied.
A function $f$ decreases on $(-2,0)$ and increases on $(0,5)$ , and is concave up for $x>0$ ; which could be $f'$ ?
Answer: An increasing line crossing the $x$ -axis at $x=0$ . 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.
If $f$ has a horizontal tangent at $x=0$ and is increasing on both sides of $0$ , which could be true about $f'(x)$ near $0$ ?
Answer: $f'(x)$ is positive for $x\ne0$ near $0$ and $f'(0)=0$ . 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.
A graph of $f$ is concave up on $(-\infty,1)$ and concave down on $(1,\infty)$ . Which could represent $f'(x)$ ?
Answer: $f'(x)$ is increasing for $x<1$ and decreasing for $x>1$ . This problem connects concavity to the derivative's monotonicity. When f is concave up on (-∞,1), the slopes of tangent lines are increasing, meaning f'(x) is increasing for x < 1. When f is concave down on (1,∞), the slopes of tangent lines are decreasing, meaning f'(x) is decreasing for x > 1. At x=1, where concavity changes, f has an inflection point and f'(x) transitions from increasing to decreasing (though f' may or may not equal zero there). Choice A incorrectly claims f' is decreasing for x < 1, which would make f concave down there instead of concave up. Remember: concave up means f' is increasing (f'' > 0), concave down means f' is decreasing (f'' < 0).
A differentiable function $f$ has $f'(x)$ crossing the $x$ -axis upward at $x=5$ (negative to positive); which could be true about $f$ at $x=5$ ?
Answer: $f$ has a local minimum at $x=5$ . This question tests your understanding of how the derivative's behavior determines extrema of the original function. When f'(x) crosses the x-axis upward at x = 5, it means f'(x) changes from negative to positive at this point. Since f'(5) = 0 (crossing the x-axis) and f' changes from negative (decreasing function) to positive (increasing function), the function f has a local minimum at x = 5. Choice A incorrectly suggests a local maximum, which would require f' to cross downward (positive to negative). When f' crosses the x-axis, always check the direction: upward crossing (- to +) indicates a local minimum, while downward crossing (+ to -) indicates a local maximum.
On $(0,6)$ , $f$ increases with decreasing slope until $x=3$ , then increases with increasing slope; which could be $f'(x)$ ?
Answer: Positive on $(0,6)$ with a local minimum at $x=3$ This question tests graph-derivative reasoning, interpreting the function's slope variations to deduce the derivative's behavior. f increasing throughout (0,6) means f'>0 everywhere, with decreasing slope until x=3 indicating f' is decreasing to a minimum there, then increasing slope means f' increases after. This describes f' positive with a local minimum at x=3, reflecting the turnaround in slope magnitude. The interval restricts analysis to (0,6) without boundary issues. A tempting distractor like choice B fails because it introduces negative values for f', implying decreasing intervals for f, contrary to f always increasing. To check sketches transferably, assess how changes in the function's steepness correspond to the monotonicity of the derivative graph.
$f'(x)<0$ for $x<1$ , $f'(1)=0$ , and $f'(x)>0$ for $x>1$ ; additionally $f''(x)>0$ for all $x$ ; which could be $f$ ?
Answer: A concave-up curve with a local minimum at $x=1$ . This problem combines first and second derivative information to determine a function's shape. Since f'(x) < 0 for x < 1, f'(1) = 0, and f'(x) > 0 for x > 1, the function f decreases before x = 1 and increases after x = 1, creating a local minimum at x = 1. Additionally, f''(x) > 0 for all x means f is concave up everywhere. A concave-up curve with a local minimum at x = 1 matches these conditions perfectly. Choice A (concave-down with maximum) contradicts both the concavity condition and the type of extremum. When both derivative conditions are given, use f' to find extrema and monotonicity, then use f'' to determine the overall shape's concavity.
A graph of $f'(x)$ is an upward-opening parabola with vertex below the $x$ -axis, crossing at $x=1$ and $x=5$ . Which could be $f$ ?
Answer: $f$ is decreasing for $x<1$ , increasing for $1<x<5$ , decreasing for $x>5$ . This question tests interpreting a parabolic derivative graph to determine the original function's behavior. An upward-opening parabola with vertex below the x-axis crosses the x-axis at two points (x=1 and x=5), where f has critical points. Since the parabola is below the x-axis between 1 and 5, f'(x) < 0 there, meaning f is decreasing. Outside this interval, the parabola is above the x-axis, so f'(x) > 0 and f is increasing. This creates a local maximum at x=1 and a local minimum at x=5. Choice A incorrectly claims f is decreasing outside [1,5] and increasing inside, reversing the derivative's actual sign pattern. When sketching from a derivative graph, remember that regions below the x-axis correspond to decreasing intervals of the original function.
A graph of $f$ has a local minimum at $x=-3$ and an inflection point at $x=1$ ; which could be $f'(x)$ ?
Answer: $f'(x)$ is zero at $x=-3$ and has a local extremum at $x=1$ 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'.
The graph of $f'(x)$ is positive for $x<0$ , negative for $0<x<4$ , and positive for $x>4$ ; which could be $f$ ?
Answer: $f$ has a local maximum at $x=0$ and a local minimum at $x=4$ . 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.
The graph of $f'(x)$ crosses the $x$ -axis at $x=-1$ and just touches it at $x=2$ ; which could be $f$ ?
Answer: $f$ has a local extremum at $x=-1$ and a point of inflection with horizontal tangent at $x=2$ This question tests graph-derivative reasoning, distinguishing between crossing and touching zeros in the derivative and their effects on the function. f' crossing at x=-1 means it changes sign there, creating a local extremum for f at x=-1. f' touching at x=2 without crossing implies f'(2)=0 but no sign change, so no extremum, but if it's a local extremum for f' itself, this indicates f''(2)=0 with sign change, yielding an inflection for f with horizontal tangent. This combination fits the described behaviors precisely. A tempting distractor like choice B fails because touching without crossing cannot produce a local extremum at x=2 for f, as that requires f' to change sign. To check sketches transferably, examine multiplicity of derivative zeros to differentiate extrema from inflections with horizontal tangents.
The graph of $f$ has local maxima at $x=-2,2$ and a local minimum at $x=0$ ; which could be $f'(x)$ ?
Answer: A cubic that crosses the $x$ -axis at $x=-2,0,2$ and is negative for $x<-2$ , positive on $(-2,0)$ , negative on $(0,2)$ , positive for $x>2$ . This problem tests your ability to connect features of a function with features of its derivative graph. Since f has local maxima at x = -2 and x = 2, the derivative f'(x) must equal zero at these points, and since f has a local minimum at x = 0, f'(0) = 0 as well. At a local maximum, f changes from increasing to decreasing, so f' changes from positive to negative; thus f' must cross from positive to negative at x = -2 and x = 2. At a local minimum, f changes from decreasing to increasing, so f' changes from negative to positive at x = 0. Choice C correctly shows a cubic that is positive on (-2, 0) and negative on (0, 2), matching these sign changes. Choice B might seem tempting as it has zeros at the right places, but an upward-opening parabola would be negative outside (-2, 2) and positive inside, giving the wrong extrema types. To verify derivative graphs, check that zeros occur at critical points and that the derivative's sign matches whether f is increasing or decreasing.

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