This question tests graph-derivative reasoning by inferring the function's concavity from the derivative's extrema. f' having a local maximum at x=-1, increasing for x<-1 and decreasing for x>-1, means f'' > 0 left (f' increasing) and f'' < 0 right (f' decreasing), so f'' changes sign at -1. This sign change indicates an inflection point for f at x=-1, where concavity shifts from up to down. The description doesn't specify f'(-1)=0 or its sign, so no guaranteed critical point or extremum for f. A tempting distractor is E, suggesting a local max for f, but if f'(-1) > 0 and doesn't change sign, f would be increasing through -1 without a max. A reliable sketch-checking strategy is to use f'' sign changes to locate inflections on f, independent of f' values.

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.

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.

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.

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.

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.

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.

This question asks you to identify which derivative graph matches the given increasing/decreasing behavior of f. When f is increasing, f'(x) > 0, and when f is decreasing, f'(x) < 0. Since f changes from increasing to decreasing at x = -1, we have a local maximum there, so f'(-1) = 0. Similarly, f changes from decreasing to increasing at x = 3, giving a local minimum, so f'(3) = 0. The derivative must be positive on (-∞,-1), negative on (-1,3), and positive on (3,∞). Option A (parabola opening down) would be positive between its roots but negative outside, which is backwards. The correct answer is a parabola opening up, which is negative between its roots -1 and 3 and positive outside, matching our required sign pattern.

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.

This question requires you to determine what derivative graph produces exactly one critical point with consistent concavity. Since f has exactly one critical point at x = 0, the derivative f'(0) = 0 and this is the only place where f' equals zero. Since f is concave down for all x, we have f''(x) < 0 everywhere, which means f' is always decreasing. A decreasing function that crosses the x-axis exactly once must go from positive to negative. Option A (positive slope line) would represent an increasing f', contradicting that f'' < 0. The correct answer is a line with negative slope crossing at x = 0, which is always decreasing (matching f'' < 0) and gives exactly one critical point.