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.

This question tests graph-derivative reasoning by connecting features of f' to concavity and inflection points in f. Since f' is differentiable, its local maximum at x = 2 with f'(2) = 0 means f' increases before 2 and decreases after, implying f'' > 0 left and f'' < 0 right. This sign change in f'' at x = 2 indicates an inflection point in f where concavity shifts from up to down. The zero in f' at 2 marks a critical point in f, but without sign change information, it does not guarantee a max or min. A tempting distractor like choice A fails because a local maximum in f would require f' to change from positive to negative at 2, which is not necessarily true given only a local max in f'. To check sketches, associate extrema in f' with inflection points in f and verify if sign changes in f' imply extrema in f.

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.

This question tests graph-derivative reasoning by analyzing how zeros and signs of f' affect the monotonicity and extrema of f. A zero in f' indicates a critical point in f, but without a sign change, there is no local extremum, and consistent positive f' means f is increasing overall. Here, f'(x) = 0 only at x = 1 but f' > 0 elsewhere ensures f is increasing everywhere, with the zero at 1 not creating an extremum due to no sign change. This describes f as strictly increasing with a horizontal tangent at 1 but no local max or min. A tempting distractor like choice B fails because a local minimum would require f' to change from negative to positive at 1, contradicting the given always positive f' except at the zero. To check sketches, examine the sign of f' around its zeros to determine if f has extrema or continues monotonically, confirming overall behavior.

This question tests graph-derivative reasoning by examining how the sign of f' determines extrema in f. Positive f' indicates f is increasing, negative f' indicates decreasing, and a sign change at a zero of f' signals a local extremum. Here, f' > 0 for x < 0 means f increasing left of 0, f'(0) = 0 provides a critical point, and f' < 0 for x > 0 means f decreasing right of 0, combining to a positive-to-negative sign change. This configuration must result in a local maximum for f at x = 0 due to the shift from increasing to decreasing. A tempting distractor like choice A fails because it suggests a local minimum, which would require a negative-to-positive sign change in f' instead. To check sketches, evaluate sign changes in f' around its zeros to determine if f has a maximum, minimum, or neither, confirming with the overall monotonicity.

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.

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.

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.

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.

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.