This question applies the second derivative test to interpret the nature of a critical point. Given f′(x)=0 at x=2 (critical point) and f″(2)<0, the second derivative test confirms that f has a local maximum at x=2 since the second derivative is negative at the critical point. Choice B might be tempting as it incorrectly applies the second derivative test, but f″<0 at a critical point always indicates a local maximum. The reliable strategy for these problems is to remember the second derivative test: if f′(c)=0 and f″(c)<0, then f has a local maximum at x=c.

This problem requires interpreting the signs of f′ and f″ to identify critical points and inflection points. The given information shows f′ changes from positive to negative at x=-1 (local maximum), then from negative to positive at x=2 (local minimum). Since f″>0 on (-3,0) and f″<0 on (0,3), there's an inflection point at x=0 where concavity changes from up to down. Choice B might seem appealing but reverses the types of extrema at x=-1 and x=2. The reliable strategy is to identify where f′ changes sign for extrema, determine the type using the first derivative test, and find inflection points where f″ changes sign.

This problem involves analyzing how the behavior of f determines the properties of f′. Since f is decreasing on (0,3), we know f′(x)<0 throughout that interval. The given information that f″(x)>0 for x<2 and f″(x)<0 for x>2 tells us that f′ increases on (0,2) then decreases on (2,3). Choice E might seem tempting as it focuses on sign changes, but f′ doesn't change sign since f is decreasing throughout. To solve these problems systematically, remember that if f is monotonic on an interval, f′ maintains its sign, and the behavior of f′ itself is determined by f″.

This problem connects the signs and behavior of f″ to properties of f′. Given f′(x)>0 on (1,5) and f″(x)=0 only at x=3 where f″ changes from negative to positive, derivative f′ has a local minimum at x=3. Since f′>0 throughout, it decreases on (1,3) then increases on (3,5) while staying positive. Choice B might be tempting as it correctly identifies f′ staying positive but incorrectly describes f′ increasing then decreasing, which is opposite to the f″ sign pattern. The systematic approach is to remember that f″<0 means f′ decreasing and f″>0 means f′ increasing.

This problem tests multi-representation reasoning by connecting second derivative information to function behavior. Since f''(x) > 0 on (-1,5), the function f is concave up throughout this interval, which means f' is increasing on (-1,5). Given f'(2) = 0 and f'' > 0 at x = 2, the second derivative test confirms f has a local minimum at x = 2. Choice E incorrectly claims f' is decreasing, but positive f'' means f' must be increasing. When f'' > 0 everywhere and f'(c) = 0, the critical point at x = c must be a local minimum by the second derivative test.

This question requires matching the signs of f′ and f″ to describe the behavior of f. Given f′(x)<0 and f″(x)<0 on (-6,0), function f is both decreasing (since f′<0) and concave down (since f″<0). Choice B might seem tempting as it correctly identifies decreasing behavior but incorrectly claims concave up when f″<0 indicates concave down. To approach these problems systematically, remember that f′<0 means decreasing and f″<0 means concave down, describing a function that slopes downward and curves downward like a frown.

This question tests multi-representation reasoning by connecting function and second derivative behavior to first derivative properties. Since f is decreasing on (-1,4), we know f'(x) < 0 throughout this interval (the derivative must be negative). Since f''(x) < 0 on (-1,4), the first derivative f' is decreasing on this interval. Combining these facts: f' is both negative and decreasing on (-1,4). Choice C incorrectly claims f' is increasing, which would require f'' > 0. When both f and f'' are negative on an interval, f' must be negative (from f decreasing) and decreasing (from f'' < 0).

This question connects the signs of f″ and the behavior of f′ to determine extrema of f. Given f″(x)<0 on (0,12) (f is concave down) and f′ crosses the x-axis once from positive to negative, f has exactly one critical point that must be a local maximum. In concave down functions, critical points where f′=0 are always local maxima. Choice B might be tempting but is impossible since concave down functions cannot have local minima at interior critical points where f′ changes sign. The systematic approach is to remember that in concave down regions, any sign change in f′ from positive to negative creates a local maximum.

This problem involves matching the signs of f′ and f″ to describe f's behavior. Given f′(x)>0 and f″(x)<0 on (0,6) with f′(x)=0 nowhere in the interval, function f is increasing and concave down throughout. The fact that f′ never equals zero means there are no critical points in the interval. Choice C might be tempting as it correctly identifies increasing behavior but incorrectly claims concave up when f″<0 indicates concave down. To approach these systematically, use f′>0 for increasing behavior and f″<0 for concave down behavior.

This problem connects the sign pattern of f″ to the behavior of f′. Given f″(x)>0 for x<-1, f″(x)<0 for -1<x<2, and f″(x)>0 for x>2, the derivative f′ increases on (-4,-1), decreases on (-1,2), then increases on (2,4). This creates a pattern where f′ has a local maximum at x=-1 and a local minimum at x=2. Choice B might seem tempting as it reverses this pattern, but the signs of f″ directly determine whether f′ is increasing or decreasing. To solve these systematically, remember that f″>0 means f′ is increasing and f″<0 means f′ is decreasing.