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 question connects the monotonicity of f and extrema of f′ to determine inflection points. Since f′(x)>0 on (-2,4), function f is increasing throughout. The fact that f′ has a local minimum at x=1 means f″ changes from negative to positive at x=1, creating an inflection point where f changes from concave down to concave up. Choice C might be tempting as it suggests a local minimum for f at x=1, but since f′>0 on both sides of x=1, f continues increasing—there's just an inflection point. The key strategy is to distinguish between extrema of f′ (which indicate inflection points for f) and extrema of f itself.

This problem connects the concavity of f to properties of critical points. Since f is concave down on (0,10), we have f″(x)<0 throughout. If f has exactly one critical point at x=6, and f is concave down everywhere, that critical point must be a local maximum (since concave down functions can only have local maxima at interior critical points). Choice B might be tempting but is impossible since concave down functions cannot have local minima at interior critical points. The key insight is that on intervals where f″<0, any critical point where f′=0 must be a local maximum.

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 applies the first derivative test with uniform concavity. The sign pattern shows f′>0 on (-∞,0) and (2,∞) but f′<0 on (0,2), indicating f increases, then decreases, then increases again, creating a local maximum at x=0 and local minimum at x=2. The condition f″(x)>0 everywhere confirms these are indeed extrema (not just horizontal tangents). Choice B might seem appealing but reverses the extrema types by misreading the sign changes. The systematic approach is to apply the first derivative test: f′ changing from positive to negative gives local maximum, negative to positive gives local minimum.

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 requires connecting the sign pattern of f′ and the sign of f″ to identify extrema and concavity. The given conditions show f′ changes from negative to positive at x=5, indicating a local minimum there by the first derivative test. Since f″(x)>0 on (2,8), function f is concave up throughout this interval. Choice B might seem appealing as it correctly identifies the concavity but incorrectly claims a local maximum when f′ changes from negative to positive, which creates a local minimum. The key strategy is to apply the first derivative test: f′ changing from negative to positive indicates a local minimum.

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.

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 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.