A differentiable function is defined on the closed interval
MATHEMATICS · Single correct · CBSE · Mathematics · Applications of Derivatives · Applications of Derivatives
Problem
A differentiable function h is defined on the closed interval [2,7]. The derivative satisfies h′(x)=0 only at x=4, yet h attains its absolute minimum at x=2. Which of the following statements best explains why the minimum occurs at x=2 instead of at x=4?
(a) Since h′(x)>0 for all x=4, the function is increasing on [2,7], so the smallest value occurs at the left endpoint x=2.
(b) Absolute extrema on a closed interval must occur at critical numbers or at the endpoints, and h(2)<h(4) even though h′(4)=0.
(c) By the Mean Value Theorem, the derivative must be zero at the point where the minimum occurs, so h′(2)=0 even if computation suggests otherwise.
(d) Because h′(4)=0 the point x=4 can be only a local extremum, and local extrema can never be absolute extrema.