Suppose is continuous on the closed interval and has a cusp at
MATHEMATICS · Single correct · CBSE · Mathematics · Applications of Derivatives · Applications of Derivatives
Problem
Suppose f is continuous on the closed interval [−2,3] and has a cusp at x=1, so that f′(1) does not exist. Which of the following best explains why x=1 must be considered when searching for the absolute extrema of f on the interval?
(a) Since f′(1) does not exist, the Candidates Test cannot be applied at x=1, so it must be discarded when searching for extrema.
(b) Because f is continuous, the Intermediate Value Theorem guarantees an extremum occurs at x=1.
(c) At a cusp, the function can change from increasing to decreasing, so an extreme value may occur there.
(d) The Mean Value Theorem requires the derivative to exist everywhere inside the interval, so x=1 must be removed before applying any extremum test.