Question

Consider the function f(x)=4−4x^(2/3) on the interval [−1,1]. Whict of the three hypotheses of Rolle's Theorem fails for this function on the inverval?

(a) f(x) is continuous on [−1,1].
(b) f(x) is differentiable on (−1,1).
(c) f(−1)=f(1)

Answer 1

The hypotheses of Rolle's Theorem that fail for the function f(x) = 4 - 4x^(2/3) on the interval [–1,1] are (b) differentiability at x = 0 and (c) the condition that f(–1) = f(1).

Step-by-step explanation:

Rolle's Theorem states that if a function f(x) is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there is at least one c in the open interval (a, b) such that f'(c) = 0. Let's check each condition for the function f(x) = 4 - 4x(2/3) on the interval –​1,1.

• (a) The function is continuous on –​1,1 since it is composed of basic algebraic operations of a continuous function.
• (b) The function may not be differentiable at x = 0, because the derivative involves the term x–(1/3), which is undefined at x = 0. So, hypothesis (b) is the one that fails.
• (c) Calculating f(–1) and f(1), we have that f(1) = 4 – 4*1(2/3) = 0 and f(–1) = 4 – 4*(–1)(2/3) = 4, hence f(–1) ≠ f(1) and the function does not meet condition (c) either.

Therefore, the hypotheses that fail for Rolle's Theorem in this case are both (b) and (c).