Question

In Exercises \(1-8,(a)\) state whether or not the function satisfies the hypotheses of the Mean Value Theorem on the given interval, and (b) if it does, find each value of \(c\) in the interval \((a, b)\) that satisfies the equation $$f(x)=x^{1 / 3} \quad \text { on }[-1,1]$$

Short answer

The function \( f(x) = x^{1/3} \) does not satisfy the hypotheses of the Mean Value Theorem on the interval [-1,1]. Therefore, there is no value of \( c \) that satisfies the Mean Value Theorem.

Step-by-step solution

Check for Continuity

The function \( f(x) = x^{1/3} \) is continuous for all real numbers, including in the closed interval [-1,1]. Therefore, the first condition of MVT is met.

Check for Differentiability

The derivative of the function is \( f'(x) = (1/3) x^{-2/3} \). However, this derivative is undefined at x = 0. Since 0 is in the open interval (-1,1), the function is not differentiable in the open interval. Therefore, the function does not satisfy both conditions of the MVT.

Apply Mean Value Theorem

Since the function is not differentiable in the open interval (-1,1), we can not apply the Mean Value Theorem. The function does not satisfy the hypotheses of the Mean Value Theorem, so we cannot find a value of \( c \) in the interval (-1,1) that satisfies the equation.