Chapter 4.2, Problem 56E

To determine

To find: the function which does not satisfy the conditions of the Mean Value Theorem in the given interval.

Answer to Problem 56E

Option D is correct.

Explanation of Solution

Given:

  $\sin x,{\sin }^{-1}x,x^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.},x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ on $\left[-1,1\right]$ .

Concept used:

Mean Value Theorem for Derivatives:

If $y=f\left(x\right)$ is continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ , then there is at least one point $c$ in $\left(a,b\right)$ at which the instantaneous rate of change equals the mean rate of change,

  $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

The functions $\sin x,{\sin }^{-1}x,x^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ are continuous on $\left[-1,1\right]$ and differentiable on $\left(-1,1\right)$ .

So, the functions $\sin x,{\sin }^{-1}x,x^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ satisfy the Hypotheses of the Mean Value Theorem on the given interval.

But the function $x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ is not differentiable on $\left(-1,1\right)$ because it is not differentiable at $x=0$ .

So, the function $x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ does not satisfy the Hypotheses of the Mean Value Theorem on the given interval.

Hence option D is correct.