Question

Given the function `f(x) = x^(2/3)`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-1,8] and find the c?

Answer 1

The Mean Value Theorem has two hypotheses:

H1 : `f` is continuous on the closed interval `[a,b]`

H2 : `f` is differentiable on the open interval `(a,b)`.

In this question, `f(x) = x^(2/3)` , `a=-1` and `b=8`.

This function is continuous on its domain, `(-oo, oo)`, so it is continuous on `[-1, 8]`

`f'(x)=2/(3root(3)x)` which exists for all `x != 0`. But `0` is in `(-1,8)`

So `f` is not differentiable on `(-1, 8)`

To determine whether there is a `c` that satisfies the conclusion, we need to determine whether there is a `c` in (-1,8)# for which

`f'(c) = (f(8)-f(-1))/(8-(-1)) = (4-1)/9 = 1/3`

We can actually solve `2/(3root(3)x) = 1/3`.

We get `x=8` which is not in the open interval `(-1,8)`. So there is no `c` that satisfies the conclusion of MVT.

For an example of a function on an interval where it does not satisfy the hypotheses, but does satisfy the conclusion, please see:
https://socratic.org/questions/given-the-function-f-x-x-1-3-how-do-you-determine-whether-f-satisfies-the-hypoth