Question

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

Answer 1

`f` does not satisfy the hypotheses on `[-1,1]` and there is no `c` that satisfies the conclusion.

Explanation:

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^2-1)/x` , `a= -1` and `b= 1`.

This function is a rational function, so it is continuous on its domain, but the domain does not include `0`. A function cannot be continuous at a point where it is not defined. So `f` is not continuous on `[-1, 1]`. The function does not satisfy the first hypothesis on this interval.
(That is, there is at least one point in `[-1,1]` at which `f` is not continuous..)

Differentiability implies continuity, so non-continuity imples non-differentiability. `f` is not differentiable at `0`. The function does not satisfy the second hypothesis on this interval.
(That is, there is at least one point in `(-1,1)` at which `f` is not differentiable.)

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

Attempting to solve

`f'(x) = (f(1)-f(-1))/(1-(-1))` leads to

`(x^2+1)/x^2 = 0`

This equation has no solution in the real numbers, so it has no solution in `(-1,1)`.

There is no `c` that satisfies the conclusion of MVT.