Question

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

Answer 1

See below

Explanation:

You determine whether it satisfies the hypotheses by determining whether `f(x) = (x^2-1)/(x-2)` is continuous on the interval `[-1,3]` and differentiable on the interval `(-1,3)`. (Those are the hypotheses of the Mean Value Theorem)

You find the `c` mentioned in the conclusion of the theorem by solving `f'(x) = (f(3)-f(-1))/(3-(-1))` on the interval `(-1,3)`.

`f` is not defined, hence not continuous, at `x = 2`.
Since `2` is in `[-1,3]`, `f` does not satisfy the hypotheses on this interval.

(There is no need to check the second hypothesis.)

To attempt to find `c` try to solve the equation `f'(x) = (f(3)-f(-1))/(3-(-1))`. Discard any solutions outside `(-1,3)`.

The equation has no solution, so there is no `c` that satisfies the conclusion of the Mean Value Theorem.

For those who think attempting to solve the equation was pointless

I suggest you consider

`f(x) = secx` on the interval `[-pi/3,(5pi)/3]`.

Or for a function for which such a `c` exists (though it is a challenge to find) try

`g(x) = x^2/(x^2-1)` on `[-2,4]`.