Question

Given the function `f(x)=abs(x-3)`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [0,6] and find the c?

Answer 1

See below.

Explanation:

This function is the same as `absx` translated 3 to the right, so `f(x)` is continuous on `RR`, hence it is continuous on `[0,6]`.

`f` is not differentiable at `3`, so it does not satisfy the second hypothesis on `[0,6]`.

Because the hypotheses are not satisfied, the Mean Value Theorem gives no information about whether there is a `c` in `(0,6)` with `f'(c) = (f(6)-f(0))/(6-0)`.

Note that `f'(x) = {(-1, " if ",x <= -3),(1," if ",x >3):}`

While

`(f(6)-f(0))/(6-0) = (3-3)/6 = 0`.

There is no `c` at which `f'(c) = (f(6)-f(0))/(6-0)`.