Question

Let f be given by the function `abs(x^2-4x)` from [-1,1], how do you find all values of c that satisfy the Mean Value Theorem?

Answer 1

This function does not satisfy the hypotheses of the mean value theorem on the interval. It is not differentiable at `0`

It also does not satisfy the conclusion of the Mean Value Theorem on that interval. There is no such `c`.

For `0 < x < 4`, we have, `f(x)-x^2-4x`, so `f'(x) = 2x-4`

for `x < 0`, we have `f(x)=-x^2+4x`, so `f'(x)=-2x+4`

`(f(1)-f(-1))/(1-(-1)) = (3-5)/2=-1`.

There is no `c` in `(-1,1)` where `f'(c)=-1`.