Question

How do you find all values of c that satisfy the mean value theorem for `f(x) = x^2` on the interval [-1,1]?

Answer 1

Since `f(1) = f(-1)` for `f(x)=x^2` the secant of
`f(x)` over the interval `[-1,1]` is `0`.

We are looking for value(s) of `c` such that
`f'(c) =` this secant.

That is (since f'(x) = 2x)
values for which
`2c = 0`

The only value satisfying this condition is `c=0`