Question

Let f(x) be a function satisfying |f(x)| ≤ x^2 for -1 ≤ x ≤ 1, how do you show that f is differentiable at x = 0 and find f’(0)?

Answer 1

The derivative exists and is zero.

Explanation:

Since `|f(x)|<=x^2` for `xin [-1,1]`, we must have
`|f(0)| <= 0^2=0`, but since it is definitely non-negative, it must be 0.

`f(0) = 0`

Now

`f^'(0) = lim_{h to 0} (f(0+h)-f(0))/h = lim_(h to 0) f(h)/h`

Thus

`|f^'(0)| = lim_{h to 0}|f(h)/h|= lim_{h to 0} |f(h)|/h`
`qquad <= lim_(h to 0) h^2/h=lim_(h to 0) h = 0`

Hence `|f^'(0)| <= 0 implies |f^'(0)| = 0`

Hence the derivative at `x=0` exists and is 0.