Show that if $f:\mathbb R\to \mathbb R$ is such that $-x^2\le f(x)\le x^2$ for all $x\in \mathbb R$, then $f$ is differentiable at $x=0$ and $f'(0)=0$.
Really confused because isnt 0 the only soltion. and How would I apply the definition of differentiable to this?
Let $h \neq 0$ be small and note that the given inequality implies that $f(0) = 0$ (take $x=0$ in the inequality). Therefore, \begin{align*} \frac{f(0+h)-f(0)}{h} = \frac{f(h)}{h}. \end{align*} Since $-x^2 \leq f(x) \leq x^2$ implies $|f(x)| \leq x^2$, it follows that \begin{align*} \left\vert\frac{f(0+h)-f(0)}{h}\right\vert = \left\vert\frac{f(h)}{h} \right\vert \leq \frac{h^2}{|h|} = |h| \to 0 \end{align*} as $h \to 0$. By definition of the derivative, we see that $f^\prime(0)$ exists and is equal to $0$.