## Cauchy Mean Value Theorem Consequence problem

Suppose that the function $f: \mathbb {R} \rightarrow \mathbb {R}$ has two derivatives, with $f(0)=f'(0)=0$ and $|f''(x)| \leq 1$ if $|x| \leq 1$. Prove that $f(x) \leq \frac {1}{2}$ if $|x| \leq 1$

Proof so far.

Suppose that $x \in \mathbb {R}$ with $|x| \leq 1$, pick $x_0 \in \mathbb {R}$, find $z \in \mathbb {R}$ strictly between $x$ and $x_0$ such that $f(x) = \frac {f''(z)}{2}(x-x_0)^2$

Can I conclude that $f''(z) \leq 1$?