Cauchy Mean Value Theorem Consequence problem

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

Proof so far.

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

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