real analysis - Proving that the function is bounded by the function of its derivative on the given interval

Suppose $h:R \longrightarrow R$ is differentiable everywhere and $h'$ is continuous on $[0,1]$, $h(0) = -2$ and $h(1) = 1$. Show that:

$|h(x)|\leq max(|h'(t)| , t\in[0,1])$ for all $x\in[0,1]$

I attempted the problem the following way:

Since $h(x)$ is differentiable everywhere then it is also continuous everywhere. $h(0) = -2$ and $h(1) = 1$ imply that h(x) should cross x-axis at some point (at least once). Denote that point by c to get $h(c) = 0$ for some $c\in[0,1]$.

$h'(x)$ continuous means that $lim[h'(x)] = h'(a)$ as $x\rightarrow a$ but then I am stuck and I don't see how what I have done so far can help me to obtain the desired inequality.

Thank you in advance!