We have f: $\Bbb R \to \Bbb R$. Suppose $|f''(x)| \le 1 $ $\forall x \in[-1,1]$. Show that over [-1,1] we have: $$|f(c+h)-f(c)-f'(c)h|\le \frac{h^2}{2}$$, for $c, c+h \in [-1,1]$.
I know that f is twice differentiable if all the partial derivatives up to and including order 2 exists and are continuous. The left side of the equation looks much like the first-order Taylor formula for approximating $f(c+h)-f(c)$ by $f'(c)h$. But I don't know how to derive the above relationship.
By Taylor: for every $c,c+h \in [-1,1]$ $$f(c+h)=f(c)+f'(c)h+\frac{f''(d)}{2}h^2$$ for some $d \in [c,c+h]$. Now applying the modulus $$|f(c+h)-f(c)-f'(c)h|=\frac{|f''(d)|}{2}h^2\le \frac{h^2}{2}$$ because $|f''(x)|\le1$ for every $x \in [-1,1]$.