Absolute of Second Derivative of f is less than or equal to M

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a twice differentiable function

1. Suppose $|f''(x)| \leq M$ on $[a, b]$. Show that for all $x \in [a, b]$ $$|f'(x)| \leq \left|\frac{f(b)-f(a)}{b-a}\right|+(b-a)M$$

2. Suppose that $|f''(x)| \leq M$ on $\mathbb{R}$ and that $\lim_{x\rightarrow \infty}{f(x)}$ exists. Show that $$\lim_{x\rightarrow \infty}{f'(x)}=0$$

I know we have to use the Mean Value theorem, as we know from that $f'(c) = \frac{f(b)-f(a)}{b-a}$, but that is for a specific value $c$ and not for all values $x \in [a, b]$.

How does this relate to the $|f''(x)| \leq M$ on $[a, b]$?

Answer 1

Answer for 1:

By MVT, it's clear that $$ \left|\frac{f(b)-f(a)}{b-a}\right| = f'(c),$$ for some $c \in (a,b)$.

Note that $$ |f'(x)| - |f'(c)| \leq |f'(x)-f'(c)| = |f''(c_0)||x-c| \leq M(b-a)\,\text{ for some } c_0 \in (x, c).$$ Add to both sides $|f'(c)|$. Hence proved.