Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(0)=0$ and $|f'(x)|\leq M$. Prove that $|f(x)|\leq M|x|$. Apply this to the function $f(x)=\sin x$.
I'm unsure of how to prove this problem. This problem is from the Mean Value Theorem section chapter. I will ask question if in doubt of the proof provided.
Mean Value Theorem: If a function $f$ is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that $$ f'(c) = \frac{f(b) - f(a)}{b-a}. $$
Apply this to your problem gives
$$ f'(c) = \frac{f(x) - f(0)}{x-0},\quad c\in(0,x) $$
$$ \implies \left| \frac{f(x)}{x}\right|=|f'(c)| \leq M \implies |f(x)|\leq M |x|.$$
Now, apply this to $\sin(x)$ and figure out what $M$.
$f(x) = \int_0^x f'(t) dt$ so $|f(x)| = |\int_0^x f'(t) dt| \le \int_0^x |f'(t)| dt \le \int_0^x M dt = M x $.
Note: this seems too easy, so I might be misapplying something or assuming something that is not necessarily true. Or I might actually be right - that sometimes happens.
Suppose you had an $x$ such that $|f(x)| > M|x|.$ See what the mean value theorem gives you applied to the interval $(0,x).$
