Let $f(t) = \sin t$. Fix $x$ such that $0 < x < $$π\over{2}$. If you were to apply the Mean Value Theorem to $f$ for $t$ in the interval $[0, x]$:
(a) Write down precisely what the conclusion of the theorem tells you.
(b) Explain why (a) allows you to immediately conclude that $\sin x < x$ for $x\in (0, {π\over{2}}$).
(c) Why is it the case that if $x \geq {π\over{2}}$ then $\sin x < x?$
Attempted Solutions:
(a) Let $f\in{C[0,x]}$. (Note that C is the set of continuous functions in the given interval). Assume that $f$ is differentiable on $(0,x)$. Then there exists $c\in{(0,x)}$ such that $f'(c)={f(b)-f(a)\over{b-a}}$ = $\sin x\over{x}$
(b) and (c) I am not sure of.
Best Answer
The theorem states that there exists $c\in (0,x)$ such that
$$\sin x-\sin 0=\sin' c \cdot (x-0)=\cos c \cdot x.$$
Since $|\cos c|<1$ for $c\in (0,\pi/2)$ we have that
$$|\sin x|= |\sin x-\sin 0|=|\cos c||x|<x.$$
Finally, if $x\ge \pi/2$ we have that
$$|\sin x|\le 1<\pi/2 \le x.$$