Establish the inequality $2/\pi < \sin x/x$ for $0 < x < \pi/2$ by showing that the function $f(x)= \sin x/x$ is strictly decreasing for $0 < x ≤ \pi/2$.
this is all i have, dunno if im on the right path
$f'(x) = \dfrac{\sin x-x\cos x}{2x} = 0$
x = 0. only critical number
I tested the interval $(0, \pi/2]$ for $1/2x$ and $\sin x-x\cos x$ and it isn't strictly decreasing. i must be doing this wrong
$f'(x) = \dfrac{x\cos x - \sin x}{x^2} = \dfrac{g(x)}{x^2}$,
and $g'(x) = \cos x - x\sin x - \cos x = -x\sin x < 0$. So $g(x) < g(0) = 0 \to f'(x) < 0$.