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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

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  • $\begingroup$ Please try to TeXify your post. It will make your question easier to read for the rest of us. $\endgroup$
    – Jose Arnaldo Bebita Dris
    Nov 6, 2014 at 23:52

1 Answer 1

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$f'(x) = \dfrac{x\cos x - \sin x}{x^2} = \dfrac{g(x)}{x^2}$,

and $g'(x) = \cos x - x\sin x - \cos x = -x\sin x < 0$. So $g(x) < g(0) = 0 \to f'(x) < 0$.

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  • $\begingroup$ how did you know how to do this? is the mean value theorem involved? $\endgroup$
    – John Smith
    Nov 7, 2014 at 0:13

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