y-sin2x-x-x-0-find-the-range-of-the-function-

Question Number 12704 by @ANTARES_VY last updated on 29/Apr/17

y = \sin 2 x - x (x \in [0; π]) find the

range of the function .

Answered by mrW1 last updated on 29/Apr/17

y = \sin 2 x - x

y^{'} = 2 \cos 2 x - 1

f o r y t o b e l o c a l m a x . o r m i n . :

y^{'} = 0 \Rightarrow 2 \cos 2 x - 1 = 0 \Rightarrow \cos 2 x = \frac{1}{2}

\Rightarrow 2 x = \frac{π}{3}, \frac{5 π}{3}

\Rightarrow x = \frac{π}{6}, \frac{5 π}{6}

a t x = π :

y = \sin (2 π) - π = - π

a t x = \frac{5 π}{6} :

y = \sin (\frac{5 π}{3}) - \frac{5 π}{6} = - \frac{\sqrt{3}}{2} - \frac{5 π}{6} = - \frac{3 \sqrt{3} + 5 π}{6} < - π

a t x = 0 :

y = \sin (0) + 0 = 0

a t x = \frac{π}{6} :

y = \sin (\frac{π}{3}) - \frac{π}{6} = \frac{\sqrt{3}}{2} - \frac{π}{6} = \frac{3 \sqrt{3} - π}{6} > 0

f o r x \in [0, π]

y \in [- \frac{3 \sqrt{3} + 5 π}{6}, \frac{3 \sqrt{3} - π}{6}]

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