y=sin2x−x (x∈[0;𝛑]) find the range of the function.


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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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