Within the interval 0≤x≤2π ,find the critical points of f(x)= sin^2 x−sin x−1 . Identify the open interval on which f is increasing and decreasing . Find the function′s local and absolute extreme values.


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Question Number 123812 by benjo_mathlover last updated on 28/Nov/20

$W i t h i n t h e i n t e r v a l 0 ⩽ x ⩽ 2 π, f i n d$ $t h e c r i t i c a l p o i n t s o f$ $f (x) = \sin^{2} x - \sin x - 1 . I d e n t i f y$ $t h e o p e n i n t e r v a l o n w h i c h f i s$ $i n c r e a s i n g a n d d e c r e a s i n g . F i n d$ $t h e {f u n c t i o n}^{'} s l o c a l a n d a b s o l u t e$ $e x t r e m e v a l u e s .$
	Answered by liberty last updated on 28/Nov/20

	$T h e f u n c t i o n f i s c o n t i n o u s o v e r [0, 2 π] a n d d i f f e r e n t i a b l e o v e r (0, 2 π)$ $s o t h e c r i t i c a l p o i n t s o c c u r a t t h e z e r o s o f f^{'} i n (0, 2 π) . W e f i n d$ $f^{'} (x) = (2 \sin x - 1) \cos x . T h e f i r s t d e r i v a t i v e i s z e r o i f o n l y i f \sin x = \frac{1}{2}$ $o r \cos x = 0 . S o t h e c r i t i c a l p o i n t s o f f i n (0, 2 π) a r e x = \frac{π}{6}; \frac{5 π}{6}; \frac{π}{2}; \frac{3 π}{2}$ $t h e y p a r t i t i o n [0, 2 π] i n t o o p e n i n t e r v a l s a s f o l l o w s$ $i n t e r v a l : (0, \frac{π}{6}) (\frac{π}{6}, \frac{π}{2}) (\frac{π}{2}, \frac{5 π}{6}) (\frac{5 π}{6}, \frac{3 π}{2}) (\frac{3 π}{2}, 2 π)$ $s i g n o f f^{'} : - + - + -$ $b e h a v i o r o f f : d e c i n c d e c i n c d e c$ $t h e r e i s a l o c a l m i n i m u m v a l u e o f f (\frac{π}{6}) = - \frac{5}{4}, a l o c a l m a x i m u m v a l u e o f$ $f (\frac{π}{2}) = - 1, a n o t h e r l o c a l m i n i m u m v a l u e o f f (\frac{5 π}{6}) = - \frac{5}{4} a n d a n o t h e r l o c a l$ $m a x i m u m v a l u e o f f (\frac{3 π}{2}) = 1 . T h e e n d p o i n t v a l u e a r e f (0) = f (2 π) = - 1 . T h e$ $a b s o l u t e m i n i m u m i n [0, 2 π] i s - \frac{5}{4} o c c u r i n g a t x = \frac{π}{6} a n d x = \frac{5 π}{6}; t h e a b s o l u t e$ $m a x i m u m i s 1 o c c u r i n g a t x = \frac{3 π}{2} .$
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