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Question Number 123799 by benjo_mathlover last updated on 28/Nov/20

Commented by benjo_mathlover last updated on 28/Nov/20

$I d e n t i f y t h e f u n c t i o n s e x t r e m e$ $v a l u e s i n g i v e n d o m a i n a n d$ $s a y w h e r e t h e y a r e a s s u m e d .$ $T e l l w h i c h o f e x t r e m e v a l u e s$ $i f a n y a r e a b s o l u t e$ $f (x) = {(x + 8)}^{2}; - \infty < x ⩽ 0$ $(a) l o c a l a n d a b s o l u t e m i n i m u m 0 a t x = - 8$ $(b) l o c a l m a x i m u m 64 a t x = 0$ $l o c a l a n d a b s o l u t e m i n i m u m 0 a t x = - 8$ $(c) l o c a l a n d a b s o l u t e m a x i m u m 64 a t x = 0$ $l o c a l a n d a b s o l u t e m i n i m u m 0 a t x = - 8$ $(d) n o l o c a l e x t r e m a, n o a b s o l u t e e x t r e m a$
Commented by benjo_mathlover last updated on 28/Nov/20

$l o o k a b o v e s i r$
	Answered by liberty last updated on 28/Nov/20

	$f i n d d e r i v a t i v e f^{'} (x) : f^{'} (x) = 2 (x + 8)$ $i t v a n i s h e s a t o n e p o i n t x_{1} = - 8 l i e i n s i d e$ $t h e i n d i c a t e d i n t e r v a l (- \infty, 0] . T o f i n d t h e$ $e x t r e m e v a l u e s o f t h e f u n c t i o n i t i s n e c e s s a r y$ $t o c o m p u t e i t s v a l u e a t t h e p o i n t x_{1} = - 8 a n d$ $a l s o a t t h e e n d - p o i n t o f t h e s e g m e n t .$ $f (- 8) = 0, f (0) = 64 a n d \lim_{x \to - \infty} f (x) = + \infty$ $i t f o l l o w s t h a t a b s o l u t e m i n i m u m 0 a t x = - 8$ $b u t n o a b s o l u t e m a x i m u m .$
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