find the relative maximum or minimum or neither at the given critical points of the function? f^′ (x)=6x(x^2 −4)^4 (x^2 −1)^2 +8x(x^2 −1)^3 (x^2 −4)^4 , x = 1, x = 2


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Question Number 159669 by zakirullah last updated on 19/Nov/21

$f i n d t h e r e l a t i v e m a x i m u m o r m i n i m u m$ $o r n e i t h e r a t t h e g i v e n c r i t i c a l$ $p o i n t s o f t h e f u n c t i o n ?$ $f^{^{'}} (x) = 6 x {(x^{2} - 4)}^{4} {(x^{2} - 1)}^{2} + 8 x {(x^{2} - 1)}^{3} {(x^{2} - 4)}^{4},$ $x = 1, x = 2$
	Answered by MohammadAzad last updated on 19/Nov/21

	$S o l u t i o n :$ $f^{^{'}} (x) = 2 x {(x^{2} - 4)}^{4} {(x^{2} - 1)}^{2} [3 + 4 (x^{2} - 1)]$ $f^{^{'}} (x) = 2 x {(x^{2} - 4)}^{4} {(x^{2} - 1)}^{2} (4 x^{2} - 1)$ $n o t i c e t h a t t h e {(x^{2} - 4)}^{4} {(x^{2} - 1)}^{2} i s ⩾ 0$ $s o w e o n l y n e e d t o w o r r y a b o u t 2 x (4 x^{2} - 1)$ $- \underset{- \frac{1}{2}}{∙} + \underset{0}{∙} - \underset{\frac{1}{2}}{∙} +$ $x = 1 a n d x = 2 a r e a f t e r \frac{1}{2} s o$ $t h e r e i s n o l o c a l e x t r e m a$ $a t x = 1 a n d x = 2$
	Commented by MohammadAzad last updated on 19/Nov/21

	Commented by zakirullah last updated on 23/Nov/21

	$a b o u n d l e o f t h a n k s s i r G$
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