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Question Number 95560 by I want to learn more last updated on 26/May/20

Commented by i jagooll last updated on 26/May/20

$y = {(x^{2} + 5)}^{- 1}$ $y^{^{'}} = \frac{- 2 x}{{(x^{2} + 5)}^{2}} ⩾ 0$ $increasing for x ⩽ 0$ $decreasing for x ⩾ 0$
	Answered by mr W last updated on 26/May/20

	$y^{'} = - \frac{2 x}{{(x^{2} + 5)}^{2}}$ $y i n c r e a s e s o r d e c r e a s e s m o s t r a p i d l y$ $m e a n s y^{'} i s m a x i m u m o r m i n i m u m,$ $t h a t m e a n s a g a i n y^{″} = 0 .$ $y^{″} = - \frac{2}{{(x^{2} + 5)}^{2}} + \frac{4 x^{2}}{{(x^{2} + 5)}^{3}} = 0$ $2 x^{2} = x^{2} + 5$ $\Rightarrow x = \pm \sqrt{5}$ $a t x = - \sqrt{5} : y^{'} > 0, y i n c r e a s e s m o s t r a p i d l y .$ $a t x = \sqrt{5} : y^{'} < 0, y d e c r e a s e s m o s t r a p i d l y .$
	Commented by I want to learn more last updated on 26/May/20

	$thanks sir$
	Commented by peter frank last updated on 26/May/20

	$thank you$
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