Given a curve y = (1/(x^2 +1)). Find the equation of tangent line with have slope of tangent minimum .


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Question Number 138085 by liberty last updated on 10/Apr/21

$G i v e n a c u r v e y = \frac{1}{x^{2} + 1} .$ $F i n d t h e e q u a t i o n o f t a n g e n t$ $l i n e w i t h h a v e s l o p e o f t a n g e n t$ $m i n i m u m .$
	Answered by EDWIN88 last updated on 10/Apr/21

	$l e t t h e e q o f t a n g e n t l i n e i s y = p x + q$ $w h e r e p = m i n i m u m v a l u e o f y^{'}$ $s t e p (1) y = {(x^{2} + 1)}^{- 1}; y^{'} = - 2 x {(x^{2} + 1)}^{- 2}$ $w e w a n t p (x) = \frac{- 2 x}{{(x^{2} + 1)}^{2}} m i n i m u m$ $p^{'} (x) = \frac{- 2 {(x^{2} + 1)}^{2} - (- 2 x) (4 x) (x^{2} + 1)}{{(x^{2} + 1)}^{4}}$ $p^{'} (x) = \frac{- 2 (x^{2} + 1) + 8 x^{2}}{{(x^{2} + 1)}^{3}} = 0; p^{'} (x) = \frac{6 x^{2} - 2}{{(x^{2} + 1)}^{3}}$ $\Rightarrow 6 x^{2} = 2; x^{2} = \frac{1}{3} o r x = \pm \frac{\sqrt{3}}{3}$ $c h e k p^{″} (x) = \frac{12 x {(x^{2} + 1)}^{3} - 6 x (6 x^{2} - 2) {(x^{2} + 1)}^{2}}{{(x^{2} + 1)}^{6}}$ $p^{″} (x) = \frac{24 x (1 - x^{2})}{{(x^{2} + 1)}^{4}} > 0 w h e n x = \frac{1}{\sqrt{3}}$ $s o m i n i m u m o f p (x) w h e n x = \frac{1}{\sqrt{3}}$ $t h e n p = \frac{- 2 (\frac{1}{\sqrt{3}})}{\frac{16}{9}} = - \frac{2 \sqrt{3}}{3} \times \frac{9}{16} = - \frac{3 \sqrt{3}}{8}$ $(2) c o n t a c t p o i n t o f t a n g e n t (\frac{1}{\sqrt{3}}, \frac{3}{4})$ $t h u s e q o f t a n g e n t l i n e h a v e t h e m i n i m u m$ $s l o p e o f c u r v e y = \frac{1}{x^{2} + 1} i s$ $3 \sqrt{3} x + 8 y = 3 \sqrt{3} (\frac{1}{\sqrt{3}}) + 8 (\frac{3}{4})$ $3 \sqrt{3} x + 8 y = 9$
	Commented by EDWIN88 last updated on 10/Apr/21

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