Question

Find the interval of strictly increasing and decreasing for the function if exist

$f(x) = \frac{x}{ {x}^{2} + 1 }$

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Answer 1

Step-by-step explanation:

Answer

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Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = \dfrac{x}{ {x}^{2} + 1}$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx} f(x) =\dfrac{d}{dx}\bigg( \dfrac{x}{ {x}^{2} + 1}\bigg)$

We know,

$\boxed{\tt{ \: \: \dfrac{d}{dx} \frac{u}{v} \: = \: \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: \: }}$

So, using this, we get

$\rm \: f'(x) = \dfrac{( {x}^{2} + 1)\dfrac{d}{dx}x - x\dfrac{d}{dx}( {x}^{2} + 1)}{ {( {x}^{2} + 1)}^{2} }$

$\rm \: f'(x) = \dfrac{( {x}^{2} + 1)(1) - x(2x - 0)}{ {( {x}^{2} + 1)}^{2} }$

$\rm \: f'(x) = \dfrac{{x}^{2} + 1 - {2x}^{2} }{ {( {x}^{2} + 1)}^{2} }$

$\rm \: f'(x) = \dfrac{1 - {x}^{2} }{ {( {x}^{2} + 1)}^{2} }$

$\rm \: f'(x) = \dfrac{(1 - x)(1 + x) }{ {( {x}^{2} + 1)}^{2} }$

So, critical points are x = - 1 and x = 1.

So, change of sign in intervals are as follow :

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Interval & \bf Sign \: of \: f'(x) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf ( - \infty , - 1) & \sf - ve \\ \\ \sf ( - 1,1) & \sf + ve \\ \\ \sf (1, \infty ) & \sf - ve \end{array}} \\ \end{gathered}$

So, for strictly increasing,

$\rm \: f'(x) > 0$

$\rm\implies \:x \: \in \: ( - 1,1)$

And, for strictly decreasing,

$\rm \: f'(x) < 0$

$\rm\implies \:x \: \in \: ( - \infty , - 1) \: \cup \: (1, \infty )$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$