$y = \frac{x}{x^{2} + 1}$ $y=x/(x^2+1)$ , find maxima and minima?

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$y = \frac{x}{x^{2} + 1}$ $y=x/(x^2+1)$ , find maxima and minima?

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Answer 1 · 2017-11-16T01:37:53

$x = 1$ $x=1$ is the maximum point on the graph and $x = - 1$ $x=-1$ is the minimum point on the graph.

Explanation:

$y = \frac{x}{x^{2} + 1}$ $y=x/(x^2+1)$
$\frac{d y}{d x} = \frac{1 - x^{2}}{{(x^{2} + 1)}^{2}}$ $dy/dx=(1-x^2)/(x^2+1)^2$ ( using quotient rule )

$For stationary points, let \frac{d y}{d x} be 0,$ $"For stationary points, let "dy/dx" be "0","$
$\frac{1 - x^{2}}{{(x^{2} + 1)}^{2}} = 0$ $(1-x^2)/(x^2+1)^2=0$

$1 - x^{2} = 0$ $1-x^2=0$

$(1 - x) (1 + x) = 0$ $(1-x)(1+x)=0$

$x = \pm 1$ $x=+-1$

$To find the second derivative, differentiate \frac{d y}{d x},$ $"To find the second derivative, differentiate "dy/dx","$
$\frac{d^{2} y}{{d x}^{2}} = \frac{2 x^{3} - 6 x}{{(x^{2} + 1)}^{3}}$ $(d^2y)/dx^2=(2x^3-6x)/(x^2+1)^3$ ( using quotient rule )

When $x = 1$ $x=1$ ,
$\frac{d^{2} y}{{d x}^{2}} = \frac{2 {(1)}^{3} - 6 (1)}{{({(1)}^{2} + 1)}^{3}}$ $(d^2y)/dx^2=(2(1)^3-6(1))/((1)^2+1)^3$
$\times . = - \frac{1}{2} > 0$ $color(white)(xx.)=-1/2>0$ ( maximum point )

When $x = - 1$ $x=-1$ ,
$\frac{d^{2} y}{{d x}^{2}} = \frac{x {(- 1)}^{3} - 6 (- 1)}{{({(- 1)}^{2} + 1)}^{3}}$ $(d^2y)/dx^2=(x(-1)^3-6(-1))/((-1)^2+1)^3$
$\times . = \frac{1}{2} < 0$ $color(white)(xx.)=1/2<0$ ( minimum point )

Check: graph{x/(x^2+1) [-10, 10, -5, 5]}

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$y = \frac{x}{x^{2} + 1}$ $y=x/(x^2+1)$ , find maxima and minima?

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Explanation:

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y=x/(x^2+1)y=xx2+1y=x/(x^2+1), find maxima and minima?

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Explanation:

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$y = \frac{x}{x^{2} + 1}$ $y=x/(x^2+1)$ , find maxima and minima?