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

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Answer 1 · 2017-11-16T01:37:53

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

$y=x/(x^2+1)$
$dy/dx=(1-x^2)/(x^2+1)^2$ ( using quotient rule )

$"For stationary points, let "dy/dx" be "0","$
$(1-x^2)/(x^2+1)^2=0$

$1-x^2=0$

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

$x=+-1$

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

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

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

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

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