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

1 Answer
Nov 16, 2017

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

Explanation:

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]}