#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]}