## f(5)=1/10## is a local max. ## f(-5)=-1/10## is a local min.

##f(x)={x}/{x^2+25}##

By ,

##f'(x)={1 cdot(x^2+25)-x cdot 2x}/{(x^2+25)^2}={25-x^2}/{(x^2+25)^2}=0##

##=> x= pm5″ “## (Critical Numbers)

First Derivative Test

##{(f’:(-) to (+)” at “x=-5 => f(-5)=-1/10″ is a local min.”),(f’:(+) to (-)” at “x=5 => f(5)=1/10″ is a local max.”):}##

Second Derivative Test

By Quotient Rule,

##f”(x)={ -2x(x^2+25)^2-(25-x^2)cdot2(x^2+25)(2x)}/{(x^2+25)^4}={2x(x^2-75)}/(x^2+25)^3##

##{(f”(-5)>0 => f(-5)=-1/10″ is a local min.”),(f”(5)<0 => f(5)=1/10″ is a local max.”):}##

I hope that this was helpful.