What are the critical points of #f(x) = x / (x^2 + 4)#?

1 Answer
Nov 24, 2016

Answer:

We have a min at #(-2,-1/4)# and a max at #(2,1/4)#

We have points of inflexions at #(0,0)# ; #(2sqrt3,sqrt3/8)# and #(-2sqrt3,-sqrt3/8)#

Explanation:

We need to differentiate a quotient

#(u/v)'=(u'v-uv')/v^2#

Here
#u=x#, #=>#, #u'=1#

#v=x^2+4#,, #=>#,#v'=2x#

Therefore,

#f'(x)=(1*(x^2+4)-x*(2x))/(x^2+4)^2#

#=(x^2+4-2x^2)/(x^2+4)^2=(4-x^2)/(x^2+4)^2#

#=((2+x)(2-x))/(x^2+4)^2#

The critical values are when #f'(x)=0#

#(2+x)(2-x)=0#

#x=-2# and #x=-2#

Let's calculate #f''(x)#

#u=4-x^2#, #=>#, #u'=-2x#

#v=(x^2+4)^2#, #=>#, #v'(x)=4x(x^2+4)#

#f''(x)=(-2x(x^2+4)^2-4x(4-x^2)(x^2+4))/(x^2+4)^4#

#=(cancel(x^2+4)(-2x(x^2+4)-4x(4-x^2)))/(cancel(x^2+4)(x^2+4)^3)#

#=(-2x^3-8x-16x+4x^3)/(x^2+4)^3#

#=(2x^3-24x)/(x^2+4)^3#

#=(2x(x^2-12))/(x^2+4)^3#

When #f''(x)=0#, we have inflexion points

#2x(x^2-12)=0# when #x=0# ; #x=sqrt12=2sqrt3# and #x=-sqrt12=-2sqrt3#

When #x=-2# , #f''(x)>0#, so we have a min

When #x=2# , #f''(x)<0#, so we have a max

graph{x/(x^2+4) [-3.895, 3.9, -1.948, 1.947]}