Question

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

Answer 1

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]}