Question

What are the values and types of the critical points, if any, of `f(x)=(x+3x^2) / (1-x^2)`?

Answer 1

There is a local maximum at `(-5.83, -2.91)` and a local minimum at `(-0.17, -0.86)`.

Explanation:

The function is

`f(x)=(x+3x^2)/(1-x^2)`

This function is a quotient of `2` derivable functions

The derivative of a quotient is

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

`u=x+3x^2`, `=>`, `u'=1+6x`

`v=1-x^2`, `=>`, `u'=-2x`

Therefore,

`f'(x)=((1+6x)(1-x^2)-(x+3x^2)(-2x))/(1-x^2)^2`

`=(1-x^2+6x-6x^3+2x^2+6x^3)/(1-x^2)^2`

`=(x^2+6x+1)/(1-x^2)^2`

The critical points are when

`f'(x)=0`

That is

`x^2+6x+1=0`

The solutions to this quadratic equation are

`x=(-6+-sqrt(36-4))/(2)=(-6+-sqrt(32))/(2)`

`=(-6+-4sqrt2)/(2)`

`=-3+-2sqrt2`

Let `x_1=-3-2sqrt2`

and

`x_2=-3+2sqrt2`

Let's build a variation chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `x_1` `color(white)(aaaa)` `-1` `color(white)(aaaaaaa)` `x_2` `color(white)(aaaa)` `1` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaa)` `+` `color(white)(aa)` `0` `color(white)(aaa)` `-` `color(white)(aaa)` `-` `color(white)(aaaa)` `0` `color(white)(aa)` `+` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaa)` `↗` `color(white)(aa)` ###color(white)(aaa)#`↘` `color(white)(aaa)` `↘` `color(white)(aaa)` ###color(white)(aa)#`↗` `color(white)(aaa)` `↗`

Therefore,

There is a local maximum at `(-5.83, -2.91)` and a local minimum at `(-0.17, -0.86)`.

graph{(x+3x^2)/(1-x^2) [-10, 10, -5, 5]}