Question

What are the values and types of the critical points, if any, of `f(x)=300/(1+.03x^2)`?

Answer 1

There is only a maximum point at `(0,300)`

Explanation:

We need

`(1/(u(x)))'=-1/((u(x)^2))*u'(x)`

Compute the derivative of `f(x)`

`f(x)=300/(1+0.03x^2)`

`f'(x)=300*(-1/((1+0.03x^2)))*0.06x`

`=-(18x)/(1+0.03x^2)`

The critical points are when

`f'(x)=0`

That is,

`-(18x)/(1+0.03x^2)=0`

`x=0` as the denominator is `AA x in RR, (1+0.03x^2)>0`

Build a variation chart

`color(white)(aaaa)` `x` `color(white)(aaaaaaa)` `-oo` `color(white)(aaaaaaa)` `0` `color(white)(aaaaaaaa)` `+oo`

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

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaa)` `↗` `color(white)(aa)` `300` `color(white)(aaaa)` `↘`

There is only a maximum point at `(0,300)`

graph{300/(1+0.03x^2) [-593.5, 640.5, -174, 443.4]}