Question

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

Answer 1

They are `x=-3 pm sqrt(6)`, which are approximately `-0.5505` and `-5.4495`.

Explanation:

By the Quotient Rule, the derivative is

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

This equals zero when `x^2+6x+3=0`. By the quadratic formula, the roots of this are

`x=(-6 pm sqrt(36-12))/2=-3 pm sqrt(4)sqrt(6)/2=-3 pm sqrt(6)`.

These are the critical values of `f`.

You can check, with the First or Second Derivative Test, that `f` has a local maximum at `-3-sqrt(6)` and a local minimum at `-3+sqrt(6)`.

Here's the graph of `f`:

graph{(x^2-3)/(x+3) [-80, 80, -40, 40]}