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

1 Answer
Dec 3, 2015

Answer:

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]}