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

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Answer 1 · 2015-12-03T18:56:05

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

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]}

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