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

Question

Ezekiel Alexander · Accepted Answer

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]}

William Abdalla · Answer

undefined The critical values of $ f(x) = \frac{x^2 - 3}{x + 3} $ are the values of $ x $ where the derivative of $ f(x) $ is equal to zero or undefined. To find these, we first find the derivative of $ f(x) $ and then solve for $ x $.

The derivative of $ f(x) $ is given by:

$$ f'(x) = \frac{(2x)(x+3) - (x^2 - 3)(1)}{(x+3)^2} $$

Simplifying, we get:

$$ f'(x) = \frac{2x^2 + 6x - x^2 + 3}{(x+3)^2} $$
$$ f'(x) = \frac{x^2 + 6x + 3}{(x+3)^2} $$

Setting the derivative equal to zero and solving for $ x $, we get:

$$ x^2 + 6x + 3 = 0 $$

This quadratic equation does not factor easily, so we can use the quadratic formula to find the solutions for $ x $:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

For $ a = 1 $, $ b = 6 $, and $ c = 3 $, we have:

$$ x = \frac{-6 \pm \sqrt{6^2 - 4(1)(3)}}{2(1)} $$
$$ x = \frac{-6 \pm \sqrt{36 - 12}}{2} $$
$$ x = \frac{-6 \pm \sqrt{24}}{2} $$
$$ x = \frac{-6 \pm 2\sqrt{6}}{2} $$
$$ x = -3 \pm \sqrt{6} $$

So, the critical values of $ f(x) $ are $ x = -3 + \sqrt{6} $ and $ x = -3 - \sqrt{6} $.