What are the critical values, if any, of #f(x)=(x-5)/(x^2+8)#?

Question

Paisley Johnson · Accepted Answer

I have done it for you.

I have used the u#-:# v rule of differentiation and we get the critical values when the dy#-:# dx is zero...

Charles Anctil · Answer

undefined To find the critical values of $ f(x) = \frac{x - 5}{x^2 + 8} $, we first need to find the derivative of $ f(x) $ with respect to $ x $ and then solve for $ x $ where the derivative equals zero or is undefined.

The derivative of $ f(x) $ is found using the quotient rule:

$$ f'(x) = \frac{(x^2 + 8)(1) - (x - 5)(2x)}{(x^2 + 8)^2} $$

Simplifying this expression, we get:

$$ f'(x) = \frac{x^2 + 8 - 2x(x - 5)}{(x^2 + 8)^2} $$
$$ f'(x) = \frac{x^2 + 8 - 2x^2 + 10x}{(x^2 + 8)^2} $$
$$ f'(x) = \frac{-x^2 + 10x + 8}{(x^2 + 8)^2} $$

To find critical values, we set the derivative equal to zero and solve for $ x $:

$$ -x^2 + 10x + 8 = 0 $$

This is a quadratic equation, we can solve it using the quadratic formula:

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

Where $ a = -1 $, $ b = 10 $, and $ c = 8 $. Substituting these values into the quadratic formula:

$$ x = \frac{-10 \pm \sqrt{10^2 - 4(-1)(8)}}{2(-1)} $$
$$ x = \frac{-10 \pm \sqrt{100 + 32}}{-2} $$
$$ x = \frac{-10 \pm \sqrt{132}}{-2} $$
$$ x = \frac{-10 \pm 2\sqrt{33}}{-2} $$
$$ x = 5 \pm \sqrt{33} $$

So, the critical values of $ f(x) $ are $ x = 5 + \sqrt{33} $ and $ x = 5 - \sqrt{33} $.