# How do you find the critical numbers for (2-x)/(x+2)^3 to determine the maximum and minimum?

May 11, 2018

At $x = - 2$, we have a global maximum of $\infty$ and a global minimum of $- \infty$. A local minimum occurs at $x = 4$.

#### Explanation:

Critical numbers are the $x$ values for which $f ' \left(x\right) = 0$. These critical points may or may not be maximums/minimums. We first need to find the derivative. We do so using the quotient rule.

$f \left(x\right) = \frac{2 - x}{x + 2} ^ 3$
$f ' \left(x\right) = \frac{\left({\left(x + 2\right)}^{3}\right) \left(- 1\right) - \left(2 - x\right) \left(3\right) {\left(x + 2\right)}^{2}}{x + 2} ^ 6$
$f ' \left(x\right) = \frac{- {\left(x + 2\right)}^{3} - 3 \left(2 - x\right) {\left(x + 2\right)}^{2}}{x + 2} ^ 6$

Set this equal to zero and find which values of $x$ satisfy the equation.

$\frac{- {\left(x + 2\right)}^{3} - 3 \left(2 - x\right) {\left(x + 2\right)}^{2}}{x + 2} ^ 6 = 0$
$- {\left(x + 2\right)}^{3} - 3 \left(2 - x\right) {\left(x + 2\right)}^{2} = 0$
$- {\left(x + 2\right)}^{3} = 3 \left(2 - x\right) {\left(x + 2\right)}^{2}$
$- \left(x + 2\right) = 3 \left(2 - x\right)$
$- x - 2 = 6 - 3 x$
$2 x = 8$
$x = 4$

Note that $x = - 2$ is also a critical number, since the denominator of $f ' \left(x\right)$ will be $0$ when $x = - 2$. Investigating this critical number, we see that as $x \to - 2$ from the right, the denominator of $f \left(x\right)$ gets very small and both the numerator and the denominator remain positive. This implies that the limit of $f \left(x\right)$ as $x \to - 2$ from the right explodes to positive infinity. Similarly, as $x \to - 2$ from the left, $f \left(x\right)$ explodes to negative infinity. Thus, the global maximum and minimum of the function are $\infty$ and $- \infty$, respectively.

Now let's investigate the critical number $x = 4$. We see that:

$f ' \left(0\right) = \frac{- 8 - 3 \left(2\right) \left(8\right)}{64} = - \frac{56}{64}$.

Also,

$f ' \left(5\right) = \frac{- {7}^{3} - 3 \left(- 3\right) \left({7}^{2}\right)}{{7}^{6}} = \frac{- 343 + 441}{{7}^{6}} = \frac{98}{7} ^ 6$.

This tells us that before $x = 4$, the function is decreasing and after $x = 4$, the function is increasing. Thus, at $x = 4$, the function must have a local minimum. 