# What are the global and local extrema of f(x)=x^3+48/x ?

Jan 22, 2018

Local: $x = - 2 , 0 , 2$
Global: $\left(- 2 , - 32\right) , \left(2 , 32\right)$

#### Explanation:

To find extrema, you just find points where $f ' \left(x\right) = 0$ or is undefined . So:

$\frac{d}{\mathrm{dx}} \left({x}^{3} + \frac{48}{x}\right) = 0$

To make this a power rule problem, we'll rewrite $\frac{48}{x}$ as $48 {x}^{-} 1$. Now:

$\frac{d}{\mathrm{dx}} \left({x}^{3} + 48 {x}^{-} 1\right) = 0$

Now, we just take this derivative. We end up with:

$3 {x}^{2} - 48 {x}^{-} 2 = 0$

Going from negative exponents to fractions again:

$3 {x}^{2} - \frac{48}{x} ^ 2 = 0$

We can already see where one of our extrema will occur: $f ' \left(x\right)$ is undefined at $x = 0$, because of the $\frac{48}{x} ^ 2$. Hence, that is one of our extrema.

Next, we solve for the other(s). To start, we multiply both sides by ${x}^{2}$, just to rid ourselves of the fraction:

$3 {x}^{4} - 48 = 0$
$\implies {x}^{4} - 16 = 0$
$\implies {x}^{4} = 16$
=> x = ±2

We have 3 places where extrema occur: $x = 0 , 2 , - 2$. To figure out what our global (or absolute) extrema are, we plug these into the original function:

So, our absolute minimum is the point $\left(- 2 , - 32\right)$, while our absolute maximum is $\left(2 , - 32\right)$.

Hope that helps :)