# What are the local extrema, if any, of f (x) = x^4-8x^2-48 ?

Dec 15, 2016

#### Answer:

Local maximum at $\left(0 , - 48\right)$
Local minimum at $\left(- 2 , - 56\right)$ and $\left(2 , - 56\right)$

#### Explanation:

$f \left(x\right) = {x}^{4} - 8 {x}^{2} - 48$

Local extrema occur when $f ' \left(x\right) = 0$

Differentiating wrt $x$ we get:

$f ' \left(x\right) = 4 {x}^{3} - 16 x$

So if $f ' \left(x\right) = 0$
$\implies 4 {x}^{3} - 16 x = 0$
$\therefore \setminus 4 x \left({x}^{2} - 4\right) = 0$
$\therefore \setminus 4 x \left({x}^{2} - {2}^{2}\right) = 0$
$\therefore \setminus 4 x \left(x + 2\right) \left(x - 2\right) = 0$
$\therefore \setminus x = 0 , \pm 2$

We can now examine the nature of these turning points by looking at the second derivative:

Differentiating $f ' \left(x\right)$ wrt $x$ we get:

$f ' ' \left(x\right) = 12 {x}^{2} - 16$

so $\left\{\begin{matrix}x = \pm 2 & \implies f ' ' \left(x\right) > 0 & \text{ie minimum" \\ x=0 & => f''(x)<0 & "ie maximum}\end{matrix}\right.$

And the values of the function at these extrema are:

$f \left(0\right) \setminus \setminus \setminus \setminus \setminus = 0 - 0 - 48 \setminus \setminus \setminus = - 48$
$f \left(\pm 2\right) = 16 - 24 - 48 = - 56$

Hence the extrema are:

Local maximum at $\left(0 , - 48\right)$
Local minimum at $\left(- 2 , - 56\right)$ and $\left(2 , - 56\right)$

We can confirm this visually by plotting the graph:
graph{x^4-8x^2-48 [-10, 10, -74.2, 74]}