# How do you graph f(x)=(x^2-4)^2 using the information given by the first derivative?

Feb 17, 2017

$f \left(x\right)$ has local minima at $x = \pm 2$ and a local maximum at $x = 0$

#### Explanation:

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

$f ' \left(x\right) = 2 \left({x}^{2} - 4\right) \cdot 2 x$ [Chain rule]
$= 4 {x}^{3} - 16 x$

For extrema $f ' \left(x\right) = 0$

$4 {x}^{3} - 16 x = 0$

$4 x \left({x}^{2} - 4\right) = 0$

$4 x \left(x + 2\right) \left(x - 2\right) = 0$

$x = 0 \mathmr{and} \pm 2$

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

$f ' ' \left(0\right) < 0 \to f \left(0\right)$ is a local maximun

$f ' ' \left(\pm 2\right) > 0 \to f \left(\pm 2\right)$ are local minimum

These results can be seen on the graph of $f \left(x\right)$ below:

graph{(x^2-4)^2 [-23.44, 22.18, -0.34, 22.45]}