# How do you find the local max and min for f(x) = x^4 - 2x^2 + 1?

Sep 26, 2016

max at (0 ,1), mins at (-1 ,0) and (1 ,0)

#### Explanation:

Differentiate f(x) and equate to zero to find $\textcolor{b l u e}{\text{critical points}}$

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

$4 {x}^{3} - 4 x = 0 \Rightarrow 4 x \left({x}^{2} - 1\right) = 0 \Rightarrow 4 x \left(x - 1\right) \left(x + 1\right) = 0$

$\Rightarrow x = 0 , x = - 1 , x = 1$

Find the y coordinates.

$f \left(0\right) = 0 - 0 + 1 = 1 \Rightarrow \left(0 , 1\right)$

$f \left(- 1\right) = 1 - 2 + 1 = 0 \Rightarrow \left(- 1 , 0\right)$

$f \left(1\right) = 1 - 2 + 1 = 0 \Rightarrow \left(1 , 0\right)$

To find if max/min use the $\textcolor{b l u e}{\text{second derivative test}}$

• If f''(a) > 0 , then minimum

• If f''(a) < 0 , then maximum

The second derivative is.

$f ' ' \left(x\right) = 12 {x}^{2} - 4$

and $f ' ' \left(0\right) = - 4 < 0 \Rightarrow \left(0 , 1\right) \text{ is a maximum}$

$f ' ' \left(- 1\right) = 8 > 0 \Rightarrow \left(- 1 , 0\right) \text{ is a minimum}$

$f ' ' \left(1\right) = 8 > 0 \Rightarrow \left(1 , 0\right) \text{ is a minimum}$
graph{x^4-2x^2+1 [-10, 10, -5, 5]}