# How do you find the local extrema of g(x)=-x^4+2x^2?

Feb 3, 2018

#### Answer:

$- 1 , 0 , 1$ are the points of local extrema of this function.

#### Explanation:

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

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

To find the points of Extrema put $g ' \left(x\right) = 0$

$\Rightarrow - 4 {x}^{3} + 4 x = 0$

$\Rightarrow - 4 \left(x\right) \left(x - 1\right) \left(x + 1\right) = 0$

Thus the points of local extrema are:-
x = 0 ;

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

and $x + 1 = 0 \Rightarrow x = - 1$

And the Graph of the Function is given below :-