# How do you find the local extrema for f(x) = 2-2x^2 on domain -1 <= x <= 1?

It has maximum for $x = 0$ and value $f \left(0\right) = 2$ which is a absolute maximum because

$f \left(x\right) \le f \left(0\right)$ for all x in $\left[- 1 , 1\right]$

Also using derivatives we find the roots of first derivative which is

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{d \left(2 - 2 {x}^{2}\right)}{\mathrm{dx}} = - 4 x$

But $\frac{\mathrm{df}}{\mathrm{dx}} = 0 \implies - 4 x = 0 \implies x = 0$

Hence point $x = 0$ is a critical value of $f \left(x\right)$.Above we show that this a maximum.

The graph of $f \left(x\right)$ is as follows