How do you find the local extrema for $f (x) = 2 - 2 x^{2}$ $f(x) = 2-2x^2$ on domain $- 1 \leq x \leq 1$ $-1 <= x <= 1$ ?

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Answer 1 · 2016-03-10T16:17:33

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

$f (x) \leq f (0)$ $f(x)<=f(0)$ for all x in $[- 1, 1]$ $[-1,1]$

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

$\frac{d f (x)}{d x} = \frac{d (2 - 2 x^{2})}{d x} = - 4 x$ $(df(x))/dx=(d(2-2x^2))/dx=-4x$

But $\frac{d f}{d x} = 0 \Rightarrow - 4 x = 0 \Rightarrow x = 0$ $(df)/dx=0=>-4x=0=>x=0$

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

The graph of $f (x)$ $f(x)$ is as follows

enter image source here

How do you find the local extrema for f(x)=2−2x2f(x) = 2-2x^2 on domain −1≤x≤1-1 <= x <= 1?