# What are the local extrema, if any, of f(x) =2-x^2?

Oct 1, 2016

$f \left(x\right)$ has a maximum value of $2$ at $x = 0$

#### Explanation:

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

$f ' \left(x\right) = - 2 x$

For a local maximum or minimum $f ' \left(x\right) = 0$

$- 2 x = 0 \to x = 0$

Hence $f \left(0\right) = 2$ is a local maximum or minimum.

Now consider $f ' ' \left(x\right) = - 2$ $< 0$ independent of $x$

So, $f \left(0\right) = 2$ is a maximum value of $f \left(x\right)$

This can be seen on the graph of $2 - {x}^{2}$ below:

graph{2-x^2 [-4.933, 4.934, -2.467, 2.465]}