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

Question

1226 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-01T02:31:09

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

$f(x) = 2-x^2$

$f'(x) = -2x$

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

$-2x = 0 -> x=0$

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

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

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

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

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

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