# Where is f(x) concave and convex?

• Aug 26th 2009, 01:24 PM
sebasto
Where is f(x) concave and convex?
Hello!

I have this function $\displaystyle f(x)=1,2x^5-4x^3$ and I need to give the intervals where the function's graph is concave and convex.

I know $\displaystyle f'(x)=0$ when $\displaystyle x=\sqrt{2}$, $\displaystyle x=-\sqrt{2}$ and $\displaystyle x=0$

• Aug 26th 2009, 01:53 PM
eXist
From what I understand of concave and convex is as follows:

Concave: Points on a graph where the slope is decreasing.
Convex: Points on a graph where the slope is increasing.

This of course is a just a simple explanation, like I said, from what I understand.

So I'd say your looking for all the points where:
$\displaystyle f''(x) > 0$ These would be the convex set.
$\displaystyle f''(x) < 0$ These would be the concave set.
• Aug 26th 2009, 01:57 PM
Defunkt
$\displaystyle f''(x) = 24x^3 - 24x = 24x(x^2-1) \Rightarrow f''(x) = 0$ at $\displaystyle x=0, x=1, x=-1$

$\displaystyle f''(2) = 24(8-2)$ and $\displaystyle f''(-2) = 24(2-8)$ , and $\displaystyle f(x)$ is continuous in $\displaystyle \mathbb{R}$ so:

$\displaystyle f(x)$ is concave at: $\displaystyle x > 1$ , $\displaystyle -1 < x < 0$ and convex at: $\displaystyle x < -1$ , $\displaystyle 0<x<1$

Woops, late :X