# How do you determine whether the function F(x)= 1/12X^4 + 1/6X^3-3X^2-2X+1 is concave up or concave down and its intervals?

Sep 13, 2015

To find intervals on which the graph of $F$ is concave up and those on which it is concave down, investigate the sign of the second derivative.

#### Explanation:

For, $F \left(x\right) = \frac{1}{12} {x}^{4} + \frac{1}{6} {x}^{3} - 3 {x}^{2} - 2 x + 1$ we have

$F ' \left(x\right) = \frac{1}{3} {x}^{3} + \frac{1}{2} {x}^{2} - 6 x - 2$ and

$F ' ' \left(x\right) = {x}^{2} + x - 6 = \left(x + 3\right) \left(x - 2\right)$

The only chance $F ' ' \left(x\right)$ has to (perhaps) change sign is at $F ' ' \left(x\right) = 0$.

Which happens at $x = - 3$ and at $x = 2$

On $\left(- \infty , - 3\right)$, both factor of $F ' '$ are negative, so $F ' '$ is positive and the graph of $f$ is concave up.

On $\left(- 3 , 2\right)$, we get $F ' ' \left(x\right)$ is negative, so the graph is concave down.

On $\left(2 , \infty\right)$, $F ' ' \left(x\right)$ is positive, so the graph is concave up.

The graph of $F$ is concave up on $\left(- \infty , - 3\right)$ and on $\left(2 , \infty\right)$ and it is concave down on $\left(- 3 , 2\right)$