# For what values of x is f(x)= 1/2x^4-6x^3-48x^2-96x+2 concave or convex?

Apr 1, 2017

$f \left(x\right)$ is convex when $x$ is $\left(- \infty , 2\right) \cup \left(8 , \infty\right)$.

$f \left(x\right)$ is concave when $x$ is $\left(- 2 , 8\right)$.

#### Explanation:

When looking for the concavity of a function, it's best to find the second derivative of the function. The second derivative, $f ' ' \left(x\right)$, tells us the concavity of the function, $f \left(x\right)$:
When $f ' ' \left(x\right) < 0$, $f \left(x\right)$ is concave down (concave)
When $f ' ' \left(x\right) > 0$, $f \left(x\right)$ is concave up (convex)

The first derivative of this function is:
$f ' \left(x\right)$ = $2 {x}^{3} - 18 {x}^{2} - 96 x - 96$.

The second derivative is:
$f ' ' \left(x\right)$ = $6 {x}^{2} - 36 x - 96$.

Then set the second derivative equal to 0, to get:
$6 {x}^{2} - 36 x - 96 = 0$
${x}^{2} - 6 x - 16 = 0$
$\left(x - 8\right) \left(x + 2\right) = 0$
$x = - 2 , 8$

This tells us that the concavity of $f \left(x\right)$ changes at $x = - 2$ and $x = 8$. When $x < - 2$, the $f ' ' \left(x\right)$ is positive. When $x > - 2$ and $x < 8$ (i.e. when $x$ is between $- 2$ and $8$) $f ' ' \left(x\right)$ is negative. When $x > 8$, $f ' ' \left(x\right)$ is positive. It can help to look at a graph:
graph{x^2-6x-16 [-8.67, 11.33, -5.04, 4.96]}

$f \left(x\right)$ is convex when $f ' ' \left(x\right)$ is positive, which is when $x$ is $\left(- \infty , 2\right) \cup \left(8 , \infty\right)$.

$f \left(x\right)$ is concave when $f ' ' \left(x\right)$ is negative, which is when $x$ is $\left(- 2 , 8\right)$.