# How do you find the interval where f is concave up and where f is concave down for f(x)= –(2x^3)–(3x^2)–7x+2?

Sep 9, 2015

Using the second derivative test, $f \left(x\right)$ is concave up when $x < - \frac{1}{2}$ and concave down when $x > - \frac{1}{2}$.

#### Explanation:

Concavity has to do with the second derivative of a function.

A function is concave up for the intervals where ${d}^{2} / {\mathrm{dx}}^{2} f \left(x\right) > 0$.
A function is concave down for the intervals where ${d}^{2} / {\mathrm{dx}}^{2} f \left(x\right) < 0$.

First, let's solve for the second derivative of the function.

The first derivative:

f(x)=–2x^3–3x^2–7x+2

d/dx(–2x^3–3x^2–7x+2)

$= - 2 \left(3\right) {x}^{2} - 3 \left(2\right) x - 7$

$= - 6 {x}^{2} - 6 x - 7$

The second derivative:

$f ' \left(x\right) = - 6 {x}^{2} - 6 x - 7$

$\frac{d}{\mathrm{dx}} \left(- 6 {x}^{2} - 6 x - 7\right)$

$= - 6 \left(2\right) x - 6$

$= - 12 x - 6$

Intervals where $f \left(x\right)$ is concave up:

$- 12 x - 6 > 0$
$- 12 x > 6$
$x < - \frac{1}{2}$

Intervals where $f \left(x\right)$ is concave down:

$- 12 x - 6 < 0$
$- 12 x < 6$
$x > - \frac{1}{2}$