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

Aug 19, 2015

Investigate the sign of the second derivative.
$f$ is concave down on $\left(- \infty , \frac{1}{2}\right)$ and concave up on $\left(\frac{1}{2} , \infty\right)$.
$\left(\frac{1}{2} , - \frac{11}{2}\right)$ is the inflection point for the graph of $f$.

#### Explanation:

$f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 12 x + 1$

$f ' \left(x\right) = 6 {x}^{2} - 6 x - 12$

$f ' ' \left(x\right) = 12 x - 6$

$f ' ' \left(x\right)$ is never undefined and is $0$ only at $x = \frac{1}{2}$

So the only $x$ value at which the concavity might change is $\frac{1}{2}$

For $x < \frac{1}{2}$ we find that $f ' ' \left(x\right) < 0$ and for $x > \frac{1}{2}$, $f ' ' \left(x\right) > 0$

So the graph of $f$ is concave down on $\left(- \infty , \frac{1}{2}\right)$ and concave up on $\left(\frac{1}{2} , \infty\right)$.

$f \left(\frac{1}{2}\right) = 2 {\left(\frac{1}{2}\right)}^{3} - 3 {\left(\frac{1}{2}\right)}^{2} - 12 \left(\frac{1}{2}\right) + 1 = - \frac{11}{2}$

$\left(\frac{1}{2} , - \frac{11}{2}\right)$ is the only inflection point for the graph of $f$.