# For what values of x is f(x)= -5x^3+x^2+4x-12  concave or convex?

Jan 8, 2016

$f \left(x\right)$ is convex on $\left(- \infty , \frac{1}{15}\right)$, concave on $\left(\frac{1}{15} , + \infty\right)$, and has a point of inflection when $x = \frac{1}{15}$.

#### Explanation:

$f \left(x\right)$ is convex when $f ' ' \left(x\right) > 0$.
$f \left(x\right)$ is concave when $f ' ' \left(x\right) < 0$.

Find $f ' ' \left(x\right)$:

$f \left(x\right) = - 5 {x}^{3} + {x}^{2} + 4 x - 12$

$f ' \left(x\right) = - 15 {x}^{2} + 2 x + 4$

$f ' ' \left(x\right) = - 30 x + 2$

The concavity could change when $f ' ' \left(x\right) = 0$. This is a possible point of inflection.

$f ' ' \left(x\right) = 0$

$- 30 x + 2 = 0$

$x = \frac{1}{15}$

Analyze the sign surrounding the point $x = \frac{1}{15}$. You can plug in test points to determine the sign.

When $x < \frac{1}{15}$, $f ' ' \left(x\right) > 0$.

When $x > \frac{1}{15}$, $f ' ' \left(x\right) < 0$.

When $x = 15$, $f ' ' \left(x\right) = 0$.

Thus, $f \left(x\right)$ is convex on $\left(- \infty , \frac{1}{15}\right)$, concave on $\left(\frac{1}{15} , + \infty\right)$, and has a point of inflection when $x = \frac{1}{15}$.

Graph of $f \left(x\right)$:

graph{-5x^3+x^2+4x-12 [-22.47, 28.85, -20.56, 5.1]}

The concavity does seem to shift very close to $x = 0$.