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

Jan 23, 2016

$f \left(x\right) = \left(x - 3\right) \left(x + 2\right) \left(3 x - 2\right)$
$\implies f \left(x\right) = \left({x}^{2} - x - 6\right) \left(3 x - 2\right)$
$\implies f \left(x\right) = 3 {x}^{3} - 5 {x}^{2} - 4 x + 12$

If $f \left(x\right)$ is a function and $f ' ' \left(x\right)$ is the second derivative of the function then,

$\left(i\right) f \left(x\right)$ is concave if $f \left(x\right) < 0$
$\left(i i\right) f \left(x\right)$ is convex if $f \left(x\right) > 0$

Here $f \left(x\right) = 3 {x}^{3} - 5 {x}^{2} - 4 x + 12$ is a function.

Let $f ' \left(x\right)$ be the first derivative.
$\implies f ' \left(x\right) = 9 {x}^{2} - 10 x - 4$

Let $f ' ' \left(x\right)$ be the second derivative.
$\implies f ' ' \left(x\right) = 18 x - 10$

$f \left(x\right)$ is concave if $f ' ' \left(x\right) < 0$
$\implies 18 x - 10 < 0$
$\implies 9 x - 5 < 0$
$\implies x < \frac{5}{9}$

Hence, $f \left(x\right)$ is concave for all values belonging to $\left(- \infty , \frac{5}{9}\right)$

$f \left(x\right)$ is convex if $f ' ' \left(x\right) > 0$.
$\implies 18 x - 10 > 0$
$\implies 9 x - 5 > 0$
$\implies x > \frac{5}{9}$

Hence, $f \left(x\right)$ is convex for all values belonging to $\left(\frac{5}{9} , \infty\right)$