# For what values of x is f(x)=(x-3)(x+2)(x-1) concave or convex?

May 27, 2018

Refer Explanation.

#### Explanation:

Given that: $f \left(x\right) = \left(x - 3\right) \left(x + 2\right) \left(x - 1\right)$
$\therefore$ $f \left(x\right) = \left({x}^{2} - x - 6\right) \left(x - 1\right)$
$\therefore$ $f \left(x\right) = \left({x}^{3} - {x}^{2} - 6 x - {x}^{2} + x + 6\right)$
$\therefore$ $f \left(x\right) = \left({x}^{3} - 2 {x}^{2} - 5 x + 6\right)$

By using second derivative test,

1. For the function to be concave downward:$f ' ' \left(x\right) < 0$
$f \left(x\right) = \left({x}^{3} - 2 {x}^{2} - 5 x + 6\right)$
$f ' \left(x\right) = 3 {x}^{2} - 4 x - 5$
$f ' ' \left(x\right) = 6 x - 4$
For the function to be concave downward:
$f ' ' \left(x\right) < 0$
$\therefore$$6 x - 4 < 0$
$\therefore$$3 x - 2 < 0$
$\therefore$$\textcolor{b l u e}{x < \frac{2}{3}}$

2. For the function to be concave upward:$f ' ' \left(x\right) > 0$
$f \left(x\right) = \left({x}^{3} - 2 {x}^{2} - 5 x + 6\right)$
$f ' \left(x\right) = 3 {x}^{2} - 4 x - 5$
$f ' ' \left(x\right) = 6 x - 4$
For the function to be concave upward:
$f ' ' \left(x\right) > 0$
$\therefore$$6 x - 4 > 0$
$\therefore$$3 x - 2 > 0$
$\therefore$$\textcolor{b l u e}{x > \frac{2}{3}}$