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

Feb 24, 2018

$f \left(x\right)$ is convex on the interval (-∞,2); $f \left(x\right)$ is concave on the interval (2, ∞)

#### Explanation:

Find the second derivative, $f ' ' \left(x\right)$:

$f ' \left(x\right) = \left(x - 3\right) \left(x - 1\right) \frac{d}{\mathrm{dx}} \left(x - 2\right) + \left(x - 3\right) \left(\frac{d}{\mathrm{dx}} \left(x - 1\right)\right) \left(x - 2\right) + \left(\frac{d}{\mathrm{dx}} \left(x - 3\right)\right) \left(x - 1\right) \left(x - 2\right)$

$f ' \left(x\right) = \left(x - 3\right) \left(x - 1\right) + \left(x - 3\right) \left(x - 2\right) + \left(x - 1\right) \left(x - 2\right)$

f''(x)=(x-3)(d/dx(x-1))+(x-1)(d/dx(x-3))+(x-3)((d/dx(x-2))+(x-2)(d/dx(x-3))+(x-1)(d/dx(x-2))+(x-2)((d/dx(x-1))

$f ' ' \left(x\right) = \left(x - 3\right) + \left(x - 1\right) + \left(x - 3\right) + \left(x - 2\right) + \left(x - 1\right) + \left(x - 2\right)$

$f ' ' \left(x\right) = 2 \left(x - 3\right) + 2 \left(x - 1\right) + 2 \left(x - 2\right)$

Set $f ' ' \left(x\right) = 0$ and solve for $x :$

$2 \left(x - 3\right) + 2 \left(x - 1\right) + 2 \left(x - 2\right) = 0$

$2 x - 6 + 2 x - 2 + 2 x - 4 = 0$

$6 x - 12 = 0$
$6 x = 12$
$x = \frac{12}{6} = 2$

The domain of $f \left(x\right)$ is (-∞,∞), as all polynomials are continuous. Let's break up the domain of $f \left(x\right)$ around the value of $x$ we've found:

(-∞,2), (2,∞)

Now, we must determine whether $f ' ' \left(x\right)$ is positive or negative in each of these intervals. If $f ' ' \left(x\right) > 0$ in an interval, $f \left(x\right)$ is concave on that interval. If $f ' ' \left(x\right) < 0$ on an interval, $f \left(x\right)$ is convex on that interval.

(-∞,2):

$f ' ' \left(0\right) = 2 \left(- 3\right) + 2 \left(- 1\right) + 2 \left(- 2\right) < 0$

$f \left(x\right)$ is convex on the interval (-∞,2)

(2,∞):

$f ' ' \left(3\right) = 2 \left(0\right) + 2 \left(2\right) + 2 > 0$

$f \left(x\right)$ is concave on the interval (2, ∞)