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

Feb 13, 2018

$\setminus$

 \qquad \quad "concave up:" \qquad \quad \quad \quad \ \ "on the interval" \quad ( 2, \infty );
$\setminus q \quad \setminus \quad \text{concave down:" \qquad \quad \ "on the interval} \setminus \quad \left(- \setminus \infty , 2\right) .$

#### Explanation:

$\setminus$

$\text{To answer the question, we want to find where the second}$
$\text{derivative of the function is positive, and where it is negative.}$

$\text{We are given:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus \left(2 x - 2\right) \left(x - 4\right) \left(3 x - 3\right) .$

$\text{Taking a second look at the function, we see we can rewrite it as:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus 2 \left(x - 1\right) \left(x - 4\right) \setminus \cdot 3 \left(x - 1\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus 6 \left(x - 4\right) {\left(x - 1\right)}^{2.}$

$\text{This makes it easier to differentiate, starting with the}$
$\text{Product Rule:}$

$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(\left(x - 4\right) \left[{\left(x - 1\right)}^{2}\right] ' + \left[\left(x - 4\right)\right] ' {\left(x - 1\right)}^{2}\right)$

$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(\left(x - 4\right) \left[2 \left(x - 1\right) \setminus \cdot 1\right] + \left[1\right] {\left(x - 1\right)}^{2}\right)$

$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(2 \left(x - 4\right) \left(x - 1\right) + {\left(x - 1\right)}^{2}\right)$

$\text{Factor out" \ ( x - 1 ) \ "from the inside:}$

$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(x - 1\right) \left[2 \left(x - 4\right) + {\left(x - 1\right)}^{1}\right]$

$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(x - 1\right) \left[2 x - 8 + x - 1\right]$

$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 6 \setminus \cdot \left(x - 1\right) \left[3 x - 9\right] \setminus = \setminus 6 \setminus \cdot \left(x - 1\right) \setminus \cdot 3 \setminus \cdot \left(x - 3\right)$

$\setminus q \quad \setminus \quad f ' \left(x\right) \setminus = \setminus 18 \left(x - 1\right) \left(x - 3\right) .$

$\text{Now calculate" \ f''(x)":}$

$\setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus \left[18 \left(x - 1\right) \left(x - 3\right)\right] '$

$\setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 18 \setminus \cdot \left[\left(x - 1\right) \left(x - 3\right)\right] '$

$\setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 18 \setminus \cdot \left[\left(x - 1\right) \left(x - 3\right) ' + \left(x - 1\right) ' \left(x - 3\right)\right]$

$\setminus q \quad \setminus \quad f ' ' \left(x\right) \setminus = \setminus 18 \setminus \cdot \left[\left(x - 1\right) \setminus \cdot 1 + 1 \setminus \cdot \left(x - 3\right)\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus 18 \setminus \cdot \left[x - 1 + x - 3\right] \setminus = \setminus 18 \setminus \cdot \left(2 x - 4\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus 18 \setminus \cdot 2 \setminus \cdot \left(x - 2\right) \setminus = \setminus 36 \setminus \cdot \left(x - 2\right)$

$\setminus$

$\therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus f ' ' \left(x\right) \setminus = \setminus 36 \setminus \cdot \left(x - 2\right) .$

$\setminus$

$\text{Now for concavity, we find where" \ f''(x) \ "is positive, and where}$
$\text{it's negative.}$

 \qquad \qquad \qquad \qquad \quad "concave up:" \qquad \quad \ \quad \ 36 \cdot ( x - 2 ) > 0;
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{concave down:} \setminus q \quad \setminus 36 \setminus \cdot \left(x - 2\right) < 0.$

$\text{There are several ways to solve such inequalities. With the}$
$\text{expression as simple as it is, probably the easiest method is to}$
$\text{do it directly.}$

$\text{As 36 is positive, we can divide through both sides of the}$
$\text{inequalities by 36, leaving us with:}$

 \qquad \qquad \qquad \qquad \quad "concave up:" \qquad \quad \quad \ \ x - 2 > 0;
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{concave down:} \setminus q \quad \setminus x - 2 < 0.$

$\text{So:}$

 \qquad \qquad \qquad \quad "concave up:" \qquad \quad \quad \ \ x > 2, \quad "or the interval" \quad ( 2, \infty );
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{concave down:" \qquad \ x < 2, \quad "or the interval} \setminus \quad \left(- \setminus \infty , 2\right) .$

$\text{This our answer.}$

$\setminus$

$\text{Summarizing:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus \left(2 x - 2\right) \left(x - 4\right) \left(3 x - 3\right) :$

 \qquad \qquad \quad "is concave up:" \qquad \quad \quad \quad \ \ "on the interval" \quad ( 2, \infty );
$\setminus q \quad \setminus q \quad \setminus \quad \text{is concave down:" \qquad \quad \ "on the interval} \setminus \quad \left(- \setminus \infty , 2\right) .$