# For what values of x is f(x)=3x^3+2x^2-x+9 concave or convex?

Feb 20, 2018

$\text{graph of" \ f(x) \ "is concave up on the interval:} \setminus q \quad \setminus \quad \left(- \frac{2}{9} , + \infty\right)$

$\text{graph of" \ f(x) \ "is concave down on the interval:} \setminus \left(- \infty , - \frac{2}{9}\right) .$

#### Explanation:

$\text{First recall the fundamental results about the concavity of the}$
$\text{graph of a function" \ f(x) ":}$

$\setminus q \quad \setminus \quad \setminus f ' ' \left(x\right) > 0 \setminus \quad \implies \setminus \quad \text{the graph of" \ \ f(x) \ \ "is concave up;}$

$\setminus q \quad \setminus \quad \setminus f ' ' \left(x\right) < 0 \setminus \quad \implies \setminus \quad \text{the graph of" \ \ f(x) \ \ "is concave down.}$

$\text{[I apologize -- with regard to concavity of the graph of a}$
$\text{function, I'm not sure I know the language: concave/convex.}$
$\text{I am used to the language: (concave up)/(concave down).}$
$\text{I hope what I provide here can help you !!]}$

$\text{So, to answer questions about the concavity of the graph of a}$
$\text{function, we need to find its second derivative first.}$

$\text{The function we are given is:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus 3 {x}^{3} + 2 {x}^{2} - x + 9.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus f ' \left(x\right) \setminus = \setminus 9 {x}^{2} + 4 x - 1.$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus \quad \setminus f ' ' \left(x\right) \setminus = \setminus 18 x + 4. \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left(1\right)$

$\text{So, we want to find where:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus f ' ' \left(x\right) > 0 \setminus q \quad \text{and} \setminus q \quad f ' ' \left(x\right) < 0.$

$\text{So, using eqn. (1), we must find where:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus \setminus 18 x + 4 > 0 \setminus q \quad \text{and} \setminus q \quad 18 x + 4 < 0.$

$\text{So, we solve these inequalities. One way to do this is to do it}$
$\text{directly, as below:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \setminus 18 x + 4 > 0 \setminus q \quad \text{and} \setminus q \quad 18 x + 4 < 0$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad 18 x > - 4 \setminus q \quad \text{and} \setminus q \quad 18 x < - 4.$

$\text{As" \ \ 18 \ \ "is positive, we can divide through both sides of these}$
$\text{inequalities by" \ 18, "without changing the order of the}$
$\text{inequality sign:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad x > - \frac{4}{18} \setminus q \quad \text{and} \setminus q \quad x < - \frac{4}{18}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad x > - \frac{2}{9} \setminus q \quad \text{and} \setminus q \quad x < - \frac{2}{9.}$

$\text{Thus, we have:}$

$x > - \frac{2}{9} \setminus \setminus \implies \setminus \setminus f ' ' \left(x\right) > 0 \setminus \implies \setminus \text{graph of" \ f(x) \ "is concave up;}$

$x < - \frac{2}{9} \setminus \setminus \implies \setminus \setminus f ' ' \left(x\right) > 0 \setminus \implies \setminus \text{graph of" \ f(x) \ "is concave down.}$

$\text{Summarizing:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{graph of" \ f(x) \ "is concave up:} \setminus q \quad \setminus x > - \frac{2}{9}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \text{graph of" \ f(x) \ "is concave down:} \setminus q \quad \setminus x < - \frac{2}{9}$

$\text{In interval notation:}$

$\text{graph of" \ f(x) \ "is concave up on the interval:} \setminus q \quad \setminus \quad \left(- \frac{2}{9} , + \infty\right)$

$\text{graph of" \ f(x) \ "is concave down on the interval:} \setminus \left(- \infty , - \frac{2}{9}\right) .$

$\text{This is the answer to our question.}$