# Find the point of inflection of the graph of the function. (If an answer does not exist, enter DNE.)?

## Find the point of inflection of the graph of the function. (If an answer does not exist, enter DNE.) f(x) = 2 − 7x^4 (x,y): DNE (answer) concave upward: DNE (answer) concave downward: need answer

Feb 24, 2018

$\setminus q \quad \setminus q \quad \text{the graph of" \ \ f(x) \ \ "is concave down on" \ \ ( -infty, infty);}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{i.e., everywhere.}$

#### Explanation:

$\text{From what I understand, you want to know about the}$
$\text{downward concavity of the graph of the function.}$
$\text{(Please correct me if I am wrong !!)}$

$\text{To investigate the concavity of a function, we will look}$
$\text{at" \ \ f''(x).}$

$\text{We are given the function:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus 2 - 7 {x}^{4.}$

$\text{We compute derivatives:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' \left(x\right) \setminus = \setminus 0 - 7 \left(4 {x}^{3}\right)$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' \left(x\right) \setminus = \setminus - 28 {x}^{3.}$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus f ' ' \left(x\right) \setminus = \setminus - 28 \left(3 {x}^{2}\right) .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus \setminus f ' ' \left(x\right) \setminus = \setminus - 84 {x}^{2.}$

$\text{Perhaps before anything else, we note that" \ \ f''(x) \ \ "is clearly}$
$\text{negative everywhere:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus f ' ' \left(x\right) \setminus < \setminus 0 , \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \text{for all} \setminus \setminus x .$

$\text{Thus, the graph of" \ \ f(x) \ \ "is concave down everywhere:}$

$\setminus q \quad \setminus q \quad \text{the graph of" \ \ f(x) \ \ "is concave down on" \ \ ( -infty, infty).}$