Consider the function f(x)X +1Find the intervals on which f(x) is concave up or down and all the inflection points. Choose all of the following that are correct.f(x) is concave up on (- , - 3) and concave down on (-3,+ x).O a.f(x) has an inflection point at x = - 3.Ob.f(x) has no inflection points.O c.f(x) is concave up on (-3,0) and (0, + ) and concave down on (- oo,-3).O d.f(x) has an inflection point at x = 0.O e.

Answered: Consider the function f(x) = x + 1…

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter3: Functions

Section3.3: More On Functions; Piecewise-defined Functions

Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...

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Question

Consider the function f(x)
X +1
Find the intervals on which f(x) is concave up or down and all the inflection points. Choose all of the following that are correct.
f(x) is concave up on (- , - 3) and concave down on (-3,+ x).
O a.
f(x) has an inflection point at x = - 3.
Ob.
f(x) has no inflection points.
O c.
f(x) is concave up on (-3,0) and (0, + ) and concave down on (- oo,-3).
O d.
f(x) has an inflection point at x = 0.
O e.

Expert Solution

Step 1

Given,

$f (x) = \frac{x + 1}{x^{2}}$

Step 2

To find an inflection point for the given function:

An inflection point is a point on the graph where it changes its concavity.

A point of inflection occurs where the second derivative equal to zero.

consider,

$f (x) = \frac{x + 1}{x^{2}}$

Differentiate with respect to x we get,

Use quotient rule, $\frac{d}{d x} \frac{u}{v} = \frac{u' v - v' u}{v^{2}}$

$\frac{d}{d x} f (x) = \frac{d}{d x} (\frac{x + 1}{x^{2}})$

$= \frac{1 (x^{2}) - 2 x (x + 1)}{{(x^{2})}^{2}} = \frac{x^{2} - 2 x^{2} - 2 x}{x^{4}} = \frac{- x^{2} - 2 x}{x^{4}}$

$f' (x) = \frac{- x^{2} - 2 x}{x^{4}}$

Again differentiate with respect to x we get,

Use quotient rule, $\frac{d}{d x} \frac{u}{v} = \frac{u' v - v' u}{v^{2}}$

$\frac{d}{d x} f' (x) = \frac{d}{d x} (\frac{- x^{2} - 2 x}{x^{4}})$

$= \frac{(- 2 x - 2) (x^{4}) - (4 x^{3}) (- x^{2} - 2 x)}{{(x^{4})}^{2}} = \frac{- 2 x^{5} - 2 x^{4} + 4 x^{5} + 8 x^{4}}{x^{8}} = \frac{2 x^{5} + 6 x^{4}}{x^{8}} = \frac{x^{4} (2 x + 6)}{x^{8}} = \frac{2 x + 6}{x^{4}}$

$f'' (x) = \frac{2 x + 6}{x^{4}}$

Put f''(x)=0 we get,

$\frac{2 x + 6}{x^{4}} = 0$

$2 x + 6 = 0 2 x = - 6 x = \frac{- 6}{2} x = - 3$

and

$x^{4} = 0 x = 0$

Therefore the inflection points are x=-3 and x=0

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