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ive extrema algebraically.
jerTest.aspx?quizme=1 & chapterld=8&sectionld=2&objectiveld=4&studyPlanAssignmentld=2437997&viewMode=0&c
William Artiaga 04/18/24 12:38 AM
This quiz: 5 point(s)
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This question: 1
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Find the
x-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema.

f(x)=5+(4+3x)^(2//3)
A. There are no relative minima. The function has a relative maximum of
◻ at
x=
◻
(Use a comma to separate answers as needed.)
B. There are no relative maxima. The function has a relative minimum of
◻ at
x=
◻
(Use a comma to separate answers as eded.)
C. The function has a relative maximum of
◻ at
x=
◻ and a relative minimum of
◻ at
x=
◻ 2.
(Use a comma to separate answers as needed.)
D. There are no relative extrema.
Next

ive extrema algebraically. $\newline$ jerTest.aspx?quizme= $1$ \& chapterld= $8$ \§ionld= $2$ \&objectiveld= $4$ \&studyPlanAssignmentld= $2437997$ \&viewMode= $0$ \&c $\newline$ William Artiaga $04$ / $18$ / $24$ $12$ : $38$ AM $\newline$ This quiz: $5$ point(s) $\newline$ possible $\newline$ This question: $1$ $\newline$ Resume later $\newline$ point(s) possible $\newline$ Submit quiz $\newline$ Find the $x$ -values of all points where the function has any relative extrema. Find the value(s) of any relative extrema. $\newline$ $f(x)=5+(4+3 x)^{2 / 3}$ $\newline$ A. There are no relative minima. The function has a relative maximum of $\square$ at $x=$ $\square$ $\newline$ (Use a comma to separate answers as needed.) $\newline$ B. There are no relative maxima. The function has a relative minimum of $\square$ at $x=$ $\square$ $\newline$ (Use a comma to separate answers as eded.) $\newline$ C. The function has a relative maximum of $\square$ at $\mathrm{x}=$ $\square$ and a relative minimum of $\square$ at $x=$ $\square$ $2$ . $\newline$ (Use a comma to separate answers as needed.) $\newline$ D. There are no relative extrema. $\newline$ Next

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Grade 7

One-step inequalities: word problems

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Q. ive extrema algebraically.

\newline

jerTest.aspx?quizme=

1

\& chapterld=

8

\§ionld=

2

\&objectiveld=

4

\&studyPlanAssignmentld=

2437997

\&viewMode=

0

\&c

\newline

William Artiaga

04

18

24

12

38

\newline

This quiz:

5

point(s)

\newline

possible

\newline

This question:

1

\newline

Resume later

\newline

point(s) possible

\newline

Submit quiz

\newline

Find the

x

-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema.

\newline

f(x)=5+(4+3 x)^{2 / 3}

\newline

A. There are no relative minima. The function has a relative maximum of

\square

x=

\square

\newline

(Use a comma to separate answers as needed.)

\newline

B. There are no relative maxima. The function has a relative minimum of

\square

x=

\square

\newline

(Use a comma to separate answers as eded.)

\newline

C. The function has a relative maximum of

\square

\mathrm{x}=

\square

and a relative minimum of

\square

x=

\square

2

\newline

(Use a comma to separate answers as needed.)

\newline

D. There are no relative extrema.

\newline

Find Derivative: To find relative extrema, we first need to find the derivative of the function $f(x) = 5 + (4 + 3x)^{\frac{2}{3}}$ .
Apply Chain Rule: Using the chain rule, the derivative $f'(x)$ is $(\frac{2}{3})(4 + 3x)^{-\frac{1}{3}} \times 3$ .
Simplify Derivative: Simplify $f'(x)$ to get $f'(x) = 2 \cdot (4 + 3x)^{-\frac{1}{3}}$ .
Find Critical Points: Set the derivative equal to zero to find critical points: $2 \times (4 + 3x)^{-\frac{1}{3}} = 0$ .
No Critical Points: Since the expression $(4 + 3x)^{-\frac{1}{3}}$ cannot be zero, there are no solutions for $x$ . Therefore, there are no critical points where the derivative is zero.
Check Derivative Existence: Check for any points where the derivative does not exist. The derivative does not exist when the inside of the cube root, $4 + 3x$ , is zero. So, set $4 + 3x = 0$ .
Solve for x: Solve for x to find $x = -\frac{4}{3}$ .
Analyze Behavior: Since the derivative does not exist at $x = -\frac{4}{3}$ , this could be a point of relative extrema. We need to analyze the behavior of the function around this point.
Check Sign of Derivative: Check the sign of the derivative before and after $x = -\frac{4}{3}$ . If the sign changes, then $x = -\frac{4}{3}$ is a relative extrema.
No Relative Extrema: For $x < -\frac{4}{3}$ , the derivative is positive, and for $x > -\frac{4}{3}$ , the derivative is also positive. Since the sign of the derivative does not change, there is no relative extrema at $x = -\frac{4}{3}$ .