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tive extrema algebraically. $\newline$ ayerTest. aspx?quizme= $1$ \&chapterld= $8$ \§ionld= $2$ \&objectiveld= $4$ \&studyPlanAssignmentld= $2437997$ \&viewMode= $0$ \&c... $\newline$ William Artiaga $04$ / $18$ / $24$ $12$ : $32$ AM $\newline$ (?) $\newline$ termine $\newline$ This quiz: $5$ point(s) 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 $\mathrm{x}=$ $\square$ $\newline$ (Use a comma to separate answers as needed.) $\newline$ C. The function has a relative maximum of $\square$ at $\mathrm{x}=$ $\square$ and a relative minimum of $\square$ at $\mathrm{x}=$ $\square$ $\newline$ (Use a comma to separate answers as needed.) $\newline$ D. There are no relative extrema. $\newline$ z Me $\newline$ Next $\newline$ Ylayer. aspx?cultureld= $8$ theme=math\&style=highered\&disableStandbylndicator=true\&assignmentHandilesLocale=true\&enablelesSession=.

Full solution

Q. tive extrema algebraically.

\newline

ayerTest. aspx?quizme=

1

\&chapterld=

8

\§ionld=

2

\&objectiveld=

4

\&studyPlanAssignmentld=

2437997

\&viewMode=

0

\&c...

\newline

William Artiaga

04

18

24

12

32

\newline

(?)

\newline

termine

\newline

This quiz:

5

point(s) 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

\mathrm{x}=

\square

\newline

(Use a comma to separate answers as needed.)

\newline

C. The function has a relative maximum of

\square

\mathrm{x}=

\square

and a relative minimum of

\square

\mathrm{x}=

\square

\newline

(Use a comma to separate answers as needed.)

\newline

D. There are no relative extrema.

\newline

z Me

\newline

\newline

Ylayer. aspx?cultureld=

8

theme=math\&style=highered\&disableStandbylndicator=true\&assignmentHandilesLocale=true\&enablelesSession=.

Find Derivative: To find relative extrema, we need to find the derivative of $f(x)$ and set it equal to zero to find critical points. $f'(x) = \frac{d}{dx} [5 + (4 + 3x)^{\frac{2}{3}}]$ Using the chain rule, $f'(x) = \left(\frac{2}{3}\right)(4 + 3x)^{-\frac{1}{3}} \times 3$
Simplify Derivative: Simplify the derivative: $\newline$ $f'(x) = \frac{2}{3} \times 3 \times (4 + 3x)^{-\frac{1}{3}}$ $\newline$ $f'(x) = 2 \times (4 + 3x)^{-\frac{1}{3}}$ $\newline$ Set the derivative equal to zero to find critical points: $\newline$ $2 \times (4 + 3x)^{-\frac{1}{3}} = 0$
Set Equal to Zero: Since multiplying by $2$ doesn't affect the equation, we can simplify it to: $\newline$ $(4 + 3x)^{-\frac{1}{3}} = 0$ $\newline$ However, there's a mistake here. A term raised to a power cannot be zero unless the term itself is zero. This means there's no solution for $x$ that will make the derivative zero.