GIVEN:
We are given the set of functions as indicated in the attached image.
Required;
To determine which pair of functions are inverses of each other.
Step-by-step solution;
We shall solve each function one after the other to determine which pair are inverses of each other and which pair is not.
Option A:
[tex]\begin{gathered} f(x)=\frac{x}{8}+3 \\ \\ Re-write\text{ }as\text{ }an\text{ }equation\text{ }with\text{ }respect\text{ }to\text{ }y: \\ \\ y=\frac{x}{8}+3 \\ \\ Next,\text{ }we\text{ }switch\text{ }places\text{ }for\text{ }x\text{ }and\text{ }y: \\ \\ x=\frac{y}{8}+3 \\ \\ We\text{ }now\text{ }solve\text{ }for\text{ }y: \\ \\ x-3=\frac{y}{8} \\ \\ Cross\text{ }multiply: \\ \\ 8(x-3)=y \\ \\ 8x-24=y \\ \\ Therefore,\text{ }we\text{ }now\text{ }write\text{ }in\text{ }function\text{ }notation: \\ \\ f^{-1}(x)=8x-24 \end{gathered}[/tex]Therefore, option A is not a pair of inverse functions.
OptionB;
[tex]\begin{gathered} f(x)=\frac{4}{x}-3 \\ \\ y=\frac{4}{x}-3 \\ \\ Switch\text{ }sides\text{ }for\text{ }x\text{ }and\text{ }y; \\ \\ x=\frac{4}{y}-3 \\ \\ Solve\text{ }for\text{ }y; \\ \\ x+3=\frac{4}{y} \\ \\ Cross\text{ }multiply: \\ \\ y=\frac{4}{x+3} \\ \\ Therefore; \\ \\ f^{-1}(x)=\frac{4}{x+3} \end{gathered}[/tex]Therefore, option B is not a pair of inverse functions.
Option C:
[tex]\begin{gathered} f(x)=10x-5 \\ \\ y=10x-5 \\ \\ Switch\text{ }sides\text{ }for\text{ }x\text{ }and\text{ }y; \\ \\ x=10y-5 \\ \\ Solve\text{ }for\text{ }y; \\ \\ x+5=10y \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }10; \\ \\ \frac{x+5}{10}=y \\ \\ Therefore: \\ f^{-1}(x)=\frac{x+5}{10} \end{gathered}[/tex]Therefore, option C is a pair of inverses.
Option D:
[tex]\begin{gathered} f(x)=\sqrt[3]{6x} \\ \\ y=\sqrt[3]{6x} \\ \\ Switch\text{ }sides\text{ }for\text{ }x\text{ }and\text{ }y; \\ \\ x=\sqrt[3]{6y} \\ \\ Solve\text{ }for\text{ }y; \\ \\ Cube\text{ }both\text{ }sides; \\ \\ x^3=(\sqrt[3]{6y})^3 \\ \\ The\text{ }exponent\text{ }cancels\text{ }out\text{ }the\text{ }radical\text{ }on\text{ }the\text{ }right\text{ }side; \\ \\ x^3=6y \\ \\ Divide\text{ }both\text{ }sides\text{ }by\text{ }6; \\ \\ \frac{x^3}{6}=y \\ \\ Therefore: \\ \\ f^{-1}(x)=\frac{x^3}{6} \end{gathered}[/tex]Therefore, option D is not a pair of inverses.
ANSWER:
The pair of functions in option C are inverses of each other.
