Question

the one-to-one functions g and h are defined as follows. g={(-7, 9), (1, 3), (6, 7), (7, -4), (8, 2)} h(x)=3x - 14 find the following. g⁻¹(7)=□ h⁻¹(x)=□ (h⁻¹ ∘ h)(5)=□

Explanation:

Part 1: Find \( g^{-1}(7) \)

Step1: Recall inverse function definition

For a one - to - one function \( g \) with ordered pair \( (a,b) \) in \( g \), \( g^{-1}(b)=a \).
We look at the set of ordered pairs for \( g=\{(-7,9),(1,3),(6,7),(7, - 4),(8,2)\} \). We want to find \( g^{-1}(7) \), so we find the ordered pair where the second element is 7. The ordered pair is \( (6,7) \), so by the definition of the inverse function, \( g^{-1}(7) = 6 \).

Part 2: Find \( h^{-1}(x) \)

Step1: Start with \( y = h(x) \)

Given \( h(x)=3x - 14 \), let \( y=3x - 14 \).

Step2: Solve for \( x \) in terms of \( y \)

Add 14 to both sides: \( y + 14=3x \).
Then divide both sides by 3: \( x=\frac{y + 14}{3}=\frac{1}{3}y+\frac{14}{3} \).

Step3: Replace \( x \) with \( h^{-1}(y) \) and \( y \) with \( x \)

Since \( h^{-1}(x) \) is the inverse of \( h(x) \), we have \( h^{-1}(x)=\frac{x + 14}{3} \) (or \( \frac{1}{3}x+\frac{14}{3} \)).

Part 3: Find \( (h^{-1}\circ h)(5) \)

Step1: Recall composition of inverse functions

For a function \( h \) and its inverse \( h^{-1} \), the composition \( (h^{-1}\circ h)(x)=x \) for all \( x \) in the domain of \( h \).
So, when \( x = 5 \), \( (h^{-1}\circ h)(5)=5 \).

Answer:

s:

• \( g^{-1}(7)=\boldsymbol{6} \)
• \( h^{-1}(x)=\boldsymbol{\frac{x + 14}{3}} \) (or \( \boldsymbol{\frac{1}{3}x+\frac{14}{3}} \))
• \( (h^{-1}\circ h)(5)=\boldsymbol{5} \)