Question

the one-to-one functions g and h are defined as follows.
$g = {(-1, 6), (2, 3), (3, -7), (8, 1)}$
$h(x) = -4x - 3$
find the following.
$g^{-1}(3) = \square$
$h^{-1}(x) = \square$
$(h^{-1} \circ h)(-1) = \square$

Explanation:

Part 1: Find \( g^{-1}(3) \)

Step1: Recall inverse of a function

For a one - to - one function \( g \) with ordered pair \( (a,b) \) in \( g \), the inverse function \( g^{-1} \) will have the ordered pair \( (b,a) \).
We are given \( g=\{(- 1,6),(2,3),(3, - 7),(8,1)\} \). We need to find the value of \( x \) such that \( g(x)=3 \). Looking at the ordered pairs in \( g \), when \( x = 2 \), \( g(2)=3 \). So, by the definition of the inverse function, if \( g(2)=3 \), then \( g^{-1}(3)=2 \).

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

Step1: Let \( y = h(x) \)

We have \( h(x)=-4x - 3 \), so let \( y=-4x - 3 \).

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

First, add 3 to both sides of the equation: \( y + 3=-4x \).
Then, divide both sides by - 4: \( x=\frac{y + 3}{-4}=-\frac{y + 3}{4}=\frac{-y - 3}{4} \).

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

Since \( h^{-1}(y)=x \), we can write \( h^{-1}(x)=-\frac{x + 3}{4} \) (or \( \frac{-x - 3}{4} \)).

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

Step1: Recall the property 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, for \( x=-1 \), \( (h^{-1}\circ h)(-1)=-1 \).

Answer:

s:

• \( g^{-1}(3)=\boldsymbol{2} \)
• \( h^{-1}(x)=\boldsymbol{-\frac{x + 3}{4}} \) (or \( \frac{-x - 3}{4} \))
• \( (h^{-1}\circ h)(-1)=\boldsymbol{-1} \)