[tex]\[
\begin{array}{l}
g=\{(-8,8),(-2,1),(4,9),(7,4)\} \\
h(x)=3x+8
\end{array}
\][/tex]

Find the following:

[tex]\[
\begin{array}{|c|}
\hline g^{-1}(4)=\square \\
\hline h^{-1}(x)=\square \\
\left(h^{-1} \circ h\right)(-4)=\square \\
\hline
\end{array}
\][/tex]



Answer :

Sure! Let's solve this step by step.

### Step 1: Finding [tex]\( g^{-1}(4) \)[/tex]

The inverse function [tex]\( g^{-1} \)[/tex] takes a [tex]\( y \)[/tex] value from the set and returns the corresponding [tex]\( x \)[/tex] value such that [tex]\( g(x) = y \)[/tex]. We need to find [tex]\( g^{-1}(4) \)[/tex].

Given the set [tex]\( g = \{(-8, 8), (-2, 1), (4, 9), (7, 4)\} \)[/tex]:

- From the set, we can see that [tex]\( g(7) = 4 \)[/tex].

So, [tex]\( g^{-1}(4) = 7 \)[/tex].

### Step 2: Finding [tex]\( h^{-1}(x) \)[/tex]

The function [tex]\( h \)[/tex] is given as [tex]\( h(x) = 3x + 8 \)[/tex]. To find the inverse function [tex]\( h^{-1}(x) \)[/tex], we need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex], where [tex]\( y = h(x) \)[/tex].

Starting from:
[tex]\[ y = 3x + 8 \][/tex]

To isolate [tex]\( x \)[/tex], we perform the following steps:
1. Subtract 8 from both sides:
[tex]\[ y - 8 = 3x \][/tex]

2. Divide both sides by 3:
[tex]\[ x = \frac{y - 8}{3} \][/tex]

Thus, the inverse function is:
[tex]\[ h^{-1}(x) = \frac{x - 8}{3} \][/tex]

### Step 3: Calculating [tex]\((h^{-1} \circ h)(-4)\)[/tex]

The composition [tex]\( (h^{-1} \circ h)(x) \)[/tex] means we first apply [tex]\( h \)[/tex] to [tex]\( x \)[/tex], and then apply [tex]\( h^{-1} \)[/tex] to the result.

Let's evaluate it step by step for [tex]\( x = -4 \)[/tex]:

1. First, calculate [tex]\( h(-4) \)[/tex]:
[tex]\[ h(-4) = 3(-4) + 8 = -12 + 8 = -4 \][/tex]

2. Next, apply [tex]\( h^{-1} \)[/tex] to the result:
[tex]\[ h^{-1}(-4) = \frac{-4 - 8}{3} = \frac{-12}{3} = -4 \][/tex]

So, [tex]\((h^{-1} \circ h)(-4) = -4\)[/tex].

### Summary of the results:
[tex]\[ g^{-1}(4) = 7 \][/tex]
[tex]\[ h^{-1}(x) = \frac{x - 8}{3} \][/tex]
[tex]\[ (h^{-1} \circ h)(-4) = -4 \][/tex]

Thus, the filled in table is:

[tex]\[ \begin{array}{|c|} \hline g^{-1}(4)=7 \\ \hline h^{-1}(x)=\frac{x-8}{3} \\ \left(h^{-1} \circ h\right)(-4)=-4 \\ \hline \end{array} \][/tex]