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The one-to-one functions [tex]g[/tex] and [tex]h[/tex] are defined as follows:

[tex]\[
\begin{array}{l}
g(x) = \frac{x - 10}{11} \\
h = \{(-9, 4), (-2, 3), (2, 8), (5, 2)\}
\end{array}
\][/tex]

Find the following:

[tex]\[
\begin{array}{|c|}
\hline g^{-1}(x) = \square \\
\left(g \circ g^{-1}\right)(-1) = \square \\
\hline h^{-1}(2) = \square \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
Let's address the questions step-by-step.

### 1. Finding [tex]\(g^{-1}(x)\)[/tex]

The function [tex]\(g(x)\)[/tex] is defined as:
[tex]\[ g(x) = \frac{x - 10}{11} \][/tex]

To find the inverse function [tex]\(g^{-1}(x)\)[/tex], we need to solve [tex]\(y = g(x)\)[/tex] for [tex]\(x\)[/tex]. So, we start by setting [tex]\(y = \frac{x - 10}{11}\)[/tex]:

[tex]\[ y = \frac{x - 10}{11} \][/tex]

Next, we solve for [tex]\(x\)[/tex]:

[tex]\[ y \cdot 11 = x - 10 \][/tex]
[tex]\[ 11y = x - 10 \][/tex]
[tex]\[ x = 11y + 10 \][/tex]

So, the inverse function [tex]\(g^{-1}(x)\)[/tex] is:

[tex]\[ g^{-1}(x) = 11x + 10 \][/tex]

### 2. Calculating [tex]\((g \circ g^{-1})(-1)\)[/tex]

[tex]\((g \circ g^{-1})(x)\)[/tex] means applying [tex]\(g\)[/tex] to the result of [tex]\(g^{-1}(x)\)[/tex]:

First, we find [tex]\(g^{-1}(-1)\)[/tex]:

[tex]\[ g^{-1}(-1) = 11(-1) + 10 \][/tex]
[tex]\[ g^{-1}(-1) = -11 + 10 \][/tex]
[tex]\[ g^{-1}(-1) = -1 \][/tex]

Now, we use this result in [tex]\(g(x)\)[/tex]:

[tex]\[ g(g^{-1}(-1)) = g(-1) \][/tex]

Since [tex]\(g(x) = \frac{x - 10}{11}\)[/tex]:

[tex]\[ g(-1) = \frac{-1 - 10}{11} \][/tex]
[tex]\[ g(-1) = \frac{-11}{11} \][/tex]
[tex]\[ g(-1) = -1 \][/tex]

Hence:
[tex]\[ (g \circ g^{-1})(-1) = -1 \][/tex]

### 3. Finding [tex]\(h^{-1}(2)\)[/tex]

The function [tex]\(h\)[/tex] is given by the set of pairs [tex]\(\{ (-9,4), (-2,3), (2,8), (5,2) \}\)[/tex]. To find the inverse function [tex]\(h^{-1}(x)\)[/tex], we swap each pair in the set, mapping each output back to its input:
[tex]\[ h^{-1} = \{ (4,-9), (3,-2), (8,2), (2,5) \} \][/tex]

We need to find [tex]\(h^{-1}(2)\)[/tex], which is the input in [tex]\(h\)[/tex] that corresponds to the output 2:

From the pair [tex]\((2, 5)\)[/tex] in [tex]\(h^{-1}\)[/tex], we see:
[tex]\[ h^{-1}(2) = 5 \][/tex]

### Final Results:

[tex]\[ \begin{array}{|c|} \hline g^{-1}(x) = 11x + 10 \\ \left(g \circ g^{-1}\right)(-1) = -1 \\ \hline h^{-1}(2) = 5 \\ \hline \end{array} \][/tex]

So, we found that:
- [tex]\( g^{-1}(x) = 11x + 10 \)[/tex]
- [tex]\((g \circ g^{-1})(-1) = -1\)[/tex]
- [tex]\(h^{-1}(2) = 5\)[/tex]
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