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

[tex]\[ g = \{(0,-2), (1,7), (4,-9), (9,4)\} \][/tex]
[tex]\[ h(x) = 3x - 10 \][/tex]

Find the following:

\begin{tabular}{|c|}
\hline
[tex]$g^{-1}(4) = -9$[/tex] \\
[tex]$h^{-1}(x) = \square$[/tex] \\
[tex]$\left(h^{-1} \circ h\right)(5) = \square$[/tex] \\
\hline
\end{tabular}

Sagot :

Let's solve the problem step by step.

1. Find [tex]\( g^{-1}(4) \)[/tex]:

- The function [tex]\( g \)[/tex] is given by the set of ordered pairs [tex]\( \{(0,-2), (1,7), (4,-9), (9,4)\} \)[/tex].
- To find the inverse function value [tex]\( g^{-1}(4) \)[/tex], we look for the pair where [tex]\( 4 \)[/tex] is the input (first element) in the function [tex]\( g \)[/tex].
- Checking the pairs in [tex]\( g \)[/tex], we find that [tex]\( g(4) = -9 \)[/tex].
- Thus, the inverse function [tex]\( g^{-1} \)[/tex] will give us the value [tex]\( g^{-1}(4) = -9 \)[/tex].

2. Find the inverse function [tex]\( h^{-1}(x) \)[/tex]:

- The function [tex]\( h(x) \)[/tex] is given by [tex]\( h(x) = 3x - 10 \)[/tex].
- To find the inverse [tex]\( h^{-1}(x) \)[/tex], we need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
- Start with the equation [tex]\( y = 3x - 10 \)[/tex].
- Add 10 to both sides: [tex]\( y + 10 = 3x \)[/tex].
- Now, divide both sides by 3: [tex]\( x = \frac{y + 10}{3} \)[/tex].
- Therefore, the inverse function [tex]\( h^{-1}(x) \)[/tex] is [tex]\( h^{-1}(x) = \frac{x + 10}{3} \)[/tex].

3. Evaluate [tex]\( \left(h^{-1} \circ h\right)(5) \)[/tex]:

- The composition [tex]\( (h^{-1} \circ h)(x) \)[/tex] means first applying [tex]\( h \)[/tex] to [tex]\( x \)[/tex], and then applying [tex]\( h^{-1} \)[/tex] to the result.
- Let's find [tex]\( h(5) \)[/tex]:
[tex]\[ h(5) = 3 \cdot 5 - 10 = 15 - 10 = 5 \][/tex]
- Now apply [tex]\( h^{-1} \)[/tex] to [tex]\( h(5) \)[/tex]:
[tex]\[ h^{-1}(5) = \frac{5 + 10}{3} = \frac{15}{3} = 5 \][/tex]
- Thus, [tex]\( \left(h^{-1} \circ h\right)(5) = 5 \)[/tex].

Summarizing the results:

[tex]\[ \begin{array}{|c|} \hline g^{-1}(4) = -9 \\ h^{-1}(x) = \frac{x + 10}{3} \\ \left(h^{-1} \circ h\right)(5) = 5 \\ \hline \end{array} \][/tex]