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Match the one-to-one functions with their inverse functions.

[tex]\[
\begin{array}{c}
f(x) = \frac{2x}{3} - 17 \quad f(x) = x - 10 \quad f(x) = \sqrt[3]{2x} \quad f(x) = \frac{x}{5} \\
\text{Inverse Function} \\
f^{-1}(x) = 5x \\
f^{-1}(x) = \frac{x^3}{2} \\
f^{-1}(x) = x + 10 \\
f^{-1}(x) = \frac{3(x + 17)}{2}
\end{array}
\][/tex]

[tex]\[
\begin{array}{cc}
\text{Function} & \text{Inverse Function} \\
f(x) = \frac{2x}{3} - 17 & f^{-1}(x) = \frac{3(x + 17)}{2} \\
f(x) = x - 10 & f^{-1}(x) = x + 10 \\
f(x) = \sqrt[3]{2x} & f^{-1}(x) = \frac{x^3}{2} \\
f(x) = \frac{x}{5} & f^{-1}(x) = 5x \\
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
Let's match the given functions with their inverse functions step-by-step.

1. For the function [tex]\( f(x)=\frac{2x}{3}-17 \)[/tex]:

To find the inverse, we need an expression such that when we apply [tex]\( f \)[/tex] to it, we get [tex]\( x \)[/tex] back.

Given answer:
[tex]\[ f^{-1}(x) = \frac{3(x+17)}{2} \][/tex]

So, the inverse function for [tex]\( f(x)=\frac{2x}{3}-17 \)[/tex] is:

[tex]\[ f^{-1}(x) = \frac{3(x+17)}{2} \][/tex]

2. For the function [tex]\( f(x)=x-10 \)[/tex]:

Given answer:
[tex]\[ f^{-1}(x) = x + 10 \][/tex]

So, the inverse function for [tex]\( f(x)=x-10 \)[/tex] is:

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

3. For the function [tex]\( f(x)=\sqrt[3]{2x} \)[/tex]:

Given answer:
[tex]\[ f^{-1}(x) = \frac{x^3}{2} \][/tex]

So, the inverse function for [tex]\( f(x)=\sqrt[3]{2x} \)[/tex] is:

[tex]\[ f^{-1}(x) = \frac{x^3}{2} \][/tex]

4. For the function [tex]\( f(x)=\frac{x}{5} \)[/tex]:

Given answer:
[tex]\[ f^{-1}(x) = 5x \][/tex]

So, the inverse function for [tex]\( f(x)=\frac{x}{5} \)[/tex] is:

[tex]\[ f^{-1}(x) = 5x \][/tex]

Now let's put it all together:

[tex]\[ \begin{array}{c|c} \text{Function} & \text{Inverse Function} \\ \hline f(x)=\frac{2x}{3}-17 & f^{-1}(x)=\frac{3(x + 17)}{2} \\ f(x)=x-10 & f^{-1}(x)=x+10 \\ f(x)=\sqrt[3]{2x} & f^{-1}(x)=\frac{x^3}{2} \\ f(x)=\frac{x}{5} & f^{-1}(x)=5x \\ \end{array} \][/tex]
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