For the function [tex] f [/tex], use the composition of functions to show that [tex] f^{-1} [/tex] is as given.

[tex]\[ f(x) = \frac{5}{7} x + 1, \quad f^{-1}(x) = \frac{7x - 7}{5} \][/tex]

To show that [tex] f^{-1}(x) = \frac{7x - 7}{5} [/tex], find [tex] (f^{-1} \circ f)(x) [/tex] and [tex] (f \circ f^{-1})(x) [/tex] and check to see that each is [tex] x [/tex].

Begin with [tex] (f^{-1} \circ f)(x) [/tex]:

[tex]\[
\begin{aligned}
(f^{-1} \circ f)(x) & = f^{-1}(f(x)) \\
& = f^{-1}\left( \frac{5}{7} x + 1 \right) \\
& = \square
\end{aligned}
\][/tex]

Definition of composition of functions:
Substitute [tex] f(x) = \frac{5}{7} x + 1 [/tex]
[tex] \square [/tex]



Answer :

To show that \(f^{-1}(x) = \frac{7x - 7}{5}\) is the inverse of \(f(x) = \frac{5}{7}x + 1\), we need to verify the following two compositions:

1. \((f^{-1} \circ f)(x) = x\)
2. \((f \circ f^{-1})(x) = x\)

We'll start with the first composition.

### 1. \((f^{-1} \circ f)(x)\)

[tex]\[ \begin{aligned} (f^{-1} \circ f)(x) & = f^{-1}(f(x)) \\ & = f^{-1}\left(\frac{5}{7}x + 1\right). \end{aligned} \][/tex]

Next, we substitute the expression for \(f(x)\) into the inverse function \(f^{-1}(y) = \frac{7y - 7}{5}\).

[tex]\[ \begin{aligned} f^{-1}\left(\frac{5}{7}x + 1\right) &= \frac{7 \left(\frac{5}{7}x + 1\right) - 7}{5}\\ & = \frac{7 \cdot \frac{5}{7}x + 7 \cdot 1 - 7}{5}\\ & = \frac{5x + 7 - 7}{5}\\ & = \frac{5x}{5}\\ & = x. \end{aligned} \][/tex]

Thus, we have shown that \((f^{-1} \circ f)(x) = x\).

### 2. \((f \circ f^{-1})(x)\)

[tex]\[ \begin{aligned} (f \circ f^{-1})(x) & = f(f^{-1}(x)) \\ & = f\left(\frac{7x - 7}{5}\right). \end{aligned} \][/tex]

Now, substitute the expression for \(f^{-1}(x)\) into the function \(f(x) = \frac{5}{7}x + 1\):

[tex]\[ \begin{aligned} f\left(\frac{7x - 7}{5}\right) &= \frac{5}{7} \left(\frac{7x - 7}{5}\right) + 1\\ & = \frac{5 \cdot (7x - 7)}{7 \cdot 5} + 1\\ & = \left(\frac{7x - 7}{7} \right) + 1\\ & = x - 1 + 1\\ & = x. \end{aligned} \][/tex]

Thus, we have shown that \((f \circ f^{-1})(x) = x\).

Having verified both compositions, we can conclude that [tex]\(f^{-1}(x) = \frac{7x - 7}{5}\)[/tex] is indeed the inverse of [tex]\(f(x) = \frac{5}{7}x + 1\)[/tex].

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