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].
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].
