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Find the inverse function of \(f\) informally. Verify that \(f\left(f^{-1}(x)\right)=x\) and \(f^{-1}(f(x))=x\). $$f(x)=\frac{x-1}{5}$$

Short Answer

Expert verified
The inverse function of \(f(x) = \frac{x - 1}{5}\) is \(f^{-1}(x) = 5x + 1\). Both identities \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\) are verified and hold true.

Step by step solution

01

Finding the Inverse Function

First, to find the inverse function, let \(y = f(x)\), then solve for \(x\). In this case, \(y = \frac{x - 1}{5}\) implies that \(x = 5y + 1\). Now, swap \(x\) and \(y\) to get the inverse function: \(f^{-1}(x) = 5x + 1\).
02

Verifying the Identity \(f(f^{-1}(x)) = x\)

Substitute \(f^{-1}(x)\) into \(f(x)\). Thus, \(f(f^{-1}(x))= f(5x + 1) = \frac{((5x + 1)-1)}{5} = x\). So \(f(f^{-1}(x)) = x\) holds true.
03

Verifying the Identity \(f^{-1}(f(x)) = x\)

Substitute \(f(x)\) into \(f^{-1}(x)\). Thus, \(f^{-1}(f(x))= f^{-1}(\frac{x - 1}{5}) = 5 * \frac{x - 1}{5} + 1 = x\). So \(f^{-1}(f(x)) = x\) holds true.

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