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Section 3-7 : Inverse Functions

5. Given \(\displaystyle f\left( x \right) = \frac{{4x}}{{5 - x}}\) find \({f^{ - 1}}\left( x \right)\) .

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Hint : Just follow the process outlines in the notes and you’ll be set to do this problem!
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For the first step we simply replace the function with a \(y\).

\[y = \frac{{4x}}{{5 - x}}\] Show Step 2

Next, replace all the \(x\)’s with \(y\)’s and all the original \(y\)’s with \(x\)’s.

\[x = \frac{{4y}}{{5 - y}}\] Show Step 3

Solve the equation from Step 2 for \(y\).

\[\begin{align*}x & = \frac{{4y}}{{5 - y}}\\ x\left( {5 - y} \right) & = 4y\\ 5x - xy & = 4y\\ 5x & = 4y + xy\\ 5x & = \left( {4 + x} \right)y\\ \frac{{5x}}{{4 + x}} & = y\end{align*}\] Show Step 4

Replace \(y\) with \({f^{ - 1}}\left( x \right)\).

\[{f^{ - 1}}\left( x \right) = \frac{{5x}}{{4 + x}}\] Show Step 5

Finally, do a quick check by computing one or both of \(\left( {f \circ {f^{ - 1}}} \right)\left( x \right)\) and \(\left( {{f^{ - 1}} \circ f} \right)\left( x \right)\) and verify that each is \(x\). In general, we usually just check one of these and well do that here.

\[\begin{align*}\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = f\left[ {{f^{ - 1}}\left( x \right)} \right] & = f\left[ {\frac{{5x}}{{4 + x}}} \right]\\ & = \frac{{4\left( {\frac{{5x}}{{4 + x}}} \right)}}{{5 - \frac{{5x}}{{4 + x}}}}\frac{{4 + x}}{{4 + x}} = \frac{{20x}}{{5\left( {4 + x} \right) - 5x}} = \frac{{20x}}{{20}} = x\end{align*}\]

The check works out so we know we did the work correctly and have inverse.