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Section 3.7 : Inverse Functions
6. Given \(\displaystyle h\left( x \right) = \frac{{1 + 2x}}{{7 + x}}\) find \({h^{ - 1}}\left( x \right)\) .
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Hint : Just follow the process outlines in the notes and you’ll be set to do this problem!
For the first step we simply replace the function with a \(y\).
\[y = \frac{{1 + 2x}}{{7 + x}}\] Show Step 2Next, replace all the \(x\)’s with \(y\)’s and all the original \(y\)’s with \(x\)’s.
\[x = \frac{{1 + 2y}}{{7 + y}}\] Show Step 3Solve the equation from Step 2 for \(y\).
\[\begin{align*}x & = \frac{{1 + 2y}}{{7 + y}}\\ x\left( {7 + y} \right) & = 1 + 2y\\ 7x + xy & = 1 + 2y\\ 7x - 1 & = 2y - xy\\ 7x - 1 & = \left( {2 - x} \right)y\\ \frac{{7x - 1}}{{2 - x}} & = y\end{align*}\] Show Step 4Replace \(y\) with \({h^{ - 1}}\left( x \right)\).
\[{h^{ - 1}}\left( x \right) = \frac{{7x - 1}}{{2 - x}}\] Show Step 5Finally, do a quick check by computing one or both of \(\left( {h \circ {h^{ - 1}}} \right)\left( x \right)\) and \(\left( {{h^{ - 1}} \circ h} \right)\left( x \right)\) and verify that each is \(x\). In general, we usually just check one of these and we’ll do that here.
\[\begin{align*}\left( {h \circ {h^{ - 1}}} \right)\left( x \right) = h\left[ {{h^{ - 1}}\left( x \right)} \right] & = h\left[ {\frac{{7x - 1}}{{2 - x}}} \right]\\ & = \frac{{1 + 2\left( {\frac{{7x - 1}}{{2 - x}}} \right)}}{{7 + \frac{{7x - 1}}{{2 - x}}}}\frac{{2 - x}}{{2 - x}} = \frac{{2 - x + 2\left( {7x - 1} \right)}}{{7\left( {2 - x} \right) + 7x - 1}} = \frac{{13x}}{{13}} = x\end{align*}\]The check works out so we know we did the work correctly and have inverse.