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Section 3.7 : Inverse Functions
4. Given \(A\left( x \right) = \sqrt[5]{{2x + 11}}\) find \({A^{ - 1}}\left( x \right)\) .
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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 \(y\).
\[y = \sqrt[5]{{2x + 11}}\] Show Step 2Next, replace all the \(x\)’s with \(y\)’s and all the original \(y\)’s with \(x\)’s.
\[x = \sqrt[5]{{2y + 11}}\] Show Step 3Solve the equation from Step 2 for \(y\).
\[\begin{align*}{x^5} & = 2y + 11\\ {x^5} - 11 & = 2y\\\frac{{{x^5} - 11}}{2} & = y\end{align*}\] Show Step 4Replace \(y\) with \({A^{ - 1}}\left( x \right)\).
\[{A^{ - 1}}\left( x \right) = \frac{{{x^5} - 11}}{2}\] Show Step 5Finally, do a quick check by computing one or both of \(\left( {A \circ {A^{ - 1}}} \right)\left( x \right)\) and \(\left( {{A^{ - 1}} \circ A} \right)\left( x \right)\) and verify that each is \(x\). In general, we usually just check one of these and well do that here.
\[\begin{align*}\left( {A \circ {A^{ - 1}}} \right)\left( x \right) = A\left[ {{A^{ - 1}}\left( x \right)} \right] & = A\left[ {\frac{{{x^5} - 11}}{2}} \right]\\ & = \sqrt[5]{{2\left( {\frac{{{x^5} - 11}}{2}} \right) + 11}} = \sqrt[5]{{{x^5} - 11 + 11}} = \sqrt[5]{{{x^5}}} = x\end{align*}\]The check works out so we know we did the work correctly and have inverse.