Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 3.7 : Inverse Functions
3. Given \(f\left( x \right) = {\left( {x - 2} \right)^3} + 1\) find \({f^{ - 1}}\left( x \right)\) .
Show All Steps Hide All Steps
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 \(y\).
\[y = {\left( {x - 2} \right)^3} + 1\] Show Step 2Next, replace all the \(x\)’s with \(y\)’s and all the original \(y\)’s with \(x\)’s.
\[x = {\left( {y - 2} \right)^3} + 1\] Show Step 3Solve the equation from Step 2 for \(y\).
\[\begin{align*}x & = {\left( {y - 2} \right)^3} + 1\\ x - 1 & = {\left( {y - 2} \right)^3}\\ {\left( {x - 1} \right)^{\frac{1}{3}}} & = y - 2\\ {\left( {x - 1} \right)^{\frac{1}{3}}} + 2 & = y\end{align*}\] Show Step 4Replace \(y\) with \({f^{ - 1}}\left( x \right)\).
\[{f^{ - 1}}\left( x \right) = {\left( {x - 1} \right)^{\frac{1}{3}}} + 2\] Show Step 5Finally, 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.
\[\begin{align*}\left( {f \circ {f^{ - 1}}} \right)\left( x \right) = f\left[ {{f^{ - 1}}\left( x \right)} \right] & = f\left[ {{{\left( {x - 1} \right)}^{\frac{1}{3}}} + 2} \right]\\ & = {\left( {{{\left( {x - 1} \right)}^{\frac{1}{3}}} + 2 - 2} \right)^3} + 1 = {\left( {{{\left( {x - 1} \right)}^{\frac{1}{3}}}} \right)^3} + 1 = x - 1 + 1 = x\end{align*}\]The check works out so we know we did the work correctly and have inverse.