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Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 3.7 : Inverse Functions
3. Given \(f\left( x \right) = {\left( {x - 2} \right)^3} + 1\) find \({f^{ - 1}}\left( x \right)\) .
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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.