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### 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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Hint : Just follow the process outlines in the notes and you’ll be set to do this problem!
Start Solution

For the first step we simply replace the function with a $$y$$.

$y = {\left( {x - 2} \right)^3} + 1$ Show Step 2

Next, 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 3

Solve 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 4

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

${f^{ - 1}}\left( x \right) = {\left( {x - 1} \right)^{\frac{1}{3}}} + 2$ 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[ {{{\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.