\[x = \sqrt[5]{{9 - 11y}}\] Show Step 3

\[\begin{align*}x & = \sqrt[5]{{9 - 11y}}\\ {x^5} & = 9 - 11y\\ & {x^5} - 9 = - 11y\\ y &= - \frac{1}{{11}}\left( {{x^5} - 9} \right)\hspace{0.25in}\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{W^{ - 1}}\left( x \right) = \frac{1}{{11}}\left( {9 - {x^5}} \right)}}\end{align*}\]

Notice that we multiplied the minus sign into the parenthesis. We did this in order to avoid potentially losing the minus sign if it had stayed out in front. This does not need to be done in order to get the inverse.

Finally, compute either \(\left( {W \circ {W^{ - 1}}} \right)\left( x \right)\) or \(\left( {{W^{ - 1}} \circ W} \right)\left( x \right)\) to verify our work.

Show Step 4

Either composition can be done so let’s do \(\left( {{W^{ - 1}} \circ W} \right)\left( x \right)\) in this case.

\[\begin{align*}\left( {{W^{ - 1}} \circ W} \right)\left( x \right) & = {W^{ - 1}}\left[ {W\left( x \right)} \right]\\ & = \frac{1}{{11}}\left( {9 - {{\left[ {\sqrt[5]{{9 - 11x}}} \right]}^5}} \right)\\ & = \frac{1}{{11}}\left( {9 - \left[ {9 - 11x} \right]} \right)\\ & = \frac{1}{{11}}\left( {11x} \right)\\ & = x\end{align*}\]

So, we got \(x\) out of the composition and so we know we’ve done our work correctly.