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Section 1.2 : Inverse Functions

2. Find the inverse for \(h\left( x \right) = 3 - 29x\). Verify your inverse by computing one or both of the composition as discussed in this section.

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Hint : Remember the process described in this section. Replace the \(h\left( x \right)\), interchange the \(x\)’s and \(y\)’s, solve for \(y\) and the finally replace the \(y\) with \({h^{ - 1}}\left( x \right)\).
Start Solution
\[y = 3 - 29x\] Show Step 2
\[x = 3 - 29y\] Show Step 3
\[\begin{align*}x - 3 & = - 29y\\ y & = - \frac{1}{{29}}\left( {x - 3} \right)\hspace{0.25in}\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{h^{ - 1}}\left( x \right) = \frac{1}{{29}}\left( {3 - x} \right)}}\end{align*}\]

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 \(\left( {h \circ {h^{ - 1}}} \right)\left( x \right)\) or \(\left( {{h^{ - 1}} \circ h} \right)\left( x \right)\) to verify our work.

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

\[\begin{align*}\left( {h \circ {h^{ - 1}}} \right)\left( x \right) & = h\left[ {{h^{ - 1}}\left( x \right)} \right]\\ & = 3 - 29\left[ {\frac{1}{{29}}\left( {3 - x} \right)} \right]\\ & = 3 - \left( {3 - x} \right)\\ & = x\end{align*}\]

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