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### Section 1.2 : Inverse Functions

7. Find the inverse for $$\displaystyle h\left( x \right) = \frac{{1 + 9x}}{{4 - x}}$$. 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
$y = \frac{{1 + 9x}}{{4 - x}}$ Show Step 2
$x = \frac{{1 + 9y}}{{4 - y}}$ Show Step 3
\begin{align*}x & = \frac{{1 + 9y}}{{4 - y}}\\ x\left( {4 - y} \right) & = 1 + 9y\\ & 4x - xy = 1 + 9y\\ 4x - 1 & = 9y + xy\\ 4x - 1 & = \left( {9 + x} \right)y\\ y & = \frac{{4x - 1}}{{9 + x}}\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{{4x - 1}}{{9 + x}}}}\end{align*}

Note that the Algebra in these kinds of problems can often be fairly messy, but don’t let that make you decide that you can’t do these problems. Messy Algebra will be a fairly common occurrence in a Calculus class so you’ll need to get used to it!

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 $$\left( {{h^{ - 1}} \circ h} \right)\left( x \right)$$ in this case. As with the previous step, the Algebra here is going to be messy and in fact will probably be messier.

\begin{align*}\left( {{h^{ - 1}} \circ h} \right)\left( x \right) & = {h^{ - 1}}\left[ {h\left( x \right)} \right]\\ & = \frac{{4\left[ {\frac{{1 + 9x}}{{4 - x}}} \right] - 1}}{{9 + \left[ {\frac{{1 + 9x}}{{4 - x}}} \right]}}\,\,\,\frac{{4 - x}}{{4 - x}}\\ & = \frac{{4\left( {1 + 9x} \right) - \left( {4 - x} \right)}}{{9\left( {4 - x} \right) + 1 + 9x}}\\ & = \frac{{4 + 36x - 4 + x}}{{36 - 9x + 1 + 9x}}\\ & = \frac{{37x}}{{37}}\\ & = x\end{align*}

In order to do the simplification we multiplied the numerator and denominator of the initial fraction by $$4 - x$$ in order to clear out some of the denominators. This in turn allowed a fair amount of simplification.

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