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### Section 3-7 : Inverse Functions

2. Given $$\displaystyle g\left( x \right) = \frac{1}{2}x + 7$$ find $${g^{ - 1}}\left( x \right)$$ .

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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 $$y$$.

$y = \frac{1}{2}x + 7$ Show Step 2

Next, replace all the $$x$$’s with $$y$$’s and all the original $$y$$’s with $$x$$’s.

$x = \frac{1}{2}y + 7$ Show Step 3

Solve the equation from Step 2 for $$y$$.

\begin{align*}x & = \frac{1}{2}y + 7\\ 2x & = y + 14\\ 2x - 14 & = y\end{align*} Show Step 4

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

${g^{ - 1}}\left( x \right) = 2x - 14$ Show Step 5

Finally, do a quick check by computing one or both of $$\left( {g \circ {g^{ - 1}}} \right)\left( x \right)$$ and $$\left( {{g^{ - 1}} \circ g} \right)\left( x \right)$$ and verify that each is $$x$$. In general, we usually just check one of these and well do that here.

$\left( {g \circ {g^{ - 1}}} \right)\left( x \right) = g\left[ {{g^{ - 1}}\left( x \right)} \right] = g\left[ {2x - 14} \right] = \frac{1}{2}\left( {2x - 14} \right) + 7 = x - 7 + 7 = x$

The check works out so we know we did the work correctly and have inverse.