Section 1.2 : Inverse Functions
1. Find the inverse for \(f\left( x \right) = 6x + 15\). 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 \(f\left( x \right)\), interchange the \(x\)’s and \(y\)’s, solve for \(y\) and the finally replace the \(y\) with \({f^{ - 1}}\left( x \right)\).
\[y = 6x + 15\]
Show Step 2
\[x = 6y + 15\]
Show Step 3
\[\begin{align*}x - 15 & = 6y\\ y & = \frac{1}{6}\left( {x - 15} \right)\hspace{0.25in}\hspace{0.25in}\,\,\,\, \to \hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{f^{ - 1}}\left( x \right) = \frac{1}{6}\left( {x - 15} \right)}}\end{align*}\]
Finally, compute either \(\left( {f \circ {f^{ - 1}}} \right)\left( x \right)\) or \(\left( {{f^{ - 1}} \circ f} \right)\left( x \right)\) to verify our work.
Show Step 3Either composition can be done so let’s do \(\left( {f \circ {f^{ - 1}}} \right)\left( x \right)\) in this case.
\[\begin{align*}\left( {f \circ {f^{ - 1}}} \right)\left( x \right) & = f\left[ {{f^{ - 1}}\left( x \right)} \right]\\ & = 6\left[ {\frac{1}{6}\left( {x - 15} \right)} \right] + 15\\ & = x - 15 + 15\\ & = x\end{align*}\]So, we got \(x\) out of the composition and so we know we’ve done our work correctly.