Verifying Inverse Functions

Show that $f (x) = \frac{1}{x + 1}$ and $f^{- 1} (x) = \frac{1}{x} - 1$ are inverses, for $x \neq 0, −1$ .

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
We must show that $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x .$

$\begin{matrix} f^{- 1} (f (x)) & = & f^{- 1} (\frac{1}{x + 1}) \\ = & \frac{1}{\frac{1}{x + 1}} - 1 \\ = & (x + 1) - 1 \\ = & x \\ f (f^{−1} (x)) & = & f (\frac{1}{x} - 1) \\ = & \frac{1}{(\frac{1}{x} - 1) + 1} \\ = & \frac{1}{\frac{1}{x}} \\ = & x \end{matrix}$
Therefore, $f (x) = \frac{1}{x + 1}$ and $f^{- 1} (x) = \frac{1}{x} - 1$ are inverses.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Show that $f (x) = \frac{x + 5}{3}$ and $f^{- 1} (x) = 3 x - 5$ are inverses.

How did it go?