Inverse functions

Example 1: a linear function

Problem Find the inverse of $f(x) = 3x - 5$ and check it.

1. Write y for $f(x)$.
   $$y = 3x - 5$$
2. Swap x and y.
   $$x = 3y - 5$$
3. Add 5 to both sides.
   $$x + 5 = 3y$$
4. Divide by 3.
   $$y = \frac{x + 5}{3}$$
5. Check: put $\frac{x + 5}{3}$ into f.
   $$3 \cdot \frac{x + 5}{3} - 5 = x$$

Answer $f^{-1}(x) = \frac{x + 5}{3}$

Example 2: a rational function

Problem Find the inverse of $f(x) = \frac{2x + 1}{x - 3}$.

1. Write $y = \frac{2x + 1}{x - 3}$, then swap x and y.
   $$x = \frac{2y + 1}{y - 3}$$
2. Multiply both sides by $y - 3$.
   $$x(y - 3) = 2y + 1$$
3. Distribute.
   $$xy - 3x = 2y + 1$$
4. Get every y term on one side and everything else on the other.
   $$xy - 2y = 3x + 1$$
5. Factor out y.
   $$y(x - 2) = 3x + 1$$
6. Divide by $x - 2$.
   $$y = \frac{3x + 1}{x - 2}$$

Answer $f^{-1}(x) = \frac{3x + 1}{x - 2}$, for $x \ne 2$

Example 3: restrict the domain first

Problem $f(x) = (x - 2)^2 + 1$ is not one-to-one, so restrict it to $x \ge 2$. Find the inverse.

1. Swap x and y in $y = (x - 2)^2 + 1$.
   $$x = (y - 2)^2 + 1$$
2. Subtract 1.
   $$x - 1 = (y - 2)^2$$
3. Take the square root. The outputs of the inverse are the old inputs, which were at least 2, so $y - 2 \ge 0$ and only the positive root fits.
   $$y - 2 = \sqrt{x - 1}$$
4. Add 2.
   $$y = 2 + \sqrt{x - 1}$$

Answer $f^{-1}(x) = 2 + \sqrt{x - 1}$, with domain $x \ge 1$

Example 4 (test-hard): an inverse value without the formula

Problem Let $f(x) = x^3 + x + 1$. Find $f^{-1}(11)$.

1. $f^{-1}(11)$ is the input that gives 11. Set up that equation instead of hunting for a formula.
   $$x^3 + x + 1 = 11$$
2. Move 11 over.
   $$x^3 + x - 10 = 0$$
3. Try small integers. $x = 2$ works.
   $$2^3 + 2 - 10 = 0$$
4. f is always increasing (both $x^3$ and x grow), so no other input gives 11.

Answer $f^{-1}(11) = 2$