Grade 11 · MCR3U · Functions

MCR3U: Inverse Functions

An inverse function undoes the original. If $f$ takes $2$ to $7$ , then $f^{-1}$ takes $7$ back to $2$ . Every skill in this topic comes from that one idea.

How It Works

Swap the inputs and outputs. Since the inverse reverses every input and output, each point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$ . That gives you the three ways to work with an inverse.

From points: switch the coordinates of every pair.

From a graph: switch the coordinates of a few key points and join them, or reflect the whole graph in the line $y = x$ . Both give the same picture.

From an equation: write $y$ for $f(x)$ , swap $x$ and $y$ , then solve for $y$ . Undo the operations in reverse order, the way you would solve any equation. For $y = 3x + 2$ , the swap gives $x = 3y + 2$ , so $y = \frac{x - 2}{3}$ . With a fraction, cross-multiply, collect every $y$ term on one side and factor $y$ out.

Domain and range trade places. The domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ . Often you can state them without finding the inverse at all.

The inverse is not always a function. Reflect a parabola in $y = x$ and you get a sideways parabola, which fails the vertical line test. The test on the original is the horizontal line test: if a horizontal line hits $f$ twice, $f^{-1}$ is not a function. The fix is to restrict the domain of $f$ to one side of the vertex, which picks one sign of the $\pm$ you get from the square root.

Read the notation carefully. $f^{-1}(x)$ means the inverse, not $\frac{1}{f(x)}$ . And to find one value like $f^{-1}(5)$ you do not need the formula: set $f(x) = 5$ and solve.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
Find the inverse of $f(x) = 4x - 7$ .
Show solution
1. Write $y = 4x - 7$ , then swap: $x = 4y - 7$ .
2. Solve for $y$ : $x + 7 = 4y$ , so $y = \frac{x + 7}{4}$ .
Answer: $f^{-1}(x) = \frac{x + 7}{4}$
Easy
$f = \{(-2, 5), (0, 1), (3, -4)\}$ . Write $f^{-1}$ , then find $f^{-1}(1)$ and $f^{-1}(5)$ .
Show solution
1. Swap every pair: $f^{-1} = \{(5, -2), (1, 0), (-4, 3)\}$ .
2. $f^{-1}(1)$ is the output paired with input $1$ : $0$ . Likewise $f^{-1}(5) = -2$ .
Answer: $f^{-1} = \{(5, -2), (1, 0), (-4, 3)\}$ ; $f^{-1}(1) = 0$ and $f^{-1}(5) = -2$ .
Medium
Find the inverse of $f(x) = (x - 3)^2 + 2$ . Restrict the domain of $f$ so the inverse is a function.
Show solution
1. Swap: $x = (y - 3)^2 + 2$ , so $(y - 3)^2 = x - 2$ and $y = 3 \pm \sqrt{x - 2}$ .
2. The $\pm$ gives two outputs for each input, so this inverse is not a function. The parabola fails the horizontal line test.
3. Restrict $f$ to the right half of the parabola, $x \ge 3$ . The inverse's outputs are then $y \ge 3$ , which keeps the $+$ branch.
Answer: With $x \ge 3$ : $f^{-1}(x) = 3 + \sqrt{x - 2}$ , with domain $x \ge 2$ and range $y \ge 3$ .
Medium
Find the inverse of $f(x) = \frac{2x + 1}{x - 3}$ and state the domain and range of both.
Show solution
1. Swap: $x = \frac{2y + 1}{y - 3}$ . Cross-multiply: $xy - 3x = 2y + 1$ .
2. Collect the $y$ terms: $xy - 2y = 3x + 1$ , so $y(x - 2) = 3x + 1$ and $y = \frac{3x + 1}{x - 2}$ .
3. $f$ excludes $x = 3$ , and its horizontal asymptote $y = 2$ is a value it never reaches. Those swap for the inverse.
Answer: $f^{-1}(x) = \frac{3x + 1}{x - 2}$ . $f$ : domain $x \ne 3$ , range $y \ne 2$ . $f^{-1}$ : domain $x \ne 2$ , range $y \ne 3$ .
Test level
For $f(x) = \sqrt{x + 4} - 1$ , find $f^{-1}$ with its domain, then find where $f$ and $f^{-1}$ intersect.
Show solution
1. $f$ has domain $x \ge -4$ and range $y \ge -1$ , so $f^{-1}$ has domain $x \ge -1$ and range $y \ge -4$ .
2. Swap: $x = \sqrt{y + 4} - 1$ , so $x + 1 = \sqrt{y + 4}$ . Square: $y = (x + 1)^2 - 4$ .
3. $f$ is increasing, so it can only meet its inverse on the line $y = x$ . Solve $\sqrt{x + 4} - 1 = x$ .
4. $\sqrt{x + 4} = x + 1$ . Square: $x + 4 = x^2 + 2x + 1$ , so $x^2 + x - 3 = 0$ and $x = \frac{-1 \pm \sqrt{13}}{2}$ .
5. Only $x = \frac{-1 + \sqrt{13}}{2}$ is at least $-1$ ; the other root came from squaring.
Answer: $f^{-1}(x) = (x + 1)^2 - 4$ for $x \ge -1$ . They meet once, at $x = y = \frac{-1 + \sqrt{13}}{2} \approx 1.30$ .

Learn It on Martian Lab

Martian Lab teaches this in the Inverse Functions lesson of its MCR3U Functions & Transformations unit. Watch it, then take the Functions & Transformations practice tests to see if it stuck.

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