AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/2/2-8)
Unit 2 · Topic 2.8
2.8 Inverse Functions
An inverse function runs a function backward: if f sends a to b, then f⁻¹ sends b back to a. A function has an inverse only where it is one-to-one, and the inverse's graph is the reflection of the original over the line y = x.
Key terms
- inverse function
- one-to-one
- reflection over y = x
- restricted domain
- f⁻¹(x)
When an inverse exists
f has an inverse on a domain if every output comes from exactly one input. Such a function is called one-to-one, or invertible.
On a graph, a one-to-one function passes the horizontal line test: no horizontal line hits the graph more than once. Functions that always increase or always decrease are one-to-one.
If a function isn't one-to-one, you can restrict its domain to a piece where it is. For example, x² isn't one-to-one on all real numbers, but it is on x ≥ 0.
What the inverse does
The inverse f⁻¹ maps each output of f back to its input: if f(a) = b, then f⁻¹(b) = a. As pairs, (a, b) on f becomes (b, a) on f⁻¹.
The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.
Composing a function with its inverse gives the identity: f(f⁻¹(x)) = x and f⁻¹(f(x)) = x, for inputs in the right domains.
Note that f⁻¹(x) does not mean 1/f(x). The −1 is notation for “inverse,” not an exponent.
Inverses from tables, graphs and formulas
Table: swap the input and output columns.
Graph: reflect the graph over the line y = x, which swaps the roles of the axes. A point (2, 7) becomes (7, 2).
Formula: write y = f(x), swap x and y, and solve for y. That y is f⁻¹(x). This undoes f's operations in reverse order.
Checking that two functions are inverses
Compose them both ways. If f(x) = 3x − 4 and g(x) = (x + 4)/3, then f(g(x)) = 3 · (x + 4)/3 − 4 = x and g(f(x)) = ((3x − 4) + 4)/3 = x. Both compositions give x, so f and g are inverses.
Notice the order. f multiplies by 3 and then subtracts 4. Its inverse undoes those steps in reverse: add 4, then divide by 3. Like taking off shoes before socks, the last step done is the first step undone.
Inverses in context
If C(t) gives the cost of a taxi ride lasting t minutes, then C⁻¹(c) gives the length of a ride that costs c dollars. The inverse answers the reverse question.
Contexts can also restrict an inverse. If t must be between 0 and 60 minutes, then C⁻¹ only makes sense for costs that C actually produces on that interval.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Finding an inverse algebraically
Find the inverse of f(x) = (2x − 1)/(x + 3).
Show the solutionHide the solution
- Step 1: Write y = (2x − 1)/(x + 3), then swap x and y: x = (2y − 1)/(y + 3).
- Step 2: Multiply both sides by (y + 3): xy + 3x = 2y − 1.
- Step 3: Collect y terms on one side: xy − 2y = −1 − 3x.
- Step 4: Factor out y: y(x − 2) = −(3x + 1), so y = −(3x + 1)/(x − 2) = (3x + 1)/(2 − x).
- Step 5: Domain check: f's range excludes 2 (its horizontal asymptote is y = 2), and f⁻¹ is undefined at x = 2, as expected.
Answer: f⁻¹(x) = (3x + 1)/(2 − x), for x ≠ 2.
- Example 2
Restricting the domain first
f(x) = (x − 2)² + 1 is not one-to-one. Restrict it to x ≥ 2 and find the inverse, with its domain.
Show the solutionHide the solution
- Step 1: On x ≥ 2, the parabola only rises, so it's one-to-one. Its range there is y ≥ 1.
- Step 2: Swap and solve: x = (y − 2)² + 1, so (y − 2)² = x − 1.
- Step 3: Take square roots: y − 2 = ±√(x − 1). Because the inverse's outputs must be ≥ 2 (the original domain), keep the + sign.
- Step 4: f⁻¹(x) = 2 + √(x − 1), with domain x ≥ 1 (the original range).
Answer: f⁻¹(x) = 2 + √(x − 1) for x ≥ 1.
- Example 3
Trap: inverse vs reciprocal
Let f(x) = 2x + 6. Find f⁻¹(10) and 1/f(10). Are they the same?
Show the solutionHide the solution
- Step 1: f⁻¹(10) asks: which input gives an output of 10? Solve 2x + 6 = 10: x = 2. (Or use f⁻¹(x) = (x − 6)/2.)
- Step 2: 1/f(10) = 1/(2 · 10 + 6) = 1/26.
- Step 3: They're completely different. The −1 in f⁻¹ never means a reciprocal.
Answer: f⁻¹(10) = 2, while 1/f(10) = 1/26. Not the same.
Common mistakes
- Reading f⁻¹(x) as 1/f(x).
- Forgetting to restrict the domain, or picking the wrong sign after a square root. The inverse's outputs must match the original function's domain.
- Swapping x and y but then solving for x instead of y.
- Not swapping domain and range: the domain of f⁻¹ is the range of f.
On the exam
- Expect to find f⁻¹(a) from a table or graph, to find an inverse formula, and to state the inverse's domain.
- This topic sets up logarithms (2.10) and inverse trig functions (3.9), which are both inverses of functions with restricted domains.
Connected topics
Videos
Check yourself
4 questions on 2.8 Inverse Functions. Pick an answer to see if you got it, and why.
Let f(x) = (x − 1)³ + 2. Which of the following is f⁻¹(x)?
Let f(x) = (x − 2)² + 1. Which of the following is the largest interval that contains x = 5 on which f is invertible?
Let f(x) = (2x + 1)/(x − 3) for x ≠ 3. Which of the following is f⁻¹(x)?
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| f(x) | 2 | 4 | 1 | 0 | 3 |
| g(x) | 3 | 0 | 4 | 1 | 2 |
Table of values
What is the value of f⁻¹(g(0))?
0 of 4 answered