A function is one-to-one when no two inputs share an output. Only these functions have inverses. If a horizontal line ever meets the graph twice, the function is not one-to-one, and its domain must be cut before an inverse exists.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Try it first
Two machines have each been run once and the output written down. Machine A follows f(x) = 2x + 7 and printed 19. Machine B follows g(x) = x² and printed 49. For each machine, which inputs could have produced that output?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
The horizontal line test
A horizontal line at height k marks every input giving output k. Two hits means the inverse would need two outputs, which no function may have.
Testing algebraically
Suppose f(a) = f(b) and see whether a = b must follow. 2a + 1 = 2b + 1 forces a = b: one-to-one. a² = b² allows a = −b: not.
Finding the inverse
Swap x and y, then solve for y. f(x) = (2x + 1)/3: x = (2y + 1)/3, 3x = 2y + 1, y = (3x − 1)/2. Check: f(f⁻¹(x)) = (2 · (3x − 1)/2 + 1)/3 = x.
Restricting the domain
y = x² fails the test, but on x ≥ 0 it passes. Cutting the domain is how the square root, arcsine and arccosine all come to exist.
The graph of an inverse
Swapping inputs and outputs swaps coordinates, so the graph of f⁻¹ is the graph of f reflected across y = x. (2, 5) on f becomes (5, 2) on f⁻¹.
Domains trade places
The domain of f becomes the range of f⁻¹, and the range of f becomes the domain of f⁻¹. x² − 4 on x ≥ 0 has range y ≥ −4, so √(x + 4) has domain x ≥ −4.
Step 2: Try It Yourself
Tap and try it out.
Find the inverse of f(x) = (2x + 1)/3 and verify it
- 1x = (2y + 1)/3swap x and y
Step 0 of 4
Step 3: In Real Life
Encryption
A code must be undoable, so its function has to be one-to-one: two messages can never encrypt to the same text. Decryption is the inverse. Secure messaging depends on that condition.
Step 4: Watch an Example
One step at a time.
Watch Dev Undo f(x) = 5 − 2x
A fresh rule, tested first for whether it can be undone at all, then swapped, solved, and checked in two different ways.
- Step 1
Dev starts with the question that comes before any algebra: can this rule be undone? Supposing f(a) = f(b) gives 5 − 2a = 5 − 2b. Taking 5 from both sides leaves −2a = −2b, and halving and changing signs leaves a = b. No two different inputs share an output, so the rule is one-to-one and an inverse exists over every real number, with nothing cut away.
Watch Maya Undo a Cube
f(x) = x³ − 5. Maya writes the opening and the swap, and leaves three decisions along the way to you.
- Step 1
Maya’s rule cubes the input and then takes 5 away.
Watch Noor Take Two Roads
f(x) = 4x − 6, undone twice: once by swapping and solving, once by reversing the chain of moves. Two valid roads have to arrive at the same rule.
- Step 1
Noor’s first road is the one written in the lesson. y = 4x − 6 becomes x = 4y − 6 after the swap, with y now the letter to find and the old output sitting where an input goes.
Step 5: Your Turn
Practice makes it stick.
The Plant Stall
Problem 1 of 2
A plant stall charges $4 for the pitch plus $2.50 for each plant, so a tray of n plants costs c(n) = 2.5n + 4 dollars. A customer’s bill comes to $36.50. How many plants were in the tray?
plants
The Repair Bill
Problem 2 of 2
A repair shop charges $60 to come out plus $45 an hour, so h hours of work costs c(h) = 45h + 60 dollars. Write the rule that turns a bill of x dollars back into a number of hours. Type the right-hand side.
Undo It
Swap x and y, then solve for y. For a single value, solving the original equation is quicker.
1 of 5
f(x) = 4x + 3. What is f⁻¹(13)?
2 of 5
f(x) = (x + 9)/2. Write f⁻¹(x). Type the right-hand side.
3 of 5
The point (−3, 10) lies on the graph of f. What is f⁻¹(10)?
4 of 5
f(x) = 4 − 3x. Write f⁻¹(x). Type the right-hand side.
5 of 5
The domain of f is x ≥ 2 and its range is y ≤ 7. What is the largest number in the domain of f⁻¹?
Whose Step?
One line on each page is a slip. Find it, then put it right.
1 of 6
Priya found the inverse of f(x) = (2x − 5)/4. One line is a slip. Which one?
Priya’s inverse of f(x) = (2x − 5)/4
- 1y = (2x − 5)/4
- 2x = (2y − 5)/4
- 34x = 2y − 5× 4
- 42x = y − 5÷ 2
- Answerf⁻¹(x) = 2x + 5+ 5
2 of 6
f(x) = (2x − 5)/4. Write f⁻¹(x) correctly. Type the right-hand side.
3 of 6
Leo found the inverse of f(x) = x² − 9 on the restricted domain x ≥ 0. One line is a slip. Which one?
Leo’s inverse of f(x) = x² − 9 on x ≥ 0
- 1f(x) = x² − 9, with x ≥ 0
- 2x = y² − 9
- 3y² = x + 9+ 9
- Answerf⁻¹(x) = ±√(x + 9)
4 of 6
f(x) = x² − 9 on the restricted domain x ≥ 0. What is f⁻¹(7)?
5 of 6
Ravi worked out f⁻¹(2) for f(x) = 2x + 6. One line is a slip. Which one?
Ravi’s f⁻¹(2) for f(x) = 2x + 6
- 1f(x) = 2x + 6 is one-to-one, so an inverse exists
- 2f⁻¹(x) = 1/(2x + 6)
- Answerf⁻¹(2) = 1/10
6 of 6
f(x) = 2x + 6. What is f⁻¹(2)?
Pick the Move
Decide which move does the job. Nothing here needs working all the way through.
1 of 2
f(x) = 9x − 4, and the only thing wanted is f⁻¹(14). Which move gets there with the least work?
2 of 2
g(x) = x² − 5 has no inverse over all real numbers. Which move gives it one, without changing the rule itself?
Same Undoing?
Judge these without solving anything. The first is settled by composing in your head; the second by what an inverse needs.
1 of 2
Select every pair whose two functions undo each other for every real number.
2 of 2
Select every statement that must be true of a function with an inverse over all real numbers.
Step 6: Quick Check
Show what you know.
Question 1 of 2
(2, 11) lies on f. What is f⁻¹(11)?
Question 2 of 2
Why must a function be one-to-one to have an inverse?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
For which number k is f(x) = kx + 8 its own inverse — that is, f(f(x)) = x for every x?
Stretch 2 of 2
Using each of the numbers 2, 3 and 6 exactly once for a, b and c in f(x) = (ax + b)/c, write a rule whose inverse is f⁻¹(x) = (6x − 2)/3. Type the right-hand side of f(x).
What You Learned
- A function has an inverse exactly when it is one-to-one.
- The horizontal line test decides it from the graph; f(a) = f(b) forcing a = b decides it algebraically.
- Swap x and y, solve for y, and verify by composing.
- Restricting the domain can rescue a function that fails.
For the grown-up
The misconception this lesson targets is reading f⁻¹ as an exponent. The small −1 looks like the one in x⁻¹, so f⁻¹(x) gets written as 1/f(x) — and the two are almost never the same rule. It is a name, not a power: f⁻¹ is the function that undoes f. Its cousin is assuming every function can be undone at all. x² cannot, until its domain is cut, because 3 and −3 have already arrived at the same output and no rule may send that output back to both. One test settles both worries: compose the candidate with the original in each order and see whether x comes back untouched.