1.7Inverse Functions
Find inverse functions algebraically and graphically, verify that f(f⁻¹(x)) = x, apply the horizontal line test, and restrict domains to create invertible functions.
Inverse functions undo each other — logarithms undo exponentials, arcsin undoes sine. Understanding inverses is essential for solving equations and is the foundation for inverse trig functions and logarithms.
Essential Question
What does it mean for two functions to be inverses of each other, and how can you find, verify, and graph a function's inverse?
Lesson Overview
The inverse of a function f, written f⁻¹, reverses the input-output relationship: if f(a) = b, then f⁻¹(b) = a. A function has an inverse only if it is one-to-one — each output comes from exactly one input. The Horizontal Line Test (HLT) checks this graphically: if every horizontal line crosses the graph at most once, the function is one-to-one and invertible. To find f⁻¹ algebraically: replace f(x) with y, swap x and y, solve for y, then write as f⁻¹(x). To verify two functions are inverses, confirm both f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. Graphically, f and f⁻¹ are reflections over the line y = x — every point (a, b) on f corresponds to (b, a) on f⁻¹. When a function fails the HLT (like f(x) = x²), you can restrict the domain to make it one-to-one (e.g., x ≥ 0), then find the inverse on that restricted domain.
Key Facts
- Notation: f⁻¹(x) means "inverse of f" — NOT 1/f(x)
- Domain/Range swap: domain of f = range of f⁻¹, and vice versa
- Composition test: f(f⁻¹(x)) = x AND f⁻¹(f(x)) = x
- Graph: f⁻¹ is the reflection of f over y = x
- One-to-one required: only one-to-one functions have inverses
Algebraic Steps
- Replace f(x) with y
- Swap x and y
- Solve for y
- Write as f⁻¹(x)
- State the domain of f⁻¹ (= range of f)
- Verify: f(f⁻¹(x)) = x
f and f⁻¹ are reflections over y = x
Graphical relationship
f⁻¹ is the reflection of f over the line y = x. Every point (a, b) on f becomes (b, a) on f⁻¹.
Coordinate swap
If f(1) = 3, then f⁻¹(3) = 1. The x and y values swap between a function and its inverse.
Horizontal Line Test (HLT)
A function has an inverse if and only if every horizontal line crosses its graph at most once.
Steps to Find f⁻¹(x) Algebraically
Replace f(x) with y
y = 2x + 6
Swap x and y
x = 2y + 6
Solve for y
y = (x − 6) / 2
Write as f⁻¹(x)
f⁻¹(x) = (x − 6) / 2
Verify: f(f⁻¹(x)) = x
f((x−6)/2) = 2·(x−6)/2 + 6 = x ✓
Verification Method
Two conditions must BOTH hold:
f(f⁻¹(x)) = x
Apply f to f⁻¹(x) — should simplify to x.
f⁻¹(f(x)) = x
Apply f⁻¹ to f(x) — should also simplify to x.
If both hold, the functions are true inverses of each other.
Worked Examples
Find f⁻¹(x) for f(x) = 3x − 7. Then verify using composition.
Replace f(x) with y: y = 3x − 7.
Swap x and y: x = 3y − 7.
Solve for y: x + 7 = 3y → y = (x + 7)/3.
Write as inverse: f⁻¹(x) = (x + 7)/3.
Verify: f(f⁻¹(x)) = 3·(x+7)/3 − 7 = (x+7) − 7 = x ✓
Verify: f⁻¹(f(x)) = (3x−7+7)/3 = 3x/3 = x ✓
Find the inverse of f(x) = (x + 2)³ − 5. State the domain and range of f⁻¹.
y = (x + 2)³ − 5.
Swap: x = (y + 2)³ − 5.
Solve: x + 5 = (y + 2)³ → ∛(x + 5) = y + 2 → y = ∛(x + 5) − 2.
f⁻¹(x) = ∛(x + 5) − 2.
f(x) = (x+2)³ − 5 has domain (−∞,∞) and range (−∞,∞).
So f⁻¹ also has domain (−∞,∞) and range (−∞,∞).
Does f(x) = x² − 4 have an inverse? If not, restrict the domain so it does, then find the inverse.
Apply HLT: a horizontal line like y = 0 crosses f(x) = x² − 4 at x = 2 and x = −2. Fails HLT.
Restrict domain to x ≥ 0 (right half of parabola). Now f is one-to-one.
y = x² − 4, x ≥ 0.
Swap: x = y² − 4, y ≥ 0.
Solve: x + 4 = y² → y = √(x + 4) (positive root since y ≥ 0).
f⁻¹(x) = √(x + 4), domain: x ≥ −4.
The graph of f passes through (−2, 5), (0, 1), (3, −4). List three points on f⁻¹.
For every point (a, b) on f, the point (b, a) is on f⁻¹.
(−2, 5) on f → (5, −2) on f⁻¹.
(0, 1) on f → (1, 0) on f⁻¹.
(3, −4) on f → (−4, 3) on f⁻¹.
Verify that f(x) = (2x − 1)/(x + 3) and g(x) = (3x + 1)/(2 − x) are inverses.
Compute f(g(x)): substitute g(x) into f.
f(g(x)) = (2·(3x+1)/(2−x) − 1) / ((3x+1)/(2−x) + 3).
Numerator: (6x+2)/(2−x) − 1 = (6x+2 − (2−x))/(2−x) = (7x)/(2−x).
Denominator: (3x+1)/(2−x) + 3 = (3x+1 + 3(2−x))/(2−x) = (7)/(2−x).
f(g(x)) = (7x/(2−x)) ÷ (7/(2−x)) = 7x/7 = x ✓
By symmetry (or similar calculation), g(f(x)) = x ✓.
Guided Practice
External Supplemental Resource
Guided Practice Video: Inverse Functions
Review how to find inverse functions algebraically and verify inverse relationships before completing the guided practice problems.
Video by The Organic Chemistry Tutor on YouTube
Watch on YouTube ↗Find f⁻¹(x) for f(x) = 5x + 3.
Hint: Replace f(x) with y, swap x and y, then solve for y. The result should be a linear function.
Find the inverse of f(x) = √(x − 4) + 1. State the domain of f⁻¹.
Hint: y = √(x−4)+1. Swap x and y: x = √(y−4)+1. Isolate the square root, then square both sides. Domain of f⁻¹ = range of f.
Does f(x) = 2x³ + 1 pass the Horizontal Line Test? Find f⁻¹(x).
Hint: x³ is strictly increasing — every horizontal line hits it once. So yes. Swap and solve: isolate the cube root.
The graph of f passes through (4, 7), (−1, 2), (0, −3). Find three points on f⁻¹ and state f⁻¹(7).
Hint: Swap x and y coordinates for each point. f⁻¹(7) is the x-value where f outputs 7.
Verify that f(x) = x³ + 2 and g(x) = ∛(x − 2) are inverses by computing f(g(x)) and g(f(x)).
Hint: Substitute g(x) into f: f(g(x)) = (∛(x−2))³ + 2. Simplify. Then compute g(f(x)) = ∛((x³+2)−2).
Key Vocabulary
Inverse Function f⁻¹
The function that reverses f: if f(a) = b, then f⁻¹(b) = a. NOT the same as 1/f(x). The notation f⁻¹ means 'inverse of f'.
One-to-One Function
A function where each output value comes from exactly one input. Required for an inverse to exist. Passes the Horizontal Line Test.
Horizontal Line Test (HLT)
A graphical test: if every horizontal line crosses the graph at most once, the function is one-to-one and has an inverse.
Composition Verification
To verify f and g are inverses: confirm f(g(x)) = x AND g(f(x)) = x. Both must hold.
Reflection over y = x
The graphical relationship between f and f⁻¹. Every point (a, b) on f corresponds to (b, a) on f⁻¹.
Domain Restriction
Limiting the domain of a non-one-to-one function to a subset where it IS one-to-one, allowing an inverse to be defined on that subset.
Interactive Practice — 5 Questions
If f(3) = 7, what is f⁻¹(7)?
Which function passes the Horizontal Line Test?
What is f⁻¹(x) for f(x) = 4x − 8?
The graph of f passes through (2, 9). Which point must be on f⁻¹?
To verify f and g are inverses, you must confirm:
Independent Practice
Independent Practice
Find f⁻¹(x) for f(x) = (x − 3)/2 + 1. Verify using f(f⁻¹(x)) = x.
Find the inverse of f(x) = x³ − 8. State the domain and range of both f and f⁻¹.
Does f(x) = x² + 6x + 9 have an inverse? If not, restrict the domain and find the inverse on that domain.
The graph of f passes through (−3, 0), (1, 4), (5, −2). List the corresponding points on f⁻¹.
Verify that f(x) = (x + 1)/3 and g(x) = 3x − 1 are inverses by computing both f(g(x)) and g(f(x)).
Common Mistakes
Writing f⁻¹(x) = 1/f(x) — treating the −1 exponent as a reciprocal.
f⁻¹(x) means the INVERSE FUNCTION, not 1/f(x). The notation is borrowed from exponents but means something completely different in function notation.
Forgetting to swap x and y — just solving for y in the original equation.
The key step is SWAPPING x and y first, then solving for y. This is what reverses the input-output relationship.
Only checking f(f⁻¹(x)) = x and skipping g(f(x)) = x.
Both compositions must equal x. For most standard functions they both work if one does, but you should verify both — especially for rational functions.
Claiming f(x) = x² has an inverse without restricting the domain.
f(x) = x² fails the HLT. You must restrict the domain (e.g., x ≥ 0) before finding an inverse. Always check the HLT first.
Math Tips
Quick check: if f(a) = b, then f⁻¹(b) = a. You can read inverse values directly from a table or graph of f — just swap the columns/coordinates.
Domain/Range swap: domain of f = range of f⁻¹, and range of f = domain of f⁻¹. Always state both when finding an inverse.
For f(x) = x (the identity function), f⁻¹(x) = x as well — it is its own inverse.
Graphical shortcut: to sketch f⁻¹, reflect the graph of f over the line y = x. Points on the line y = x stay fixed.
Strictly increasing or strictly decreasing functions always pass the HLT and always have inverses — no domain restriction needed.