One-to-one functions, horizontal line test, finding inverses algebraically, domain and range swap, graphical reflection, cancellation equations, domain restriction, and inverse trig — everything in Stewart Chapter 2.7.
The foundational requirement: a function must be one-to-one before its inverse is also a function.
A function f is one-to-one if different inputs always produce different outputs. In symbols: whenever a and b are in the domain of f and a is not equal to b, it follows that f(a) is not equal to f(b). Equivalently: if f(a) equals f(b), then a must equal b.
One-to-One Examples
NOT One-to-One (on full domain)
A function is one-to-one if and only if every horizontal line intersects its graph at most once. This is the graphical version of the definition.
A line y = k hits f(x) = 3x + 1 exactly once for any k — passes the test.
The line y = 4 hits f(x) = x squared at both x = 2 and x = −2 — fails the test.
When a function fails the test, you must restrict the domain to a region where it does pass before finding an inverse.
If f is one-to-one, its inverse f-inverse reverses every input-output pair.
Let f be a one-to-one function with domain A and range B. Then its inverse function, written f⁻¹, has domain B and range A and is defined by:
In other words: if f sends x to y, then f-inverse sends y back to x. The notation f⁻¹ does NOT mean 1 divided by f(x). It is the inverse function, not a reciprocal.
Use this swap as a quick check: the domain restriction you state for f⁻¹ should equal the range of the original f.
These two equations must BOTH hold. Checking only one composition is not sufficient.
The swap-and-solve method works for any invertible function. Memorize these six steps.
Verify the function is one-to-one
Apply the horizontal line test mentally or graphically. If the function is not one-to-one on its full domain, state the domain restriction you will use before continuing.
Replace f(x) with y
Rewrite the equation as y equals the formula. This sets up the algebra cleanly.
Swap x and y
Replace every x in the equation with y and every y with x. This is the key step that reverses the function.
Solve the new equation for y
Use algebra to isolate y on the left side. For linear functions this is one or two steps; for rational functions you may need to collect y terms and factor.
Write the result as f-inverse of x
Replace y with f⁻¹(x). State the domain of f⁻¹, which equals the range of the original f.
Verify with both cancellation equations
Compute f(f⁻¹(x)) and simplify. Compute f⁻¹(f(x)) and simplify. Both must equal x.
Five complete examples covering linear, quadratic (restricted), radical, rational, and exponential function types.
f(x) = 5x − 3. Find f⁻¹(x) and verify.
Step 1: Linear functions with nonzero slope are always one-to-one. No restriction needed.
Step 2: Write y = 5x − 3
Step 3: Swap x = 5y − 3
Step 4: Solve for y → x + 3 = 5y → y = (x + 3) / 5
f⁻¹(x) = (x + 3) / 5, domain: all real numbers
Verification:
f(f⁻¹(x)) = f((x+3)/5) = 5 · (x+3)/5 − 3 = (x+3) − 3 = x ✓
f⁻¹(f(x)) = f⁻¹(5x−3) = (5x−3+3)/5 = 5x/5 = x ✓
f(x) = x² + 4, x ≥ 0. Find f⁻¹(x).
Step 1: x squared fails the horizontal line test on all reals. Restricting to x ≥ 0 makes it one-to-one (right branch of parabola only).
Step 2: y = x² + 4
Step 3: Swap x = y² + 4
Step 4: x − 4 = y² → y = ±√(x−4)
Since domain of f was x ≥ 0, the range of f⁻¹ must be y ≥ 0. Take the positive root.
f⁻¹(x) = √(x − 4), domain: x ≥ 4
f(x) = √(2x + 1). Find f⁻¹(x).
Step 1: Square root functions are one-to-one on their natural domain. Domain of f: 2x+1 ≥ 0, so x ≥ −1/2.
Step 2: y = √(2x + 1)
Step 3: Swap x = √(2y + 1)
Step 4: Square both sides → x² = 2y + 1 → y = (x² − 1) / 2
f⁻¹(x) = (x² − 1) / 2, domain: x ≥ 0
f(x) = (3x + 2) / (x − 1). Find f⁻¹(x).
Step 1: Rational functions of this form (Moebius transformations) are one-to-one when defined. Domain of f: x ≠ 1.
Step 2: y = (3x + 2) / (x − 1)
Step 3: Swap x = (3y + 2) / (y − 1)
Step 4: Multiply both sides by (y − 1):
x(y − 1) = 3y + 2
xy − x = 3y + 2
xy − 3y = x + 2 (collect y terms)
y(x − 3) = x + 2 (factor out y)
y = (x + 2) / (x − 3)
f⁻¹(x) = (x + 2) / (x − 3), domain: x ≠ 3
f(x) = e⁽₀²ˣ − 3. Find f⁻¹(x).
Step 1: Exponential functions are strictly increasing and one-to-one.
Step 2: y = e⁽₀²ˣ − 3
Step 3: Swap x = e⁽₀²ʏ − 3
Step 4: x + 3 = e⁽₀²ʏ → ln(x + 3) = 0.2y → y = 5 ln(x + 3)
f⁻¹(x) = 5 ln(x + 3), domain: x > −3
The graph of f⁻¹ is the mirror image of the graph of f across the line y = x.
If the point (a, b) lies on the graph of f, then the point (b, a) lies on the graph of f⁻¹. This is because f(a) = b means f⁻¹(b) = a by definition.
Geometrically: fold the paper along the diagonal line y = x. The two graphs land exactly on top of each other. The line y = x is the axis of symmetry between f and f⁻¹.
If a function's graph passes through the line y = x, the graph of its inverse also passes through the same point, since (a, a) reflected is still (a, a).
| Function f | Point on f | Point on f⁻¹ | f⁻¹(x) |
|---|---|---|---|
| f(x) = 2x + 1 | (0, 1) | (1, 0) | (x−1)/2 |
| f(x) = x², x≥0 | (3, 9) | (9, 3) | √x |
| f(x) = e^x | (0, 1) | (1, 0) | ln(x) |
| f(x) = x³ | (2, 8) | (8, 2) | x^(1/3) |
The two cancellation equations are the most important tool for verifying and using inverses.
Apply f-inverse first (getting into the domain of f), then apply f. Valid for all x in the domain of f⁻¹ (= range of f).
Apply f first (getting into the domain of f⁻¹), then apply f-inverse. Valid for all x in the domain of f.
Students often verify only f(g(x)) = x and conclude f and g are inverses. This is not enough. You must verify BOTH f(g(x)) = x AND g(f(x)) = x. On restricted domains, one composition may simplify to x while the other does not — especially when domain restrictions involve absolute values or square roots.
When a function fails the horizontal line test, restrict its domain to a region where it is one-to-one.
There is usually more than one valid domain restriction. Convention and context guide the choice:
f(x) = (x − 2)² + 1. Find f⁻¹ using the standard restriction x ≥ 2.
The vertex is at (2, 1). Restrict to x ≥ 2 (right half of parabola).
y = (x − 2)² + 1
Swap: x = (y − 2)² + 1
x − 1 = (y − 2)²
√(x − 1) = y − 2 (take positive root since y ≥ 2)
f⁻¹(x) = √(x − 1) + 2, domain: x ≥ 1
Range of f = [1, ∞) confirms domain of f⁻¹ = [1, ∞). ✓
Sine, cosine, and tangent are periodic and fail the horizontal line test on their full domains. To define their inverses, we must restrict each to a specific interval.
Restricted domain for sin
Range of sin on that domain
Domain of arcsin
Range of arcsin
arcsin(x) = y means sin(y) = x and −π/2 ≤ y ≤ π/2
sin(arcsin(x)) = x for −1 ≤ x ≤ 1
arcsin(sin(x)) = x for −π/2 ≤ x ≤ π/2
Key values:
arcsin(0) = 0 arcsin(1/2) = π/6 arcsin(√2/2) = π/4
arcsin(√3/2) = π/3 arcsin(1) = π/2 arcsin(−1) = −π/2
Restricted domain for cos
Range of cos on that domain
Domain of arccos
Range of arccos
arccos(x) = y means cos(y) = x and 0 ≤ y ≤ π
cos(arccos(x)) = x for −1 ≤ x ≤ 1
arccos(cos(x)) = x for 0 ≤ x ≤ π
Key values:
arccos(1) = 0 arccos(√3/2) = π/6 arccos(√2/2) = π/4
arccos(1/2) = π/3 arccos(0) = π/2 arccos(−1) = π
Restricted domain for tan
Range of tan on that domain
Domain of arctan
Range of arctan
arctan(x) = y means tan(y) = x and −π/2 < y < π/2
tan(arctan(x)) = x for all real x
arctan(tan(x)) = x only for −π/2 < x < π/2
Key values and limits:
arctan(0) = 0 arctan(1) = π/4 arctan(√3) = π/3
arctan(x) → π/2 as x → ∞ arctan(x) → −π/2 as x → −∞
| Function | Notation | Domain | Range | Restriction on sin/cos/tan |
|---|---|---|---|---|
| arcsin(x) | sin⁻¹(x) | [−1, 1] | [−π/2, π/2] | x in [−π/2, π/2] |
| arccos(x) | cos⁻¹(x) | [−1, 1] | [0, π] | x in [0, π] |
| arctan(x) | tan⁻¹(x) | all reals | (−π/2, π/2) | x in (−π/2, π/2) |
Range vs. Domain Alert
Arccos has range [0, π] while arcsin has range [−π/2, π/2]. This means arcsin and arccos give different answers even for the same input. For example: arcsin(1/2) = π/6 but arccos(1/2) = π/3. Students frequently mix these up on exams.
The most important inverse pair in precalculus: exponentials and logarithms undo each other.
logₛ(b^x) = x for all real x
b^(logₛ(x)) = x for x > 0
ln(e^x) = x for all real x
e^(ln x) = x for x > 0
Start with f(x) = b^x and find its inverse algebraically:
y = b^x
Swap: x = b^y
Apply log base b to both sides: logₛ(x) = logₛ(b^y) = y
f⁻¹(x) = logₛ(x) ✓
The domain of b^x is all reals, so the range of logₛ(x) is all reals. The range of b^x is (0, ∞), so the domain of logₛ(x) is (0, ∞). The swap is consistent.
Graphically, the curves y = b^x and y = logₛ(x) are reflections of each other over the line y = x. Both curves pass through the point (1, 1) if b = e (they intersect on y = x when x = 1).
| f(x) | f⁻¹(x) | Domain of f⁻¹ | Cancellation Check |
|---|---|---|---|
| mx + b (m≠0) | (x−b)/m | all reals | m·(x−b)/m + b = x ✓ |
| x², x≥0 | √x | [0, ∞) | (√x)² = x for x≥0 ✓ |
| x³ | x^(1/3) | all reals | (x^(1/3))³ = x ✓ |
| 1/x (x≠0) | 1/x | x≠0 | 1/(1/x) = x ✓ |
| b^x | logₛ(x) | (0, ∞) | logₛ(b^x) = x ✓ |
| e^x | ln(x) | (0, ∞) | ln(e^x) = x ✓ |
| sin(x), x∈[−π/2,π/2] | arcsin(x) | [−1, 1] | sin(arcsin x) = x ✓ |
| cos(x), x∈[0,π] | arccos(x) | [−1, 1] | cos(arccos x) = x ✓ |
| tan(x), x∈(−π/2,π/2) | arctan(x) | all reals | tan(arctan x) = x ✓ |
Confusing f⁻¹(x) with 1/f(x)
f⁻¹(x) means the inverse function, not the reciprocal. The reciprocal is written [f(x)]⁻¹ or 1/f(x). These are different unless f(x) = x.
Forgetting to state the domain restriction
When restricting the domain of a quadratic or trig function before inverting, you must state the restriction explicitly. The inverse is incomplete without it.
Taking both ± roots without eliminating one
When you get y = ±√(something), you must decide which sign applies based on the restricted domain. The domain of the original function determines the range of the inverse.
Checking only one cancellation equation
Always verify BOTH f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. On restricted domains they may differ.
Forgetting domain of arccos is different from arcsin
Both have domain [−1, 1] but different ranges: arcsin gives [−π/2, π/2], arccos gives [0, π]. Same input, different output.
Applying arcsin(sin(x)) = x for all x
arcsin(sin(x)) = x only when x is in [−π/2, π/2]. Outside that interval, the result is different: e.g., arcsin(sin(π)) = arcsin(0) = 0, not π.
Try these before checking the solutions. Each covers a different function type from this guide.
#1Find the inverse of f(x) = (x + 5) / 3 and verify with both cancellation equations.
Hint: Linear function — straightforward swap.
#2Find the inverse of f(x) = x² − 6x + 9 restricted to x ≥ 3. State the domain of f⁻¹.
Hint: Factor as (x−3)², vertex at (3, 0). Restrict to right branch.
#3Find the inverse of f(x) = (x + 1) / (2x − 5). State any domain restrictions.
Hint: Rational function. Cross-multiply, collect y terms, factor.
#4Evaluate arcsin(√3/2), arccos(−1/2), and arctan(−1) without a calculator.
Hint: Use the unit circle and the standard ranges of each function.
#5Simplify: sin(arctan(x)) for any real x. Express as an algebraic expression with no trig.
Hint: Draw a right triangle with angle arctan(x). Label opposite = x, adjacent = 1, hypotenuse = √(1+x²).
#6Is f(x) = x² + 2x − 3 one-to-one on its full domain? If not, give two valid domain restrictions that would make it one-to-one.
Hint: Complete the square to find the vertex. Restrict to each branch.
A function f is one-to-one (injective) if each output value is produced by exactly one input value. Formally: if f(a) = f(b) then a = b. Graphically, every horizontal line intersects the graph at most once — this is the horizontal line test. Only one-to-one functions have inverses that are also functions.
Step 1: Replace f(x) with y. Step 2: Swap x and y in the equation. Step 3: Solve the resulting equation for y. Step 4: Write the result as f inverse of x. Step 5: State the domain of the inverse (which equals the range of the original function). Step 6: Verify by checking that f composed with its inverse gives x, and the inverse composed with f also gives x.
The two cancellation equations are: f(f-inverse(x)) = x for every x in the domain of f-inverse, and f-inverse(f(x)) = x for every x in the domain of f. These equations capture the idea that the inverse function exactly undoes the original function. They are used both to verify inverses and to simplify expressions involving composed inverse functions.
Quadratic functions like f(x) = x squared fail the horizontal line test on their full domain — a horizontal line above the vertex hits the parabola twice. To get an invertible function, we restrict the domain to one branch. The standard restriction is x greater than or equal to 0, which gives f-inverse(x) = the square root of x. The restricted domain choice should be stated explicitly when finding the inverse of a quadratic.
Arcsin: domain is [-1, 1], range is [-pi/2, pi/2]. Arccos: domain is [-1, 1], range is [0, pi]. Arctan: domain is all real numbers, range is (-pi/2, pi/2). These restricted ranges come from the necessary domain restrictions placed on sin, cos, and tan to make them one-to-one before computing their inverses.
The exponential function f(x) = b^x and the logarithm g(x) = log base b of x are inverse functions of each other for any base b greater than 0 and not equal to 1. The cancellation equations are: log base b of (b^x) = x for all real x, and b raised to the log base b of x = x for x greater than 0. Their graphs are reflections of each other over the line y = x.
The graph of f-inverse is the reflection of the graph of f over the line y = x. To obtain it, swap all x and y coordinates: if the point (a, b) lies on the graph of f, then (b, a) lies on the graph of f-inverse. Folding the paper along the line y = x maps one graph onto the other.
Interactive practice problems with private tutoring — step-by-step solutions for every inverse function type in Stewart Chapter 2.7.
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