How to Find the Inverse of a Function

To find the inverse of a function, you need to know how to find the inverse of a function using the algebraic swap-and-solve method. Write the function as y = f(x), swap x and y, solve for y, and rename y as f⁻¹(x). The inverse undoes the original: if f(a) = b, then f⁻¹(b) = a. The graph of the inverse is the reflection of the original across the line y = x. The inverse exists as a function only when the original is one-to-one, meaning it passes the horizontal line test. A quadratic like f(x) = x² is not one-to-one over all real numbers; restrict its domain to find an inverse.

The Four Steps And Their Traps

The four steps are mechanical but each carries a trap. Step one: replace f(x) with y. Step two: swap every x and y. Step three: solve for y. Step four: rename y as f⁻¹(x). The most common mistake is stopping after step two. Swapping variables without solving gives the relation, not the function. Always complete the solve. Verification is optional but reliable: confirm f(f⁻¹(x)) = x and f⁻¹(f(x)) = x for all x in the appropriate domains.

Linear functions are the easiest inverse test. Take f(x) = 2x + 3. Write y = 2x + 3. Swap: x = 2y + 3. Solve: subtract 3 to get x - 3 = 2y, then divide by 2 to get y = (x - 3)/2. Rename: f⁻¹(x) = (x - 3)/2. Verify: f(f⁻¹(x)) = 2*((x - 3)/2) + 3 = x - 3 + 3 = x. The inverse of any linear function f(x) = ax + b is f⁻¹(x) = (x - b)/a, provided a ≠ 0.

For f(x) = -4x + 7, the inverse is (x - 7)/-4, or (7 - x)/4. The domain of the inverse equals the range of the original. Since a linear function with nonzero slope has domain ℝ and range ℝ, its inverse also has domain and range ℝ. No domain restriction is needed. If a = 0, the function is constant and has no inverse.

Working Example: f(x) = a/(bx + c) + d

A rational function of the form f(x) = a/(bx + c) + d has a vertical asymptote at x = -c/b and a horizontal asymptote at y = d. Its inverse is also rational. Find the inverse by following the four steps. For f(x) = 2/(x - 3) + 1, write y = 2/(x - 3) + 1. Swap: x = 2/(y - 3) + 1. Solve: subtract 1 from both sides: x - 1 = 2/(y - 3). Multiply both sides by (y - 3): (x - 1)(y - 3) = 2. Divide by (x - 1): y - 3 = 2/(x - 1). Add 3: y = 2/(x - 1) + 3. Rename: f⁻¹(x) = 2/(x - 1) + 3.

The inverse has its own vertical asymptote at x = 1 (the original horizontal asymptote) and horizontal asymptote at y = 3 (the original vertical asymptote). The domain of the inverse is all real numbers except x = 1. The range of the inverse is all real numbers except y = 3. A rational function where a is large and the denominator small can produce steep slopes; solve carefully to avoid algebraic errors when multiplying by denominators.

Working Example: f(x) = √(2x - 4)

Radical functions often involve square roots. The domain of the original must be restricted so that the radicand is non-negative.Write y = √(2x - 4). Swap: x = √(2y - 4). Square both sides: x² = 2y - 4. Solve: add 4: x² + 4 = 2y, then divide by 2: y = (x² + 4)/2. Rename: f⁻¹(x) = (x² + 4)/2. The domain of the inverse is x ≥ 0, which equals the range of the original function (since √ returns non-negative values).

If the original function had a constant added, like f(x) = 2√(x + 1) + 3, the algebra extends. Write y = 2√(x + 1) + 3. Swap: x = 2√(y + 1) + 3. Subtract 3: x - 3 = 2√(y + 1). Divide by 2: (x - 3)/2 = √(y + 1). Square: ((x - 3)/2)² = y + 1. Subtract 1: y = ((x - 3)/2)² - 1.

Inverse of an Exponential Function

Exponential functions of the form f(x) = a·bˣ are one-to-one for b > 0, b ≠ 1. Their inverses are logarithmic functions. For f(x) = 3·2ˣ, write y = 3·2ˣ. Swap: x = 3·2ʸ. Divide by 3: x/3 = 2ʸ. Take log base 2: y = log₂(x/3). Rename: f⁻¹(x) = log₂(x/3). The domain of the inverse is x/3 > 0, or x > 0. The range is all real numbers. Verify: f(f⁻¹(x)) = 3·2^(log₂(x/3)) = 3·(x/3) = x.

Inverse of a Logarithmic Function

Logarithmic functions f(x) = log_b(x) have inverses that are exponential functions. For f(x) = ln(x) (natural log), write y = ln(x). Swap: x = ln(y). Exponentiate: eˣ = y. Rename: f⁻¹(x) = eˣ. The domain of the original is x > 0, range is ℝ. The domain of the inverse is ℝ, range is y > 0. The inverse undoes the logarithm exactly. For base 10, f(x) = log(x) has inverse f⁻¹(x) = 10ˣ.

OpenStax Precalculus 2e (Section 1.7) confirms: the inverse of f(x) = aˣ is f⁻¹(x) = log_a(x) for a > 0, a ≠ 1. The inverse of f(x) = log_a(x) is f⁻¹(x) = aˣ. These are the fundamental pairs for exponential and logarithmic families.

Working Example: f(x) = x³ + 2

Cubic functions are one-to-one if they are strictly increasing or decreasing, which all odd-degree polynomials with positive leading coefficient are. For f(x) = x³ + 2, write y = x³ + 2. Swap: x = y³ + 2. Subtract 2: x - 2 = y³. Take cube root: y = ∛(x - 2). Rename: f⁻¹(x) = ∛(x - 2). Domain and range of both function and inverse are ℝ. No domain restriction is needed.

For a general cubic like f(x) = ax³ + bx² + cx + d, the inverse is not always expressible with elementary algebraic operations. Use the swap-and-solve method only when the cubic can be rearranged as a pure cube (after completing the cube on the linear term). For f(x) = 2x³ - 1, write y = 2x³ - 1. Swap: x = 2y³ - 1. Add 1: x + 1 = 2y³. Divide by 2: (x + 1)/2 = y³. Cube root: y = ∛((x + 1)/2). Rename: f⁻¹(x) = ∛((x + 1)/2). The cubic inverse is simpler than the quadratic inverse because no domain restriction is required.

Mistakes fall into five categories. First, failing the one-to-one test. A function like f(x) = x² fails the horizontal line test. To find an inverse, restrict the domain to x ≥ 0 or x ≤ 0. The inverse of the restricted quadratic is f⁻¹(x) = √x for the non-negative branch. Without restriction, no single inverse function exists.

Confusing Notation And Algebra Errors

Second, confusing f⁻¹(x) with 1/f(x). The superscript -1 denotes the inverse function, not the multiplicative reciprocal. f⁻¹(x) ≠ 1/f(x). This distinction traps many newcomers. Third, algebra errors when solving rational functions.Rushing the steps produces sign errors.

Domain Restrictions And Function Properties

Fourth, ignoring domain and range swaps. The domain of f⁻¹ is the range of f.If you write f⁻¹(x) = x² + 1 without noting the domain restriction, the inverse is incomplete. Fifth, assuming all functions have inverses. Absolute value function f(x) = |x| is not one-to-one and has no inverse without splitting into branches. Constant functions have no inverses either.

Here is a mistakes checklist for quick reference: (1) Did you check one-to-one? (2) Did you swap x and y completely? (3) Did you solve fully? (4) Did you rename? (5) Did you state the domain of the inverse? (6) Did you verify with composition? (7) Did you restrict the domain of the original if needed?