Inverse Function Calculator — Linear and Rational f(x) = (ax + b)/(cx + d)

How it works

y = (ax + b)/(cx + d) → swap x and y → x(cy + d) = ay + b → f⁻¹(x) = (dx − b)/(−cx + a) · linear case (c = 0): f⁻¹(x) = (dx − b)/a · an inverse exists when ad − bc ≠ 0

To invert a function, write y = f(x), swap x and y, and solve for y. For the rational (Möbius) form (ax + b)/(cx + d) this always gives (dx − b)/(−cx + a), and for a line it gives (x − b)/a after dividing by d. The function is one-to-one exactly when ad − bc ≠ 0; otherwise it is a constant and has no inverse. The domain of f excludes the zero of the denominator, x = −d/c, and its range excludes the horizontal asymptote y = a/c — these swap roles for f⁻¹ (OpenStax Precalculus §3.7).

Worked example

f(x) = (2x + 1)/(x − 3), so a = 2, b = 1, c = 1, d = −3. Swap and solve: x(y − 3) = 2y + 1 → xy − 2y = 3x + 1 → f⁻¹(x) = (3x + 1)/(x − 2). ad − bc = −6 − 1 = −7 ≠ 0, so the inverse exists. Domain of f: x ≠ 3, range of f: y ≠ 2. Check: f(4) = 9/1 = 9 and f⁻¹(9) = 28/7 = 4.

Sources: OpenStax Precalculus 2e — 3.7 Inverse Functions · Paul’s Online Math Notes — Inverse Functions

Frequently asked questions