Domain and Range of Inverse Functions

The domain of an inverse function is the range of the original function. The range of the inverse is the domain of the original. This swap is the single rule you need to find either one without a calculator. For f(x)=2x+3, the domain is all reals, range is all reals, so f⁻¹(x)=(x-3)/2 also has domain all reals and range all reals. The rule holds for every one-to-one function. If you do not know the original's range, you cannot state the inverse's domain. That is where most mistakes start.

Before you can state the domain of an inverse, you must know the original function's range. For a linear function, this is trivial: range equals domain if the slope is non‑zero.That means the domain of f⁻¹(x)=√x is [0,∞). If you skip finding the range of f, you will guess the inverse's domain incorrectly. Always start by asking: what y‑values does the original actually output?

The inverse of f(x)=√x is f⁻¹(x)=x², but only if the original's domain is [0,∞). That gives the original a range of [0,∞), so the inverse's domain is [0,∞). The failure case: if you try f⁻¹(−1) you get 1, which is a real number, but it is not the inverse because f(1) never outputs −1. The inverse is only defined on the original's range. For f(x)=x³, no restriction is needed because cubics are one‑to‑one. The range is all reals, so the inverse f⁻¹(x)=∛x has domain all reals and range all reals.

For f(x)=eˣ, the range is (0,∞), so the inverse f⁻¹(x)=ln(x) has domain (0,∞). Feed a negative number into ln(x) and you get an error. For f(x)=ln(x), the domain is (0,∞), so the inverse f⁻¹(x)=eˣ has domain all reals. The base does not matter: f(x)=2ˣ has range (0,∞), so f⁻¹(x)=log₂(x) has domain (0,∞). The relationship is always the same: the exponential's output interval becomes the logarithm's input interval.

For f(x)=1/x, the original's domain is all reals except 0, and its range is also all reals except 0. The inverse is the same function, so the domain of f⁻¹ is all reals except 0. For f(x)=(ax+b)/(cx+d), find the horizontal asymptote of the original. That asymptote value is the one number the original cannot output, so it becomes the one number excluded from the inverse's domain. For example, f(x)=1/(x−2)+3 has a horizontal asymptote at y=3, so the range of f is all reals except 3, and the domain of f⁻¹ is all reals except 3.

A function that is not one‑to‑one does not have an inverse function. The horizontal line test tells you this: if a horizontal line crosses the graph more than once, the inverse would fail the vertical line test. The fix is to restrict the original's domain so that it becomes one‑to‑one.The restricted domain [0,∞) becomes the inverse's range, and the original's range [0,∞) becomes the inverse's domain. For trigonometric functions, the restrictions are standard: arcsin uses [−π/2,π/2], arccos uses [0,π], arctan uses (−π/2,π/2). If the problem does not give a restricted domain, state that the function is not invertible.

Let f(x)=x²−4, restricted to x≥0. The range of f is [−4,∞). Therefore the domain of f⁻¹ is [−4,∞). To find the inverse, solve y=x²−4 for x: x=√(y+4), so f⁻¹(x)=√(x+4). The range of the inverse is [0,∞), which matches the restricted domain of the original. If you try f⁻¹(−5), you get an error because −5 is not in the domain.

Let f(x)=2ˣ. The range is (0,∞), so the domain of f⁻¹ is (0,∞). The inverse is f⁻¹(x)=log₂(x). The range of f⁻¹ is all reals, which matches the domain of f. If you attempt log₂(−1), the calculator returns an error or a complex number. The domain restriction (0,∞) is not optional; it is the result of the swap rule.

Let f(x)=1/(x−1)+2. The vertical asymptote is at x=1, so the domain is all reals except 1. The horizontal asymptote is at y=2, so the range is all reals except 2. Therefore the domain of f⁻¹ is all reals except 2, and the range of f⁻¹ is all reals except 1. If you input x=2 into the inverse, it is undefined. The inverse function is f⁻¹(x)=1/(x−2)+1.

Let f(x)=sin(x) with domain restricted to [−π/2,π/2]. The range of this restricted sine is [−1,1]. Therefore the domain of f⁻¹(x)=arcsin(x) is [−1,1], and its range is [−π/2,π/2]. If you input x=2 into arcsin, you get an error because 2 is outside the domain. The principal value range is fixed by convention; OpenStax Precalculus 2e (Section 6.4) confirms these intervals.

Students confuse the domain of the inverse with the domain of the original. The domain of f⁻¹ is never the domain of f unless the function is its own inverse (like f(x)=1/x). The swap rule is the only fact you need. Write it down before you start: domain of f⁻¹ = range of f. If you memorize that, you will never fail a domain‑of‑inverse question.