Inverse Function Calculator With Steps
This inverse function calculator finds f⁻¹(x) for the kinds of functions you meet in algebra and precalculus. Type a function such as 2x + 3, (2x + 1)/(x − 3), (x − 1)^3, e^(2x) or ln(x + 4) and it returns the inverse in simplified form. It shows the standard method: write y = f(x), swap x and y, then solve for y. Every answer is checked by plugging test values into f⁻¹(f(x)) to confirm you get x back. When a function is not one-to-one, such as x² or sin x, the tool tells you which branch it used and how to restrict the domain.
Inverse Function Calculator
Find the inverse of a function f(x), with the swap-and-solve steps and a numeric check.
How to use the Inverse Function Calculator
- Type your function using x, like (2x+1)/(x-3).
- Optionally enter an x value to check.
- Read f⁻¹(x) and the steps.
- Check the notes for any domain restriction.
Inverse function method
y = f(x) → swap x and y → solve x = f(y) for y → f⁻¹(x) = y. For f(x) = (ax + b)/(cx + d), f⁻¹(x) = (dx − b)/(a − cx).
Features
- Linear, rational (ax+b)/(cx+d), powers and roots
- Exponentials and logarithms in any base
- Inverse trig with principal branches
- Swap-and-solve steps
- Numeric check of f⁻¹(f(x)) = x
Frequently asked questions
How do you find the inverse of a function?
Replace f(x) with y, swap x and y, and solve the new equation for y. The result is f⁻¹(x). For f(x) = 2x + 3, x = 2y + 3 gives y = (x − 3)/2.
How do I know if a function has an inverse?
A function has an inverse only if it is one-to-one, meaning no two inputs give the same output. On a graph, every horizontal line must cross it at most once.
What is the inverse of a rational function like (2x + 1)/(x − 3)?
Swap and solve: x(y − 3) = 2y + 1, so xy − 2y = 3x + 1 and y = (3x + 1)/(x − 2).
Why does x² not have a true inverse?
Both 3 and −3 square to 9, so x² is not one-to-one. If you restrict it to x ≥ 0, its inverse is √x.
How can I check an inverse?
Compose them. If f⁻¹(f(x)) = x and f(f⁻¹(x)) = x for values in the domain, the inverse is correct. The graphs are also mirror images across the line y = x.