Inverse Function Calculator
Calculate inverse function instantly with our math tool. Shows detailed work, formulas used, and multiple solution methods.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
y = f(x) => x = f(y) => solve for y = f-1(x)
To find the inverse, swap x and y in the original equation and solve for y. The inverse function undoes the original: f(f-1(x)) = x and f-1(f(x)) = x. The graphs of f and f-1 are reflections across the line y = x.
Worked Examples
Example 1: Finding Inverse of a Linear Function
Problem:Find the inverse of f(x) = 3x + 7 and verify by computing f(f-1(4)).
Solution:Step 1: Write y = 3x + 7\nStep 2: Swap x and y: x = 3y + 7\nStep 3: Solve for y: 3y = x - 7, so y = (x - 7) / 3\nInverse: f-1(x) = (x - 7) / 3\n\nVerification: f-1(4) = (4 - 7) / 3 = -1\nf(f-1(4)) = f(-1) = 3(-1) + 7 = 4 = original input. Verified!
Result:f-1(x) = (x - 7)/3 | Verified: f(f-1(4)) = 4
Example 2: Inverse of Exponential Function
Problem:Find the inverse of f(x) = 2^x + 5 and evaluate the inverse at x = 13.
Solution:Step 1: Write y = 2^x + 5\nStep 2: Swap: x = 2^y + 5\nStep 3: Solve: 2^y = x - 5, so y = log_2(x - 5)\nInverse: f-1(x) = log_2(x - 5)\n\nDomain of inverse: x > 5\nEvaluate: f-1(13) = log_2(13 - 5) = log_2(8) = 3\nVerify: f(3) = 2^3 + 5 = 13. Correct!
Result:f-1(x) = log_2(x - 5) | f-1(13) = 3 | Domain: (5, infinity)
Frequently Asked Questions
What does it mean for a function to be one-to-one?
A function is one-to-one (also called injective) if every output value corresponds to exactly one input value. In other words, no two different inputs produce the same output. Graphically, a function is one-to-one if and only if it passes the horizontal line test, meaning every horizontal line intersects the graph at most once. This property is essential for inverse functions because if two inputs map to the same output, the inverse would not know which input to return. Functions like f(x) = x^2 are not one-to-one on their full domain because f(2) = f(-2) = 4, but they can be made one-to-one by restricting the domain to x >= 0.
How do the domain and range relate between a function and its inverse?
The domain and range swap between a function and its inverse. The domain of the original function becomes the range of the inverse function, and the range of the original function becomes the domain of the inverse function. This relationship is logical because inverse functions swap inputs and outputs. For example, if f(x) = sqrt(x) has domain [0, infinity) and range [0, infinity), its inverse f-1(x) = x^2 (restricted to x >= 0) has the same domain and range in this case. For f(x) = e^x with domain (-infinity, infinity) and range (0, infinity), the inverse ln(x) has domain (0, infinity) and range (-infinity, infinity).
What is the graphical relationship between a function and its inverse?
A function and its inverse are always reflections of each other across the line y = x. This geometric relationship comes directly from the fact that inverse functions swap x and y coordinates. Every point (a, b) on the graph of f corresponds to the point (b, a) on the graph of f-1, and reflecting any point across y = x swaps its coordinates. This means that if you fold the coordinate plane along the line y = x, the graph of f would land exactly on the graph of f-1. This property provides a useful visual check for verifying that you have correctly found an inverse function, and it explains why the inverse of an increasing function is also increasing.
How do you find the inverse of an exponential function?
The inverse of an exponential function is a logarithmic function, and vice versa. To find the inverse of f(x) = a^x, swap x and y to get x = a^y, then take the logarithm base a of both sides to get y = log_a(x). So f-1(x) = log_a(x). For the natural exponential f(x) = e^x, the inverse is f-1(x) = ln(x). If the exponential is shifted, like f(x) = 2^x + 3, swap to get x = 2^y + 3, solve: 2^y = x - 3, then y = log_2(x - 3). The domain of the inverse becomes (3, infinity) because you can only take the logarithm of a positive number. This exponential-logarithm relationship is one of the most important inverse pairs in mathematics.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy