Inverse Function

Inverse Function

The function obtained by switching the x- and y-variables in a function. The inverse of function f is written f^-1.

Note: The new relation obtained by reversing the x- and y-values of a function is not necessarily a function itself. The new relation is only a function if the original function is one-to-one.

Example finding f⁻¹(x) for f(x)=∛(6−x): swap variables, solve for y; result is f⁻¹(x)=6−x³.

Key Formula

\begin{gathered}\text{If } y = f(x), \text{ then } x = f^{-1}(y)\\f\!\left(f^{-1}(x)\right) = x \quad \text{and} \quad f^{-1}\!\left(f(x)\right) = x\end{gathered}

Where:

$f(x)$ = The original function
$f^{-1}(x)$ = The inverse function, which undoes the action of f
$x$ = An input value in the domain of the respective function
$y$ = The output value, where y = f(x)

Worked Example

Problem: Find the inverse of f(x) = 3x + 6.

Step 1: Replace f(x) with y to write the equation in terms of x and y.

y = 3x + 6

Step 2: Swap x and y. This reflects the idea that inverse functions reverse inputs and outputs.

x = 3y + 6

Step 3: Solve for y. Subtract 6 from both sides.

x - 6 = 3y

Step 4: Divide both sides by 3.

y = \frac{x - 6}{3}

Step 5: Write the result using inverse notation.

f^{-1}(x) = \frac{x - 6}{3}

Answer: The inverse function is f⁻¹(x) = (x − 6)/3. You can verify: f(f⁻¹(x)) = 3·((x − 6)/3) + 6 = x − 6 + 6 = x. ✓

Another Example

This example shows a non-linear function where the domain must be restricted to make the original function one-to-one. Without the restriction x ≥ 0, f(x) = x² would not have an inverse that is a function, because both x = 2 and x = −2 give the same output.

Problem: Find the inverse of f(x) = x² for x ≥ 0.

Step 1: Replace f(x) with y.

y = x^2, \quad x \ge 0

Step 2: Swap x and y.

x = y^2, \quad y \ge 0

Step 3: Solve for y by taking the square root of both sides. Since y ≥ 0, you take only the positive root.

y = \sqrt{x}

Step 4: Write in inverse notation.

f^{-1}(x) = \sqrt{x}

Answer: The inverse function is f⁻¹(x) = √x, defined for x ≥ 0.

Frequently Asked Questions

What is the difference between f⁻¹(x) and 1/f(x)?

The notation f⁻¹(x) means the inverse function — it reverses the operation of f. It does not mean the reciprocal 1/f(x). For example, if f(x) = 2x, then f⁻¹(x) = x/2, but 1/f(x) = 1/(2x). These are completely different expressions. The superscript −1 in f⁻¹ is not an exponent; it is a special notation reserved for inverse functions.

How do you tell if a function has an inverse?

A function has an inverse that is also a function only if it is one-to-one, meaning each output corresponds to exactly one input. You can check this graphically with the horizontal line test: if every horizontal line crosses the graph at most once, the function is one-to-one and its inverse exists. If a function fails this test, you may still find an inverse by restricting the domain.

What does the graph of an inverse function look like?

The graph of f⁻¹(x) is the reflection of the graph of f(x) across the line y = x. This makes sense because forming an inverse swaps every (a, b) point to (b, a), and reflecting across y = x is exactly the geometric operation that swaps coordinates.

Inverse Function f⁻¹(x) vs. Reciprocal Function 1/f(x)

	Inverse Function f⁻¹(x)	Reciprocal Function 1/f(x)
Meaning	Reverses the operation of f; maps outputs back to inputs	Divides 1 by the output of f
Notation	f⁻¹(x) — the −1 is not an exponent	[f(x)]⁻¹ or 1/f(x)
Example with f(x) = 2x + 4	f⁻¹(x) = (x − 4)/2	1/f(x) = 1/(2x + 4)
Key property	f(f⁻¹(x)) = x	f(x) · [1/f(x)] = 1
Existence requirement	f must be one-to-one	f(x) ≠ 0

Why It Matters

Inverse functions appear throughout algebra, precalculus, and calculus whenever you need to "undo" an operation — for instance, using logarithms to undo exponentials or using arcsine to undo sine. In calculus, the inverse function theorem connects the derivative of a function to the derivative of its inverse, which is essential for differentiating logarithmic and inverse trigonometric functions. Beyond math class, inverse functions model real-world reversals like converting Celsius to Fahrenheit and back, or decoding an encoded message.

Common Mistakes

Mistake: Confusing f⁻¹(x) with 1/f(x) and treating the −1 as an exponent.

Correction: The notation f⁻¹ specifically denotes the inverse function, not the reciprocal. If you want the reciprocal, write 1/f(x) or [f(x)]⁻¹ to avoid ambiguity.

Mistake: Forgetting to check whether the original function is one-to-one before finding the inverse.

Correction: If a function is not one-to-one, swapping x and y produces a relation that is not a function. Always apply the horizontal line test first, or restrict the domain so the function becomes one-to-one (as with f(x) = x² restricted to x ≥ 0).

Related Terms

Function — The original object whose inverse you find
One-to-One Function — Required property for an inverse to exist
Relation — Swapping variables always gives a relation
Variable — The x and y values that get swapped
Horizontal Line Test — Test for whether an inverse function exists
Composition of Functions — Used to verify f(f⁻¹(x)) = x
Domain — Domain of f becomes range of f⁻¹ and vice versa