To find the inverse of a function, you basically “undo” what the original function does, step by step, and then rewrite the result as a new function.

How to Find the Inverse of a Function (Quick Scoop)

Inverse functions are like perfectly matched undo buttons: one takes you from input to output, the other walks you back from output to input. If fff sends xxx to yyy, then the inverse sends that same yyy back to xxx.

Core Idea (What an Inverse Is)

  • A function f(x)f(x)f(x) takes an input xxx and gives an output yyy.
  • The inverse function, usually written f−1(x)f^{-1}(x)f−1(x), swaps the roles: it takes what used to be the output and returns the original input.
  • In symbols, if y=f(x)y=f(x)y=f(x), then x=f−1(y)x=f^{-1}(y)x=f−1(y).
  • Function and inverse “cancel” each other:
    • f(f−1(x))=xf(f^{-1}(x))=xf(f−1(x))=x
    • f−1(f(x))=xf^{-1}(f(x))=xf−1(f(x))=x

A useful way to picture this: if a machine doubles a number and adds 3, the inverse machine will subtract 3 and then halve the result.

Step‑by‑Step Recipe (Algebraic Method)

A very standard method taught in algebra courses is:

  1. Start with the function written as an equation.
    • If it’s given as f(x)f(x)f(x), first rewrite it as yyy.
    • Example: f(x)=3x−2f(x)=3x-2f(x)=3x−2 becomes y=3x−2y=3x-2y=3x−2.
  1. Swap xxx and yyy.
    • Replace every yyy with xxx, and every xxx with yyy.
    • Using the example: y=3x−2y=3x-2y=3x−2 becomes x=3y−2x=3y-2x=3y−2.
  1. Solve this new equation for yyy.
    • Now treat it like a normal algebra problem where you isolate yyy.
    • From x=3y−2x=3y-2x=3y−2:
      • Add 2: x+2=3yx+2=3yx+2=3y
      • Divide by 3: y=x+23y=\dfrac{x+2}{3}y=3x+2​.
  1. Rename that yyy as f−1(x)f^{-1}(x)f−1(x).
    • The final expression is your inverse.
    • So f−1(x)=x+23f^{-1}(x)=\dfrac{x+2}{3}f−1(x)=3x+2​.

That’s the core algorithm you can use for most algebraic functions—linear, many rational ones, radicals, etc.

Quick Examples (Different Types)

1. Linear function

Let f(x)=2x+5f(x)=2x+5f(x)=2x+5.

  1. Write as yyy: y=2x+5y=2x+5y=2x+5.
  2. Swap: x=2y+5x=2y+5x=2y+5.
  3. Solve for yyy:
    • x−5=2yx-5=2yx−5=2y
    • y=x−52y=\dfrac{x-5}{2}y=2x−5​.
  4. So f−1(x)=x−52f^{-1}(x)=\dfrac{x-5}{2}f−1(x)=2x−5​.

You can test: plug a number into fff, then plug that result into f−1f^{-1}f−1; you should get back the starting number.

2. Rational function

Let f(x)=−2x−5f(x)=\dfrac{-2}{x-5}f(x)=x−5−2​.

  1. y=−2x−5y=\dfrac{-2}{x-5}y=x−5−2​.
  2. Swap: x=−2y−5x=\dfrac{-2}{y-5}x=y−5−2​.
  3. Solve for yyy:
    • Multiply both sides by (y−5)(y-5)(y−5): x(y−5)=−2x(y-5)=-2x(y−5)=−2.
    • Distribute: xy−5x=−2xy-5x=-2xy−5x=−2.
    • Move terms: xy=5x−2xy=5x-2xy=5x−2.
    • Divide by xxx (assuming x≠0x\neq 0x=0): y=5x−2xy=\dfrac{5x-2}{x}y=x5x−2​.
  1. So f−1(x)=5x−2xf^{-1}(x)=\dfrac{5x-2}{x}f−1(x)=x5x−2​.

3. Squared function with restricted domain

The function f(x)=x2f(x)=x^2f(x)=x2 on all real numbers does not have an inverse function, because it fails the horizontal line test (one output comes from two different inputs, like 22=42^2=422=4 and (−2)2=4(-2)^2=4(−2)2=4).

But if we restrict the domain to x≥0x\ge 0x≥0, then it becomes one‑to‑one and does have an inverse.

  • For f(x)=x2f(x)=x^2f(x)=x2, x≥0x\ge 0x≥0:
    • Write y=x2y=x^2y=x2.
    • Swap: x=y2x=y^2x=y2.
    • Solve for yyy: y=xy=\sqrt{x}y=x​ with y≥0y\ge 0y≥0.
    • So f−1(x)=xf^{-1}(x)=\sqrt{x}f−1(x)=x​ (with appropriate domain and range restrictions).

This shows how domain restriction is sometimes essential for an inverse to exist as a function.

Domain, Range, and the “Swap”

A powerful viewpoint: the inverse literally swaps domain and range.

  • The domain of fff becomes the range of f−1f^{-1}f−1.
  • The range of fff becomes the domain of f−1f^{-1}f−1.

For example, if the original function’s domain is “all real numbers” and its range is “y≥0y\ge 0y≥0”, then the inverse’s domain will be “x≥0x\ge 0x≥0” and its range “all real numbers”. On a graph, the inverse is the reflection of the original function across the line y=xy=xy=x.

When a Function Has an Inverse

Not every function has an inverse function over its entire domain. To have an inverse function:

  • The function must be one‑to‑one : no two different inputs share the same output.
  • Graphically, it must pass the horizontal line test (any horizontal line cuts the graph at most once).
  • If it fails, you can sometimes restrict the domain so it passes the test and then define an inverse on that restricted domain.

How to Check You Did It Right

Once you think you’ve found the inverse f−1(x)f^{-1}(x)f−1(x):

  1. Compute f(f−1(x))f(f^{-1}(x))f(f−1(x)).
  2. Compute f−1(f(x))f^{-1}(f(x))f−1(f(x)).
  3. If both simplify to xxx (on the relevant domains), your inverse is correct.

For instance, with f(x)=2x+5f(x)=2x+5f(x)=2x+5 and f−1(x)=x−52f^{-1}(x)=\dfrac{x-5}{2}f−1(x)=2x−5​:

  • f(f−1(x))=2(x−52)+5=x−5+5=xf(f^{-1}(x))=2\left(\dfrac{x-5}{2}\right)+5=x-5+5=xf(f−1(x))=2(2x−5​)+5=x−5+5=x.
  • f−1(f(x))=(2x+5)−52=2x2=xf^{-1}(f(x))=\dfrac{(2x+5)-5}{2}=\dfrac{2x}{2}=xf−1(f(x))=2(2x+5)−5​=22x​=x.

Mini FAQ Style “Forum” Take

Q: I keep mixing up the steps. Is there a simple memory trick?
Think “rename, swap, solve, rename”:

  • Rename f(x)f(x)f(x) to yyy
  • Swap xxx and yyy
  • Solve for yyy
  • Rename yyy to f−1(x)f^{-1}(x)f−1(x).

Q: Why is it called f−1f^{-1}f−1? Is that a reciprocal?
No—f−1f^{-1}f−1 is not 1f\dfrac{1}{f}f1​. It’s “the inverse under composition,” meaning fff followed by f−1f^{-1}f−1 gets you back to where you started.

Q: What if I can’t solve for yyy?
Then an inverse function might not exist in a nice algebraic form, or the function may not be one‑to‑one without restricting the domain.

Very Short TL;DR

To find the inverse of a function:

  1. Replace f(x)f(x)f(x) with yyy.
  2. Swap xxx and yyy.
  3. Solve the resulting equation for yyy.
  4. Rewrite yyy as f−1(x)f^{-1}(x)f−1(x), keeping in mind domain and range.

Information gathered from public forums or data available on the internet and portrayed here.