Inverse functions (verifying with composition, domain restrictions)
Inverse Functions
Inverse Functions is a topic in Functions in the Common Core State Standards. It is aligned to Standard F-BF.4, which requires students to find inverse functions, verify them by composition, and read them as reflections across the line y = x.
An inverse function \(f^{-1}\) undoes \(f\), so \(f(f^{-1}(x))=x\); its graph is the reflection of \(f\) across the line \(y=x\), and only one-to-one functions have one.
Theory
The inverse \(f^{-1}\) of a function \(f\) undoes what \(f\) does: if \(f\) sends \(a\mapsto b\), then \(f^{-1}\) sends \(b\mapsto a\). Formally,
Because inputs and outputs trade places, the domain and range swap, and the graphs of \(f\) and \(f^{-1}\) are mirror images across the line \(y=x\).
The notation \(f^{-1}\) means the inverse function, not the reciprocal: \(f^{-1}(x)\neq \dfrac{1}{f(x)}\).
The defining relationship and the domain/range swap:
How to find an inverse function
- Replace \(f(x)\) with \(y\).
- Swap \(x\) and \(y\).
- Solve the new equation for \(y\).
- Write \(y=f^{-1}(x)\), and state any domain restriction.
- Verify with \(f(f^{-1}(x))=x\) and \(f^{-1}(f(x))=x\).
Write \(y=f(x)\), swap \(x\) and \(y\), then solve for \(y\).
| \(y\) | \(=\) | \(2x+3\) |
| \(x\) | \(=\) | \(2y+3\) |
| \(x-3\) | \(=\) | \(2y\) |
| \(y\) | \(=\) | \(\dfrac{x-3}{2}\) |
So \(f^{-1}(x)=\dfrac{x-3}{2}\).
Inverses satisfy \(f(g(x))=x\) and \(g(f(x))=x\). Check both:
| \(f(g(x))\) | \(=\) | \(2\!\left(\dfrac{x-3}{2}\right)+3\) |
| \(=\) | \((x-3)+3=x\) | |
| \(g(f(x))\) | \(=\) | \(\dfrac{(2x+3)-3}{2}\) |
| \(=\) | \(\dfrac{2x}{2}=x\) |
Both compositions return \(x\), so they are inverses.
Swap and solve, undoing the cube with a cube root.
| \(y\) | \(=\) | \(x^3-1\) |
| \(x\) | \(=\) | \(y^3-1\) |
| \(x+1\) | \(=\) | \(y^3\) |
| \(y\) | \(=\) | \(\sqrt[3]{x+1}\) |
So \(f^{-1}(x)=\sqrt[3]{x+1}\).
On all of \(\mathbb{R}\), \(x^2\) fails the horizontal line test. Restrict to \(x\ge 0\), where it is one-to-one.
| \(y\) | \(=\) | \(x^2\quad(x\ge 0)\) |
| \(x\) | \(=\) | \(y^2\) |
| \(y\) | \(=\) | \(\sqrt{x}\) |
So \(f^{-1}(x)=\sqrt{x}\), with domain \(x\ge 0\) — the range of the restricted \(f\).
Common pitfalls
Frequently asked questions
What is an inverse function?
A function \(f^{-1}\) that undoes \(f\): if \(f(a)=b\) then \(f^{-1}(b)=a\), and \(f(f^{-1}(x))=x\).
How do you find the inverse of a function?
Write \(y=f(x)\), swap \(x\) and \(y\), solve for \(y\), and write \(y=f^{-1}(x)\). Then verify by composition.
How do you know a function has an inverse?
It must be one-to-one — pass the horizontal line test. If a horizontal line meets the graph more than once, restrict the domain first.
Does f to the minus 1 mean one over f?
No. \(f^{-1}\) is the inverse function, not the reciprocal: \(f^{-1}(x)\neq \dfrac{1}{f(x)}\).
Why are f and its inverse reflections across y = x?
Because the inverse swaps each point \((a,b)\) for \((b,a)\), and swapping coordinates reflects a point across the line \(y=x\).