Inverse functions (verifying with composition, domain restrictions)
Inverse Functions
Inverse Functions is a topic in Functions in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.2(E), P.2(H), which requires students to determine and analyze inverse functions using multiple representations.
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 \(y=x\).