Inverse functions

One to One Functions

A one‑to‑one correspondence exists when all elements of set A map to exactly one element of set B, and vice versa.

This is a one‑to‑one correspondence:

This is a function, but not a one‑to‑one correspondence:

When a one‑to‑one correspondence exists, the function that maps from set B to set A is called the inverse of \(f\).

The inverse of a function \(f(x)\) is denoted \(f^{-1}(x)\).

For each element: \[ f(a) = b \quad \text{and} \quad f^{-1}(b) = a \]

The domain of \(f\) is the range of \(f^{-1}\). The range of \(f\) is the domain of \(f^{-1}\).

\[f^{-1}(f(x)) = f(f^{-1}(x)) = x \]

Inverse a Function

To inverse a function (which must be one‑to‑one), reflect it in the line \(y = x\).

Example

Find the inverse of the function \(f(x) = 2x + 2\).

The graph of \(f(x) = 2x + 2\) is the line \(y = 2x + 2\).

\(f(0) = 2\), so the point \((0,2)\) lies on the line. \(f(-1) = 0\), so the point \((-1,0)\) lies on the line.

Reflecting in the line \(y = x\) swaps the coordinates:

\((0,2) \rightarrow (2,0)\) \((-1,0) \rightarrow (0,-1)\)

The equation of the reflected line can now be found:

\[ m = \frac{0 - (-1)}{2 - 0} = \frac{1}{2} \] \[ \text{Point } (2,0) \text{ is on the line} \] \[ y - b = m(x - a) \] \[ y - 0 = \frac12 (x - 2) \] \[ \Rightarrow\; y = \frac12 x - 1 \] \[ \text{Given } f(x) = 2x + 2,\; \text{its inverse } f^{-1}(x) = \frac12 x - 1 \]

Which looks like this:

Algebraic Method

Example

Find the inverse of the function \(f(x) = 2x + 2\).

\[ f^{-1}(f(x)) = f(f^{-1}(x)) = x \]

Substitute into \(f(x) = 2x + 2\):

\[ 2(f^{-1}(x)) + 2 = x \]

Rearrange to make \(f^{-1}(x)\) the subject:

\[ 2f^{-1}(x) = x - 2 \] \[ f^{-1}(x) = \frac{x - 2}{2} = \frac{1}{2}x - 1 \]

Alternatively

Start with the graph equation \(y = 2x + 2\) and make \(x\) the subject:

\[ y = 2x + 2 \] \[ y - 2 = 2x \] \[ x = \tfrac12 (y - 2) \]

Swap the letters (reflection in \(y = x\)):

\[ y = \tfrac12(x - 2) = \tfrac12 x - 1 \]

So \(f^{-1}(x) = \tfrac12 x - 1\).

Example

\[ f(x) = 2x^2 + 4,\quad -3 \le x \le 3,\quad x \in \mathbb{Z} \] \[ \text{a) State the range.} \] \[ \text{b) Does an inverse function exist?} \]

\[ f(x) = 2x^2 + 4 \] \[ f(-3) = 2(-3)^2 + 4 = 2 \cdot 9 + 4 = 18 + 4 = 22 \] \[ f(-2) = 2(-2)^2 + 4 = 2 \cdot 4 + 4 = 8 + 4 = 12 \] \[ f(-1) = 2(-1)^2 + 4 = 2 \cdot 1 + 4 = 2 + 4 = 6 \] \[ f(0) = 2(0)^2 + 4 = 0 + 4 = 4 \] \[ f(1) = 2(1)^2 + 4 = 2 + 4 = 6 \] \[ f(2) = 2(2)^2 + 4 = 8 + 4 = 12 \] \[ f(3) = 2(3)^2 + 4 = 18 + 4 = 22 \] \[ \text{Range is } \{4, 6, 12, 22\} \]

\[ \text{An inverse function does not exist, since it is not a one-to-one correspondence.} \]

Example

\[ f(x) = 2x^2 + 4,\quad x \in \mathbb{R} \] \[ \text{Does an inverse function exist?} \]

\[ y = 2x^2 + 4 \] \[ \Rightarrow\; y - 4 = 2x^2 \] \[ \Rightarrow\; \frac{y - 4}{2} = x^2 \] \[ \Rightarrow\; x = \sqrt{\frac{y - 4}{2}} \]

\[ \text{Swap } x \text{ and } y \text{ (reflection in the line } y = x\text{)} \] \[ \sqrt{\frac{x - 4}{2}} = y \] \[ y = \pm \sqrt{\frac{x - 4}{2}} \] \[ \text{An inverse function does not exist, since it is not a one-to-one correspondence.} \]