Inverse of the Function <em>y = ax + q</em>
Inverse of the function \(y=ax+q\)
Example
Question
Determine the inverse function of \(p(x) = -3x + 1\) and sketch the graphs of \(p(x)\) and \(p^{-1}(x)\) on the same system of axes.
Determine the inverse of the given function
- Interchange \(x\) and \(y\) in the equation.
- Make \(y\) the subject of the new equation.
- Express the new equation in function notation.
Therefore, \(p^{-1}(x) = -\cfrac{x}{3} + \cfrac{1}{3}\).
Sketch the graphs of the same system of axes
The graph of \(p^{-1}(x)\) is the reflection of \(p(x)\) about the line \(y = x\). This means that every point on the graph of \(p(x)\) has a mirror image on the graph of \(p^{-1}(x)\).
To determine the inverse function of \(y=ax+q\):
\[\begin{array}{rll} &(1) \quad \text{Interchange } x \text{ and } y: & x = ay+q \\ &(2) \quad \text{Make } y \text{ the subject of the equation}: & x - q = ay \\ && \cfrac{x}{a} - \cfrac{q}{a} = \cfrac{ay}{a} \\ &&\therefore y = \cfrac{1}{a}x-\cfrac{q}{a} \end{array}\]
Therefore the inverse of \(y=ax+q\) is \(y=\cfrac{1}{a}x-\cfrac{q}{a}\). If a linear function is invertible, then its inverse will also be linear.
Example
Question
Determine and sketch the inverse of the function \(f(x) = 2x - 3\). State the domain, range and intercepts.
Determine the inverse of the given function
- Interchange \(x\) and \(y\) in the equation.
- Make \(y\) the subject of the new equation.
- Express the new equation in function notation.
Therefore, \(f^{-1}(x) = \cfrac{x}{2} + \cfrac{3}{2}\).
Sketch the graphs on the same system of axes
The graph of \(f^{-1}(x)\) is the reflection of \(f(x)\) about the line \(y = x\).
Determine domain, range and intercepts
\begin{align*} \text{Domain of } f &: \{ x: x \in \mathbb{R} \} \\ \text{Range of } f &: \{ y: y \in \mathbb{R} \} \\ \text{Intercepts of } f &: (0;-3) \text{ and } \left(\cfrac{3}{2}; 0 \right) \\ & \\ \text{Domain of } f^{-1} &: \{ x: x \in \mathbb{R} \} \\ \text{Range of } f^{-1} &: \{ y: y \in \mathbb{R} \} \\ \text{Intercepts of } f^{-1} &: \left(0;\cfrac{3}{2}\right) \text{ and } (-3;0) \end{align*}Notice that the intercepts of \(f\) and \(f^{-1}\) are mirror images of each other. In other words, the \(x\)- and \(y\)-values have “swapped” positions. This is true of every point on the two graphs.
Domain and range
For a function of the form \(y=ax+q\), the domain is \(\{x:x\in ℝ\}\) and the range is \(\{y:y\in ℝ\}\). When a function is inverted the domain and range are interchanged. Therefore, the domain and range of the inverse of an invertible, linear function will be \(\{x:x\in ℝ\}\) and \(\{y:y\in ℝ\}\) respectively.
Intercepts
The general form of an invertible, linear function is \(y=ax+q \enspace (a \ne 0)\) and its inverse is \(y=\cfrac{1}{a}x-\cfrac{q}{a}\).
The \(y\)-intercept is obtained by letting \(x=0\):
\begin{align*} y &= \cfrac{1}{a}(0)-\cfrac{q}{a} \\ y &= -\cfrac{q}{a} \end{align*}This gives the point \((0; -\cfrac{q}{a})\).
The \(x\)-intercept is obtained by letting \(y=0\):
\begin{align*} 0 &= \cfrac{1}{a}x -\cfrac{q}{a} \\ \cfrac{q}{a} &= \cfrac{1}{a}x \\ q &= x \end{align*}This gives the point \((q; 0)\).
It is interesting to note that if \(f(x)=ax+q \enspace (a \ne 0)\), then \({f}^{-1}(x)=\cfrac{1}{a}x-\cfrac{q}{a}\) and the \(y\)-intercept of \(f(x)\) is the \(x\)-intercept of \({f}^{-1}(x)\) and the \(x\)-intercept of \(f(x)\) is the \(y\)-intercept of \({f}^{-1}(x)\).
This lesson is part of:
Functions III