2.4 Linear functions | Functions

2.4 Linear functions (EMCF9)

Inverse of the function \(y=ax+q\) (EMCFB)

Worked example 4: Inverse of the function \(y=ax+q\)

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.

\begin{align*} \text{Let } y & = -3x + 1 \\ \text{Interchange } x \text{ and } y: \quad x & = -3y + 1 \\ x - 1 & = -3y \\ -\frac{1}{3}(x - 1) & = y \\ \therefore y &= -\frac{x}{3} + \frac{1}{3} \end{align*}

Therefore, \(p^{-1}(x) = -\frac{x}{3} + \frac{1}{3}\).

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 \(y=ax+q\):

\[\begin{array}{rll} &(1) \quad \text{Interchange } x \text{ and } y: & x = ay+q \\ &(2) \quad \text{Make } y \text{ the subject of the equation}: & x - q = ay \\ && \frac{x}{a} - \frac{q}{a} = \frac{ay}{a} \\ &&\therefore y = \frac{1}{a}x-\frac{q}{a} \end{array}\]

Therefore the inverse of \(y=ax+q\) is \(y=\frac{1}{a}x-\frac{q}{a}\). If a linear function is invertible, then its inverse will also be linear.

Worked example 5: Inverses - domain, range and intercepts

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.

\begin{align*} \text{Let } y & = 2x - 3 \\ \text{Interchange } x \text{ and } y: \quad x & = 2y - 3 \\ x + 3 & = 2y \\ \frac{1}{2}(x + 3) & = y \\ \therefore y &= \frac{x}{2} + \frac{3}{2} \end{align*}

Therefore, \(f^{-1}(x) = \frac{x}{2} + \frac{3}{2}\).

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(\frac{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;\frac{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 \(\left\{x:x\in ℝ\right\}\) and the range is \(\left\{y:y\in ℝ\right\}\). 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 \(\left\{x:x\in ℝ\right\}\) and \(\left\{y:y\in ℝ\right\}\) respectively.

Intercepts

The general form of an invertible, linear function is \(y=ax+q \enspace (a \ne 0)\) and its inverse is \(y=\frac{1}{a}x-\frac{q}{a}\).

The \(y\)-intercept is obtained by letting \(x=0\):

\begin{align*} y &= \frac{1}{a}(0)-\frac{q}{a} \\ y &= -\frac{q}{a} \end{align*}

This gives the point \(\left(0; -\frac{q}{a}\right)\).

The \(x\)-intercept is obtained by letting \(y=0\):

\begin{align*} 0 &= \frac{1}{a}x -\frac{q}{a} \\ \frac{q}{a} &= \frac{1}{a}x \\ q &= x \end{align*}

This gives the point \((q; 0)\).

It is interesting to note that if \(f\left(x\right)=ax+q \enspace (a \ne 0)\), then \({f}^{-1}\left(x\right)=\frac{1}{a}x-\frac{q}{a}\) and the \(y\)-intercept of \(f\left(x\right)\) is the \(x\)-intercept of \({f}^{-1}\left(x\right)\) and the \(x\)-intercept of \(f\left(x\right)\) is the \(y\)-intercept of \({f}^{-1}\left(x\right)\).

Video: 28B3

Inverse of the function \(y = ax + q\)

Textbook Exercise 2.3

Given \(f\left(x\right)=5x + 4\), find \({f}^{-1}\left(x\right)\).

\begin{align*} f(x): y & = 5x + 4 \\ f^{-1}(x): x & = 5y + 4 \\ \therefore x - 4 & = 5y \\ y & = \frac{1}{5}x - \frac{4}{5}\\ \therefore f^{-1}(x) & = \frac{1}{5}x - \frac{4}{5} \end{align*}

Is the relation a function? Explain your answer.

It is a function. Every \(x\)-value relates to only one \(y\)-value, it is a one-to-one relation.

Identify the domain and range.

\begin{align*} \text{Domain } & \{ x: x \in \mathbb{R} \} \\ \text{Range } & \{ y: y \in \mathbb{R} \} \end{align*}

Determine \(f^{-1}(x)\).

\begin{align*} f(x): y & = -3x-7 \\ f^{-1}(x): x & = -3y-7 \\ \therefore x + 7 & = -3y \\ y & = - \frac{1}{3}x - \frac{7}{3} \\ \therefore f^{-1}(x) & = - \frac{1}{3}x - \frac{7}{3} \end{align*}

Sketch the graph of the function \(f\left(x\right)=3x-1\) and its inverse on the same system of axes. Indicate the intercepts and the axis of symmetry of the two graphs.

\begin{align*} y & = 3x - 1 \\ x & = 3y - 1\\ \therefore 3y & = x + 1 \\ y & = \frac{1}{3}x + \frac{1}{3} \end{align*}

The intercepts are:

\begin{align*} f(x): x = 0, y = -\text{1}\\ y=0, x = \frac{1}{3}\\ f^{-1}(x): x = 0, y = \frac{1}{3}\\ y = 0, x = -\text{1} \end{align*}

\(T\left( \frac{4}{3}; 3 \right)\) is a point on \(f\) and \(R\) is a point on \(f^{-1}\). Determine the coordinates of \(R\) if \(R\) and \(T\) are symmetrical.

\(R\left( 3; \frac{4}{3} \right)\)

Explain why the line \(y = x\) is an axis of symmetry for a function and its inverse.

To reflect a function about the \(y\)-axis, we replace every \(x\) with \(-x\). Similarly, to reflect a function about the \(x\)-axis, we replace every \(y\) with \(-y\). To reflect a function about the line \(y=x\), we replace \(x\) with \(y\) and \(y\) with \(x\), which is how we determine the inverse.

Will the line \(y = -x\) be an axis of symmetry for a function and its inverse?

No it will not.

Given \({f}^{-1}\left(x\right)=-2x+4\), determine \(f\left(x\right)\).

\begin{align*} f^{-1}(x): y & = -2x + 4 \\ f(x): x & = -2y + 4 \\ \therefore 2y & = -x + 4 \\ y & = -\frac{1}{2}x + 2\\ \therefore f(x) & = -\frac{1}{2}x + 2 \end{align*}

Calculate the intercepts of \(f(x)\) and \(f^{-1}(x)\).

Consider \(f(x)\):

\begin{align*} \text{Let } y & = -\frac{1}{2}x + 2 \\ \text{Let } x=0: \quad y & = -\frac{1}{2}(0) + 2 \\ y & = 2 \\ \therefore y &-\text{intercept is } (0;2) \\ & \\ \text{Let } y=0: \quad 0 & = -\frac{1}{2}x + 2 \\ -2 & = -\frac{1}{2}x \\ x & = 4 \\ \therefore x &-\text{intercept is } (4;0) \end{align*}

Consider \(f^{-1}(x)\):

\begin{align*} \text{Let } y & = - 2x+4 \\ \text{Let } x=0: \quad y & = - 2(0) + 4\\ y & = 4 \\ \therefore y &-\text{intercept is } (0;4) \\ & \\ \text{Let } y=0: \quad 0 & = - 2x+4\\ 2x & = 4 \\ x & = 2 \\ \therefore x &-\text{intercept is } (2;0) \end{align*}

Therefore the intercepts for \(f(x)\) are \((4;0)\) and \((0;2)\) and the intercepts for \(f^{-1}(x)\) are \((2;0)\) and \((0;4)\).

Determine the coordinates of \(T\), the point of intersection of \(f(x)\) and \(f^{-1}(x)\).

To find the point on intersection, we let \(f(x) = f^{-1}(x)\):

\begin{align*} -\frac{1}{2}x + 2 & = -2x + 4 \\ -\frac{1}{2}x + 2x & = 4 - 2 \\ \frac{3}{2}x & = 2 \\ \therefore x & = \frac{4}{3} \\ \text{And } y & = -2 \left( \frac{4}{3} \right) + 4 \\ & = - \frac{8}{3} + 4 \\ & = \frac{4}{3} \end{align*}

This gives the point \(T \left( \frac{4}{3};\frac{4}{3} \right)\).

Sketch the graphs of \(f\) and \(f^{-1}\) on the same system of axes. Indicate the intercepts and point \(T\) on the graph.

Is \(f^{-1}\) an increasing or decreasing function?

Decreasing function. The function values decrease as \(x\) increases.

2.3 Inverse functions

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2.5 Quadratic functions