# 4.1 Graphs of inverse functions  (Page 2/2)

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## Intercepts

The general form of the inverse function of the form $y=ax+q$ is $y=\frac{1}{a}x-\frac{q}{a}$ .

By setting $x=0$ we have that the $y$ -intercept is ${y}_{int}=-\frac{q}{a}$ . Similarly, by setting $y=0$ we have that the $x$ -intercept is ${x}_{int}=q$ .

It is interesting to note that if $f\left(x\right)=ax+q$ , 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 ${f}^{-1}\left(x\right)$ .

## Exercises

1. Given $f\left(x\right)=2x-3$ , find ${f}^{-1}\left(x\right)$
2. Consider the function $f\left(x\right)=3x-7$ .
1. Is the relation a function?
2. If it is a function, identify the domain and range.
3. Sketch the graph of the function $f\left(x\right)=3x-1$ and its inverse on the same set of axes.
4. The inverse of a function is ${f}^{-1}\left(x\right)=2x-4$ , what is the function $f\left(x\right)$ ?

## Inverse function of $y=a{x}^{2}$

The inverse relation, possibly a function, of $y=a{x}^{2}$ is determined by solving for $x$ as:

$\begin{array}{ccc}\hfill y& =& a{x}^{2}\hfill \\ \hfill {x}^{2}& =& \frac{y}{a}\hfill \\ \hfill x& =& ±\sqrt{\frac{y}{a}}\hfill \end{array}$

We see that the inverse ”function” of $y=a{x}^{2}$ is not a function because it fails the vertical line test. If we draw a vertical line through the graph of ${f}^{-1}\left(x\right)=±\sqrt{x}$ , the line intersects the graph more than once. There has to be a restriction on the domain of a parabola for the inverse to also be a function. Consider the function $f\left(x\right)=-{x}^{2}+9$ . The inverse of $f$ can be found by witing $f\left(y\right)=x$ . Then

$\begin{array}{ccc}\hfill x& =& -{y}^{2}+9\hfill \\ \hfill {y}^{2}& =& 9-x\hfill \\ \hfill y& =& ±\sqrt{9-x}\hfill \end{array}$

If we restrict the domain of $f\left(x\right)$ to be $x\ge 0$ , then $\sqrt{9-x}$ is a function. If the restriction on the domain of $f$ is $x\le 0$ then $-\sqrt{9-x}$ would be a function, inverse to $f$ .

## Exercises

1. The graph of ${f}^{-1}$ is shown. Find the equation of $f$ , given that the graph of $f$ is a parabola. (Do not simplify your answer)
2. $f\left(x\right)=2{x}^{2}$ .
1. Draw the graph of $f$ and state its domain and range.
2. Find ${f}^{-1}$ and, if it exists, state the domain and range.
3. What must the domain of $f$ be, so that ${f}^{-1}$ is a function ?
3. Sketch the graph of $x=-\sqrt{10-{y}^{2}}$ . Label a point on the graph other than the intercepts with the axes.
1. Sketch the graph of $y={x}^{2}$ labelling a point other than the origin on your graph.
2. Find the equation of the inverse of the above graph in the form $y=...$ .
3. Now sketch the graph of $y=\sqrt{x}$ .
4. The tangent to the graph of $y=\sqrt{x}$ at the point A(9;3) intersects the $x$ -axis at B. Find the equation of this tangent and hence or otherwise prove that the $y$ -axis bisects the straight line AB.
4. Given: $g\left(x\right)=-1+\sqrt{x}$ , find the inverse of $g\left(x\right)$ in the form ${g}^{-1}\left(x\right)=...$ .

## Inverse function of $y={a}^{x}$

The inverse function of $y=a{x}^{2}$ is determined by solving for $x$ as follows:

$\begin{array}{ccc}\hfill y& =& {a}^{x}\hfill \\ \hfill log\left(y\right)& =& log\left({a}^{x}\right)\hfill \\ & =& xlog\left(a\right)\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}x& =& \frac{log\left(y\right)}{log\left(a\right)}\hfill \end{array}$

The inverse of $y={10}^{x}$ is $x={10}^{y}$ , which we write as $y=log\left(x\right)$ . Therefore, if $f\left(x\right)={10}^{x}$ , then ${f}^{-1}=log\left(x\right)$ .

The exponential function and the logarithmic function are inverses of each other; the graph of the one is the graph of the other, reflected in the line $y=x$ . The domain of the function is equal to the range of the inverse. The range of the function is equal to the domain of the inverse.

## Exercises

1. Given that $f\left(x\right)={\left(\frac{1}{5}\right)}^{x}$ , sketch the graphs of $f$ and ${f}^{-1}$ on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is $f$ and which is ${f}^{-1}$ .
2. Given that $f\left(x\right)={4}^{-x}$ ,
1. Sketch the graphs of $f$ and ${f}^{-1}$ on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is $f$ and which is ${f}^{-1}$ .
2. Write ${f}^{-1}$ in the form $y=...$ .
3. Given $g\left(x\right)=-1+\sqrt{x}$ , find the inverse of $g\left(x\right)$ in the form ${g}^{-1}\left(x\right)=...$
1. Sketch the graph of $y={x}^{2}$ , labeling a point other than the origin on your graph.
2. Find the equation of the inverse of the above graph in the form $y=...$
3. Now, sketch $y=\sqrt{x}$ .
4. The tangent to the graph of $y=\sqrt{x}$ at the point $A\left(9;3\right)$ intersects the $x$ -axis at $B$ . Find the equation of this tangent, and hence, or otherwise, prove that the $y$ -axis bisects the straight line $AB$ .

## End of chapter exercises

1. Sketch the graph of $x=-\sqrt{10-{y}^{2}}$ . Is this graph a function ? Verify your answer.
2. $f\left(x\right)=\frac{1}{x-5}$ ,
1. determine the $y$ -intercept of $f\left(x\right)$
2. determine $x$ if $f\left(x\right)=-1$ .
3. Below, you are given 3 graphs and 5 equations.
1. $y={log}_{3}x$
2. $y=-{log}_{3}x$
3. $y={log}_{3}\left(-x\right)$
4. $y={3}^{-x}$
5. $y={3}^{x}$
Write the equation that best describes each graph.
4. Given $g\left(x\right)=-1+\sqrt{x}$ , find the inverse of $g\left(x\right)$ in the form ${g}^{-1}\left(x\right)=...$
5. Consider the equation $h\left(x\right)={3}^{x}$
1. Write down the inverse in the form ${h}^{-1}\left(x\right)=...$
2. Sketch the graphs of $h\left(x\right)$ and ${h}^{-1}\left(x\right)$ on the same set of axes, labelling the intercepts with the axes.
3. For which values of $x$ is ${h}^{-1}\left(x\right)$ undefined ?
1. Sketch the graph of $y=2{x}^{2}+1$ , labelling a point other than the origin on your graph.
2. Find the equation of the inverse of the above graph in the form $y=...$
3. Now, sketch $y=\sqrt{x}$ .
4. The tangent to the graph of $y=\sqrt{x}$ at the point $A\left(9;3\right)$ intersects the $x$ -axis at $B$ . Find the equation of this tangent, and hence, or otherwise, prove that the $y$ -axis bisects the straight line $AB$ .

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