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
Given
$f\left(x\right)=2x-3$ , find
${f}^{-1}\left(x\right)$
Consider the function
$f\left(x\right)=3x-7$ .
Is the relation a function?
If it is a function, identify the domain and range.
Sketch the graph of the function
$f\left(x\right)=3x-1$ and its inverse on the same set of axes.
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:
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
${f}^{-1}\left(x\right)=\pm \sqrt{x}$ , 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
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
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)
$f\left(x\right)=2{x}^{2}$ .
Draw the graph of
$f$ and state its domain and range.
Find
${f}^{-1}$ and, if it exists, state the domain and range.
What must the domain of
$f$ be, so that
${f}^{-1}$ is a function ?
Sketch the graph of
$x=-\sqrt{10-{y}^{2}}$ . Label a point on the graph other than the intercepts with the axes.
Sketch the graph of
$y={x}^{2}$ labelling a point other than the origin on your graph.
Find the equation of the inverse of the above graph in the form
$y=...$ .
Now sketch the graph of
$y=\sqrt{x}$ .
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.
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:
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
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
${f}^{-1}$ .
Given that
$f\left(x\right)={4}^{-x}$ ,
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
${f}^{-1}$ .
Write
${f}^{-1}$ in the form
$y=...$ .
Given
$g\left(x\right)=-1+\sqrt{x}$ , find the inverse of
$g\left(x\right)$ in the form
${g}^{-1}\left(x\right)=...$
Answer the following questions:
Sketch the graph of
$y={x}^{2}$ , labeling a point other than the origin on your graph.
Find the equation of the inverse of the above graph in the form
$y=...$
Now, sketch
$y=\sqrt{x}$ .
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
$AB$ .
End of chapter exercises
Sketch the graph of
$x=-\sqrt{10-{y}^{2}}$ . Is this graph a function ? Verify your answer.
$f\left(x\right)={\displaystyle \frac{1}{x-5}}$ ,
determine the
$y$ -intercept of
$f\left(x\right)$
determine
$x$ if
$f\left(x\right)=-1$ .
Below, you are given 3 graphs and 5 equations.
$y={log}_{3}x$
$y=-{log}_{3}x$
$y={log}_{3}(-x)$
$y={3}^{-x}$
$y={3}^{x}$
Write the equation that best describes each graph.
Given
$g\left(x\right)=-1+\sqrt{x}$ , find the inverse of
$g\left(x\right)$ in the form
${g}^{-1}\left(x\right)=...$
Consider the equation
$h\left(x\right)={3}^{x}$
Write down the inverse in the form
${h}^{-1}\left(x\right)=...$
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.
For which values of
$x$ is
${h}^{-1}\left(x\right)$ undefined ?
Sketch the graph of
$y=2{x}^{2}+1$ , labelling a point other than the origin on your graph.
Find the equation of the inverse of the above graph in the form
$y=...$
Now, sketch
$y=\sqrt{x}$ .
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
$AB$ .
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Rafiq
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Rafiq
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