The general form of the inverse function of the form
is
.
By setting
we have that the
-intercept is
. Similarly, by setting
we have that the
-intercept is
.
It is interesting to note that if
, then
and the
-intercept of
is the
-intercept of
and the
-intercept of
is the
-intercept of
.
Exercises
Given
, find
Consider the function
.
Is the relation a function?
If it is a function, identify the domain and range.
Sketch the graph of the function
and its inverse on the same set of axes.
The inverse of a function is
, what is the function
?
Inverse function of
The inverse relation, possibly a function, of
is determined by solving for
as:
We see that the inverse ”function” of
is not a function because it fails the vertical line test. If we draw a vertical line through the graph of
, 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
. The inverse of
can be found by witing
. Then
If we restrict the domain of
to be
, then
is a function. If the restriction on the domain of
is
then
would be a function, inverse to
.
Exercises
The graph of
is shown. Find the equation of
, given that the graph of
is a parabola. (Do not simplify your answer)
.
Draw the graph of
and state its domain and range.
Find
and, if it exists, state the domain and range.
What must the domain of
be, so that
is a function ?
Sketch the graph of
. Label a point on the graph other than the intercepts with the axes.
Sketch the graph of
labelling a point other than the origin on your graph.
Find the equation of the inverse of the above graph in the form
.
Now sketch the graph of
.
The tangent to the graph of
at the point A(9;3) intersects the
-axis at B. Find the equation of this tangent and hence or otherwise prove that the
-axis bisects the straight line AB.
Given:
, find the inverse of
in the form
.
Inverse function of
The inverse function of
is determined by solving for
as follows:
The inverse of
is
, which we write as
. Therefore, if
, then
.
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
.
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
, sketch the graphs of
and
on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is
and which is
.
Given that
,
Sketch the graphs of
and
on the same system of axes indicating a point on each graph (other than the intercepts) and showing clearly which is
and which is
.
Write
in the form
.
Given
, find the inverse of
in the form
Answer the following questions:
Sketch the graph of
, labeling a point other than the origin on your graph.
Find the equation of the inverse of the above graph in the form
Now, sketch
.
The tangent to the graph of
at the point
intersects the
-axis at
. Find the equation of this tangent, and hence, or otherwise, prove that the
-axis bisects the straight line
.
End of chapter exercises
Sketch the graph of
. Is this graph a function ? Verify your answer.
,
determine the
-intercept of
determine
if
.
Below, you are given 3 graphs and 5 equations.
Write the equation that best describes each graph.
Given
, find the inverse of
in the form
Consider the equation
Write down the inverse in the form
Sketch the graphs of
and
on the same set of axes, labelling the intercepts with the axes.
For which values of
is
undefined ?
Sketch the graph of
, labelling a point other than the origin on your graph.
Find the equation of the inverse of the above graph in the form
Now, sketch
.
The tangent to the graph of
at the point
intersects the
-axis at
. Find the equation of this tangent, and hence, or otherwise, prove that the
-axis bisects the straight line
.