OpenStax-CNX module: m39282
1
Functions and Graphs: Graphs of
∗
Inverse Functions (Grade 12)
Free High School Science Texts Project
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 3.0
1 Graphs of Inverse Functions
In earlier grades, you studied various types of functions and understood the eect of various parameters in
the general equation. In this section, we will consider
inverse functions.
An inverse function is a function which "does the reverse" of a given function. More formally, if
function with domain
X,
then
f −1
is its inverse function if and only if for every
x∈X
f −1 (f (x)) == x
A simple way to think about this is that a function, say
x-value into f (x).
sin
y -value
into
sin
y1
=
(1)
y = f (x), gives you a y -value if you substitute an
x-value was used to get a particular y -value
There are some things which can complicate this for example,
x-values
that give you a peak as the function oscillates. This
function would be tricky to dene because if you substitute the peak
into it you won't know which of the
=
f −1 (x).
function, there are many
means that the inverse of a
y
is a
The inverse function tells you tells you which
when you substitue the
think about a
f
we have:
x-values
y -value
was used to get the peak.
f (x)
we have a function
f (x1 )
we substitute a specific x − value into the function to get a specific y − value
consider the inverse function
x
=
f −1 (y)
x
=
f −1 (y)
(2)
substituting the specific y − value into the inverse should return the specific x − value
=
f −1 (y1 )
=
x1
This works both ways, if we don't have any complications like in the case of the
sin
function, so we can
write:
f −1 (f (x)) = f f −1 (x) = x
∗ Version
1.1: Aug 1, 2011 2:35 am -0500
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(3)
OpenStax-CNX module: m39282
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For example, if the function
x → 3x + 2
is given, then its inverse function is
x→
(x−2)
3 . This is usually
written as:
: x → 3x + 2
f
f −1
(4)
(x−2)
3
x→
:
The superscript "-1" is not an exponent.
If a function
If
f
f
has an inverse then
that is a horizontal line
real
f
is said to be invertible.
is a real-valued function, then for
y=k
f
to have a valid inverse, it must pass the horizontal line test,
placed anywhere on the graph of
f
must pass through
f
exactly once for all
k.
It is possible to work around this condition, by dening a multi-valued function as an inverse.
If one represents the function
of a straight line,
f −1 ,
f
Algebraically, one computes the inverse
for
x,
xy -coordinate system, the inverse function of the equation
f across the line y = x.
function of f by solving the equation
graphically in a
is the reection of the graph of
and then exchanging
y
and
x
y = f (x)
(5)
y = f −1 (x)
(6)
to get
Khan academy video on inverse functions - 1
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Figure 1
1.1 Inverse Function of
The inverse function of
y = ax + q
y = ax + q
is determined by solving for
y
=
ax + q
ax
=
y−q
x
=
y−q
a
=
1
ay
−
x
as:
(7)
q
a
q
a.
The inverse function of a straight line is also a straight line, except for the case where the straight line
Therefore the inverse of
y = ax + q
is
y = a1 x −
is a perfectly horizontal line, in which case the inverse is undened.
For example, the straight line equation given by
y = 2x − 3
has as inverse the function,
y = 12 x + 32 .
The
graphs of these functions are shown in Figure 2. It can be seen that the two graphs are reections of each
other across the line
y = x.
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OpenStax-CNX module: m39282
Figure 2:
3
The graphs of the function
f (x) = 2x − 3
and its inverse
f −1 (x) = 21 x +
3
. The line
2
y=x
is shown as a dashed line.
1.1.1 Domain and Range
We have seen that the domain of a function of the form
y = ax+q
is
{x : x ∈ R} and the range is {y : y ∈ R}.
Since the inverse function of a straight line is also a straight line, the inverse function will have the same
domain and range as the original function.
1.1.2 Intercepts
y = ax + q is y = a1 x − aq .
yint = − aq . Similarly, by setting y = 0
The general form of the inverse function of the form
By setting
x-intercept
is
x = 0 we
xint = q .
have that the
y -intercept
is
we have that the
f (x) = ax + q , then f −1 (x) = a1 x − aq and the y -intercept
x-intercept of f (x) is the y -intercept of f −1 (x).
It is interesting to note that if
x-intercept
of
f
−1
(x)
and the
of
f (x)
is the
1.2 Exercises
1. Given
f (x) = 2x − 3,
f −1 (x)
f (x) = 3x − 7.
nd
2. Consider the function
a. Is the relation a function?
b. If it is a function, identify the domain and range.
f (x) = 3x − 1 and its inverse on the same
f −1 (x) = 2x − 4, what is the function f (x)?
3. Sketch the graph of the function
4. The inverse of a function is
1.3 Inverse Function of
set of axes.
y = ax2
The inverse relation, possibly a function, of
y = ax2
is determined by solving for
y
=
x2
=
x =
http://cnx.org/content/m39282/1.1/
x
as:
ax2
y
a
p
± ay
(8)
OpenStax-CNX module: m39282
Figure 3:
4
The function
f (x) = x2
and its inverse
√
f −1 (x) = ± x.
The line
y=x
is shown as a dashed
line.
y = ax2 is not a√function because it fails the vertical line test. If
−1
graph of f
(x) = ± x, the line intersects the graph more than once.
We see that the inverse function of
we draw a vertical line through the
There has to be a restriction on the domain of a parabola for the inverse to also be a function. Consider the
function
f (x) = −x2 + 9.
If we restrict the
of
f
is
x≤0
then
The inverse of
f
can be found by witing
x
=
y2
=
f (y) = x.
Then
−y 2 + 9
9−x
√
y = ± 9−x
√
9 − x is
domain of f (x) to be x ≥ 0, then
√
− 9 − x would be a function, inverse to f .
(9)
a function. If the restriction on the domain
Khan academy video on inverse functions - 2
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Figure 4
Khan academy video on inverse functions - 3
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Figure 5
1.4 Exercises
1. The graph of
f −1
is shown. Find the equation of
simplify your answer)
http://cnx.org/content/m39282/1.1/
f,
given that the graph of
f
is a parabola. (Do not
OpenStax-CNX module: m39282
5
Figure 6
2.
f (x) = 2x2 .
a. Draw the graph of
b. Find
f −1
f
and state its domain and range.
and, if it exists, state the domain and range.
c. What must the domain of
3. Sketch the graph of
x=−
p
f
be, so that
10 −
f −1
is a function ?
y 2 . Label a point on the graph other than the intercepts with the
axes.
4.
a. Sketch the graph of
y = x2
labelling a point other than the origin on your graph.
b. Find the equation of the inverse of the above graph in the form
c. Now sketch the graph of
y=
√
x. √
y = x
d. The tangent to the graph of
y = ....
x-axis at B. Find the
y -axis bisects the straight line AB.
g −1 (x) = ....
at the point A(9;3) intersects the
equation of this tangent and hence or otherwise prove that the
5. Given:
g (x) = −1 +
1.5 Inverse Function of
The inverse function of
√
x,
nd the inverse of
g (x)
in the form
y = ax
y = ax2
is determined by solving for
x
as follows:
y
=
ax
log (y)
=
log (ax )
(10)
= xlog (a)
∴
The inverse of
f
−1
y = 10x
is
x = 10y ,
x
=
log(y)
log(a)
which we write as
y = log (x).
Therefore, if
f (x) = 10x ,
then
= log (x).
Figure 7:
The function
f (x) = 10x
and its inverse
f −1 (x) = log (x).
The line
y = x
is shown as a
dashed line.
The exponential function and the logarithmic function are inverses of each other; the graph of the one is
the graph of the other, reected 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.
1.6 Exercises
1 x
−1
, sketch the graphs of f and f
on the same system of axes indicating a point
5
−1
on each graph (other than the intercepts) and showing clearly which is f and which is f
.
1. Given that
f (x) =
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OpenStax-CNX module: m39282
2. Given that
6
f (x) = 4−x ,
a. Sketch the graphs of
f
f −1
and
on the same system of axes indicating a point on each graph
(other than the intercepts) and showing clearly which is
b. Write
3. Given
f
−1
in the form
g (x) = −1 +
√
x,
f
and which is
f −1 .
y = ....
nd the inverse of
g (x)
in the form
g −1 (x) = ...
4. Answer the following questions:
a. Sketch the graph of
y = x2 ,
labeling a point other than the origin on your graph.
b. Find the equation of the inverse of the above graph in the form
c. Now, sketch
y=
√
x.
d. The tangent to the graph of
y =
√
x
at the point
A (9; 3)
y = ...
intersects the
equation of this tangent, and hence, or otherwise, prove that the
y -axis
x-axis
at
B.
Find the
bisects the straight line
AB .
2 End of Chapter Exercises
1. Sketch the graph of
2.
f (x) =
p
x = − 10 − y 2 .
Is this graph a function ? Verify your answer.
1
x−5 ,
a. determine the
b. determine
x
if
y -intercept of f (x)
f (x) = −1.
3. Below, you are given 3 graphs and 5 equations.
Figure 8
a.
b.
c.
d.
e.
y
y
y
y
y
= log 3 x
= −log 3 x
= log 3 (−x)
= 3−x
= 3x
Write the equation that best describes each graph.
4. Given
g (x) = −1 +
5. Consider the
√
x, nd the inverse
x
equation h (x) = 3
a. Write down the inverse in the form
b. Sketch the graphs of
h (x)
and
−1
h
of
g (x)
in the form
h−1 (x) = ...
(x) on the same
g −1 (x) = ...
set of axes, labelling the intercepts with the
axes.
c. For which values of
6.
a. Sketch the graph of
x
is
h−1 (x)
2
undened ?
y = 2x + 1,
labelling a point other than the origin on your graph.
b. Find the equation of the inverse of the above graph in the form
c. Now, sketch
y=
√
x.
d. The tangent to the graph of
y =
√
x
at the point
A (9; 3)
y = ...
intersects the
equation of this tangent, and hence, or otherwise, prove that the
AB .
http://cnx.org/content/m39282/1.1/
y -axis
x-axis
at
B.
Find the
bisects the straight line