39.1 Introduction
In grades 10 and 11 you have learnt about linear functions and quadratic
functions as well as the
hyperbolic functions and exponential functions and many more. In grade 12 you
are expected
to demonstrate the ability to work with various types of functions and relations
including the
inverses of some functions and generate graphs of the inverse relations of
functions, in particular
the inverses of:
39.2 Definition of a Function
A function is a relation for which there is only one value of y corresponding to
any value of x .
We sometimes write y = f(x), which is notation meaning ’y is a function of x’.
This definition
makes complete sense when compared to our real world examples — each person has
only one
height, so height is a function of people; on each day, in a specific town,
there is only one average
temperature.
However, some very common mathematical constructions are not functions. For
example, consider
the relation x2 + y2 = 4. This relation describes a circle
of radius 2 centred at the origin,
as in figure 39.1. If we let x = 0, we see that y2 = 4 and thus
either y = 2 or y = −2. Since
there are two y values which are possible for the same x value, the relation x2
+ y2 = 4 is not
a function.
There is a simple test to check if a relation is a function, by looking at its
graph. This test
is called the vertical line test . If it is possible to draw any vertical line (a
line of constant x)
which crosses the relation more than once, then the relation is not a
function. If more than one
intersection point exists, then the intersections correspond to multiple values
of y for a single
value of x.
We can see this with our previous example of the circle by looking at its graph
again in Figure
39.1.
We see that we can draw a vertical line, for example the dotted line in the
drawing, which cuts
the circle more than once. Therefore this is not a function.
39.2.1 Exercises
1. State whether each of the fol lowing equations are functions or not:
A x + y = 4
Figure 39.1: Graph of y2 + x2 = 4
2. The table gives the average per capita income, d, in a
region of the country as a function
of the percent unemployed, u. Write down the equation to show that income is a
function
of the persent unemployed.
| u |
1 |
2 |
3 |
4 |
| d |
22500 |
22000 |
21500 |
21000 |
39.3 Notation used for Functions
In grade 10 you were introduced to the notation used to ”name” a function. In a
function
y = f(x), y is called the dependent variable, because the value of y depends on
what you
choose as x. We say x is the independent variable, since we can choose x to be
any number.
Similarly if g(t) = 2t + 1, then t is the independent variable and g is the
function name. If
f(x) = 3x−5 and you are ask to de termine f (3), then you have to work out the
value for f(x)
when x = 3. For example,
39.4 Graphs of Inverse Functions
In earlier grades, you studied various types of functions and understood the
effect 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 f is a function with domain X, then f -1 is its inverse function
if and only if for every x ∈ X
we have:
f -1(f(x)) = f(f -1(x)) = x (39.1)
For example, if the function x → 3x+2 is given, then its
inverse function is x →
.
This is usually written as:
The superscript ”-1” is not an exponent.
If a function f has an inverse then f is said to be invertible.
If f is a real-valued function, then for f to have a valid inverse, it must pass
the horizontal line
test, that is a horizontal line y = k placed anywhere on the graph of f must
pass through f
exactly once for all real k.
It is possible to work around this condition, by defining a multi-valued
function as an inverse.
If one represents the function f graphically in a xy-coordinate system, then the
graph of f -1 is
the reflection of the graph of f across the line y = x.
Algebraically, one computes the inverse function of f by solving the equation
y = f(x)
for x, and then ex changing y and x to get
y = f -1(x)
39.4.1 Inverse Function of y = ax + q
The inverse function of y = ax + q is determined by solving for x as:
Therefore the inverse of y = ax + q is
.
The inverse function of a straight line is also a straight line.
For example, the straight line equation given by y = 2x − 3 has as inverse the
function, y =
. The graphs of these functions are shown in
Figure 39.2. It can be seen that the two
graphs are reflections of each other across the line y = x.
Figure 39.2: The function f(x) = 2x − 3 and its inverse
.The
line y = x is
shown as a dashed line.
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.
Intercepts
The general form of the inverse function of the form y = ax + q is
.
By setting x = 0 we have that the y- intercept is
. Similarly, by setting y = 0 we have
that the x-intercept is xint = q.
It is interesting to note that if f(x) = ax + q, then
and
the y-intercept of
f (x) is the x-intercept of f -1(x) and the x-intercept of
f(x) is the y-intercept of f -1(x).
39.4.2 Exercises
1. Given f(x) = 2x − 3, find f -1(x)
2. Consider the function f(x) = 3x − 7.
A Is the relation a function?
B Identify the domain and range.
3. Sketch the graph of the function f(x) = 3x − 1 and its inverse on the same
set of axes.
4. The inverse of a function is f -1(x) = 2x − 4, what is the
function f(x)?
39.4.3 Inverse Function of y = ax2
The inverse function of y = ax2 is determined by solving for x as:
We see that the inverse function of y = ax2 is
not a function because it fails the vertical line
test. If we draw a vertical line through the graph of f -1(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(x) = −x2 +9. The
inverse of f can be found by
witing f(y) = x. Then
If x ≥ 0, then 
is a function. If the
restriction on the domain of f is x ≤ 0 then −
would be a function.
39.4.4 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)
Figure 39.3: The function f(x) = x2 and its
inverse f -1(x) = ±.
The line y = x is shown
as a dashed line.
2. f(x) = 2x2.
A Draw the graph of f and state its domain and range.
B Find f -1 and state the domain and range.
C What must the domain of f be, so that f -1 is a function ?
3. Sketch the graph of x = −
. 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 y = . . ..
C Now sketch the y = .
D The tangent to the graph of y =
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.
5. Given: g(x) = −1 +
, find the inverse of
g(x) in the form g-1(x).
39.4.5 Inverse Function of y = ax
The inverse function of y = ax2 is determined by solving for x as:
The inverse of y = 10x is x = 10y,
which we write as y = logx. Therefore, if f(x) = 10x, then
f -1 = logx.
Figure 39.4: 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, 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.
39.4.6 Exercises
1. Given that f(x) = [ 
]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(x) = 4-x,
A Sketch the graphs of f and f -1on 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.
B Write f -1 in the form y = . . ..
3. Given g(x) = −1 +
, find the inverse of
g(x) in the form g-1(x) = . . .
4. 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 y = . . .
C Now, sketch y = .
D The tangent to the graph of y =
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.
39.5 End of Chapter Exercises
1. Sketch the graph of x = −
. Is this graph a
function ? Verify your answer.
2. 
,
A determine the y-intercept of f(x)
B determine x if f(x) = −1.
3. Below, you are given 3 graphs and 5 equations.
Write the equation that best describes each graph.
4. The graph of y = f(x) is shown in the diagram below.
A Find the value of x such that f(x) = 0.
B Evaluate f(3) + f(−1).
5. Given g(x) = −1 +
, find the inverse of g(x) in the form g-1(x)
= . . .
6. Given the equation h(x) = 3x
A Write down the inverse in the form h-1(x) = ...
B Sketch the graphs of h(x) and h-1(x) on teh same set of axes,
labelling the intercepts
with the axes.
C For which values of x is h-1(x) undefined ?
7. 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 y = . . .
C Now, sketch y = .
D The tangent to the graph of y =
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.
