# Functions ```Refresher Unit 8: Basic Functions
8.1 Concept of a function
•
A function is the mathematical expression for the way the value of one quantity
(called the dependent variable) depends on the value of another variable (called the
independent variable).
•
Often the letter x is used to denote the independent variable and the letter y to denote
the dependent variable. The fact that y is a function of x (that is, it depends on x) is
written as
y = f ( x)
as a short hand. When we know the formula for f(x) we can calculate the value of y
for any given value of x
Example 8.1
If y = f ( x) = x3 + 2 find
(a) f (2)
(b) f (−2)
1
(c) f  
2
(a)
f (2) is the value of y when x = 2 so that
f (2) = (2)3 + 2 = 10
(b)
f (−2) is the value of y when x = –2 so that
f (−2) = (−2)3 + 2 = −8 + 2 = −6
(c)
( )
f 1 is the value of y when x = 1 so that
2
2
( ) ( )
3
f 1 = 1 +2 = 1 +2 = 21
2
2
8
8
Example 8.2
An isosceles trapezium has the dimensions, expressed in cms, shown in Figure 8.1. Find
the formula which relates the enclosed area A cm2 to the height x cm. Evaluate A for x =
1, 2, 3, 4 and 5.
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10
x
x
2x
2x
Figure 8.1 Trapezium of Example 8.2
Re-drawing Figure 8.1, we can simplify the geometry
10
2x
x
Hence the area is given by
A = (10 + 2 x) x
Calculating the values of A using the values of x gives us the following table
x
A
1
12
2
28
3
48
4
72
5
100
In this example x is the independent variable and A is the dependent variable, and
f ( x) = (10 + 2 x) x
Practice Questions 8.1
1. If f ( x) =
x2 + 1
find
x −1
(a) f (3)
(b) f (−1)
(c) f (0)
2. A car costs &pound;1200 to run in its first year and subsequently the annual cost
increases by &pound;100 each year. How much does it cost to run the second year?,
third year?, fourth year? Write down a formula for the cost &pound;C in the x-th year.
3. A cake manufacturer found that the weekly sales for a certain product depended
on its selling price. Experience showed that the maximum number that could be
sold per week was 20000 but that the actual number sold decreased by 100 for
every 1p increase in price. The total product cost consisted of a set-up cost of
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&pound;200 plus 50p for each product made. Find he formula that relates the weekly
profit y to the selling price x.
(Assume sales match production)
8.2 Graph of a function
•
Often the easiest way of understanding the properties of a function is to draw its
graph using coordinate axes as introduced in Unit 7; with the independent variable
measured along the horizontal x-axis and the dependent variable along the vertical yaxis.
•
The graph of the function y = f ( x) intercepts the x-axis when f ( x) = 0 . Such
points are called the zeroes of the function f ( x) or the solutions of the equation
f ( x) = 0 .
•
When drawing a graph by hand, we begin by drawing up a table of values and then
plotting the corresponding points (x, y) on graph paper.
Example 8.3
Sketch the graph of the function (see Example 8.2) with formula
y = (10 + 2 x) x
From the table constructed in the answer to Example 8.2 we have
x
y
1
12
2
28
3
48
4
72
5
100
We also notice that for negative values of x we have
x
y
0
0
–1
–8
–2
–12
–3
–12
–4
–8
–5
0
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Plotting these values on graph paper we obtain the graph in Figure 8.2
Figure 8.2 Graph of y = (10 + 2 x) x
(Comment: In the context of Example 8.2 we are only interested in positive values
of x, since it denotes a length)
•
Sketching a graph by hand will not be as accurate as using a computer package but will
usually give adequate information about a function. In Example 8.3 we see from the
graph of Figure 8.2 that the function increases rapidly once the values of x are away
from the origin (0, 0)
Practice Questions 8.2
1. Draw up a table of values for the function with formula
y = 1100 + 100 x
for x values 0 to 10 in steps of 1. Hence sketch the graph of the function
(See Practice Question 8.1 (2)).
2. Draw up a table of values for the function given by
y = 25000 x − 100 x 2 − 1020000
for x values 25 to 200 in steps of 25
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In the context of Practice Question 8.1 (3), how many items should the cake
manufacturer produce in order to maximise his profit.
8.3 Linear functions
•
A linear function has the basic formula
y = ax + b
where a and b are constants. We saw in Unit 7 that the graph of this function is a
straight line. The constant a is the slope (or gradient) of the line and the constant b is
the intercept on the y-axis. The function has a zero at x = −(b / a ).
•
To determine the values of a and b, we need two values of the function; that is, we
need to know the value of y at x = x1 and x = x2 (say). If
y = y1 at x = x1 and y = y2 at x = x2 , then, using (7.3), the function is given by
y=
y2 − y1
( x − x1 ) + y1
x2 − x1
Example 8.4
Find the equation of the linear function which takes the values 1260 where
x = 1250 and 1345 where x = 1500.
Here y1 = 1260 and x1 = 1250, and y2 = 1345 and x2 = 1500. Putting these
values into formula (8.1) gives
1345 − 1260
( x − 1250) + 1260
1500 − 1250
85
=
( x − 1250) + 1260
250
=0.34 x + 835
y=
Thus the equation of the linear function is
y = 0.34x + 835
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Practice Questions 8.3
1. Find the formulae of the linear functions such that:
(a) y = 0 where x = 32, y = 100 where x = 220
(b) y = 8 where x = 10, y = 12 where x = 25
2. The total cost, &pound;C, of manufacturing a batch of articles is suspected to be related to the
number of articles in the batch, x, by a linear model of the form
C = ax + b
where a and b are constants. The following returns were noted
x
C
20
214
40
248
60
282
80
316
(a) By plotting the graph of C against x show that these figures fit the above model and
find the values of a and b. What is the significance of a and b in this problem?
(b) The selling price is fixed at &pound;5.30 per article. Using the same axes and scales, plot
another graph relating the total selling price, &pound;S, to the total number of articles sold.
(i) the number of articles to be made which, if all sold,
would just cover the cost of manufacture.
(ii) the number of articles to be made which, if all sold, would
yield a profit of &pound;72.
3. If the retail price of a particular item produced by a firm is &pound;P then the number of items
the firm is prepared to supply at this price is given by the equation
S=
1
P −9
2
whilst the number D of items that customers will buy at this price is given by the
equation
1
D = 36 − P
3
(a) Using the same axes draw graphs of D and S against P. Indicate the point
corresponding to the situation when the supply S equals the demand D (that is,
the market is in equilibrium) and determine the number of articles produced in
this case. Find also the selling price corresponding to the equilibrium state of the
market.
(b) When the government levies a tax of &pound;10 per item sold, the customers are
concerned only with the new total retail price &pound;P, whereas the suppliers react as
though the price was &pound;(P – 10).
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(i)
(ii)
What is the adjusted equation relating S to P?
On the same axes as before draw a graph of this adjusted function
relating P to S.
(iii) How many items are produced when the market is now in a state of
equilibrium and what is the corresponding selling price?
(iv) Who really pays the tax? What percentage is passed on to
the customer?
•
A quadratic function has the basic formula
y = f ( x) = ax 2 + bx + c
(8.2)
where a, b, and c are constants. The graph of this function is a parabola but several
different cases occur depending on the values of the constants a, b and c.
• In Unit 3, Section 3.3, we saw that the equation
ax 2 + bx + c = 0
(8.3)
has solutions
x=
−b &plusmn; b 2 − 4ac
2a
Thus:
(a) If b 2 &gt; 4ac then equation (8.3) will have two distinct solutions. So the
quadratic function (8.2) will have two distinct zeros,
and its graph will cut the x-axis twice.
(b) If b 2 = 4ac then equation (8.3) will have two equal solutions.
So the quadratic function (8.2) will have a repeated zero and its graph
will ‘touch’ the x-axis (that is, the x-axis is a tangent to the graph).
(c) If b 2 &lt; 4ac then equation (8.3) will have no real solutions. So the
quadratic function (8.2) has no real zeros and its graph will not cut the xaxis.
•
For a &gt; 0 the graph of the quadratic function (8.2) will be ‘cup’ shaped. Graphs,
corresponding to cases (a)-(c), are illustrated in Figure (8.3).
b2 &gt; 4ac
b2 = 4ac
(two roots)
(repeated root)
b2 &lt; 4ac
(no roots)
Figure 8.3: Graphs y = ax 2 + bx + c, a &gt; 0
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NOTE:
In each case the function has a minimum value at
x = x* = −
b
2a
From (8.2) the minimum value of the function is
b2
y = f (x ) = c −
4a
*
•
*
For a&lt; 0 the graph of the quadratic function (8.2) will be ‘cap’ shaped.
Graphs, corresponding to cases (a)-(c), are illustrated in Figure (8.4).
b2 &gt; 4ac
(two roots)
b2 = 4ac
(repeated root)
b2 &lt; 4ac
(no roots)
Figure 8.4 Graphs y = ax 2 + bx + c, a &lt; 0
NOTE
In each case the function has a maximum value at
b
2a
From (8.2) the maximum value of the function is
x = x* = −
y * = f ( x* ) = c −
b2
4a
Example 8.5
Determine the nature of the quadratic function with formula
(a) y = 15 x 2 − 3x + 25
(b) y = 12 + 7 x − 2 x 2
(a) In this case a =15, b = –3 and c =25
Since a &gt; 0, we have a ‘cup’ type curve.
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Since b2 = (–3)2 = 9 and 4ac = 4&times;15&times;25 &gt; 9 we have b2 &lt; 4ac and the curve does not
cut the x-axis.
(b) In this case a = –2, b = 7 and c = 12.
Since a &lt; 0, we have a ‘cap’ type curve.
Since b2 = (7)2 = 49 and 4ac = 4&times;(–2)&times;(12) &lt; 49 we have b2 &gt; 4ac
and the curve cuts the x–axis twice.
Example 8.6
y = f ( x) = 12 x − 11 − 4 x 2
Decide whether the graph of f(x) has a minimum or a maximum value, and evaluate
it.
In this case a = -4, b = 12 and c = –11.
Since a &lt; 0 we have a ‘cap’ type curve so the function has a maximum
value.
Maximum value attained when
x = x* = −
b
(12) 3
=−
=
2a
(−8) 2
Thus maximum value of function is
( )
( )
( )
y* = f 3 = 12 3 − 11 − 4 3
2
2
2
2
= 18 − 11 − 9 = −2
Practice Questions 8.4
1. Determine the nature of the quadratic functions with formula:
(a) y = x 2 − 3x − 28
(b) y = x 2 − 7 x
(c) y = 2 x 2 − 5 x + 13
(d) y = 5 x − 7 − x 2
(e) y = 1 + x(2 − 3x)
(f) y = 4 x 2 − 4 x + 1
y = f ( x) = 2 x 2 − 11x + 12
Decide whether the graph of f(x) has a minimum or a maximum value,
and evaluate it.
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3. A gun is situated at a point O on horizontal ground at sea level. It fires a shell which
reaches a maximum height of 1000 m and lands at A, also at sea level, a horizontal
distance of 10 km from O. The path of the shell is a parabola. Choosing suitable
units, sketch the path (vertical height against horizontal distance), and show that an
appropriate function f1 ( x) to describe it is
f1 ( x) = 40 x(10 − x)
where O is the origin of co-ordinates and x is the horizontal distance covered.
The gun is transported to a point P at the top of a mountain 960 m above sea level,
and fired in an identical manner. The shell lands at a point B at sea level. Sketch this
path, derive a second function f 2 ( x) to describe it, and show that
f 2 ( x) = 40(2 + x)(12 − x)
Deduce the horizontal distance of B from P
4. An agent for a block of 50 flats knows that he can just fill all the flats if he charges a
rent of &pound;240 per month. From experience he knows that for each increase of &pound;6 per
month, one flat becomes vacant.
(a) Show that the monthly income I is given by
I = (240 + 6 x)(50 − x)
where x is the number of vacant flats.
(b) Sketch a graph of I against x.
(c) How many flats should be left vacant for maximum income, and what rent
should be charged?
(d) The agent discovers that the cost of maintaining an occupied flat is &pound;140 per
month, whereas an unoccupied flat costs only &pound;80 per month. Show that the
monthly profit P is given by
P = 5000 + 120 x − 6 x 2
Find the number of flats occupied for maximum profit.
8.5 Cubic functions
•
A cubic function has the basic formula
y = f ( x) = ax 3 + bx 2 + cx + d
where a ,b ,c and d are constants.
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•
The graphs of these functions always cut the x-axis at least once, indicating that a
cubic function has at least one zero. Many different cases are possible depending on
the values of the constants a, b, c and d. For example, for a &gt; 0, we may have any of
the options in Figure 8.5
Figure 8.5: Possible forms of graph of cubic
y = ax 3 + bx 2 + cx + d , a &gt; 0
Example 8.7
Sketch the graph of the cubic function with formula
y = x3 + 3x 2 − 3x + 4
This is easiest performed using a graphics calculator or a computer program. Alternatively a
short table of values will help:
x -4
–3
–2
–1
0
1
2
3
4
y 0
13
14
9
4
5
18
49
104
Plotting these points enables us to sketch the graph as in Figure 8.6
Figure 8.6: Sketch of y = x 3 + 3x 2 − 3 x + 4
Practice Questions 8.5
1. By drawing up a short table of values sketch the graphs of the following cubic functions.
(a) y = x 3 − 2 x 2 − 7 x − 4
(b) y = 4 x 3 − 8 x 2 − 3 x + 5
2. The cubic function f ( x) is defined by
f ( x) = 54 x 3 − 270 x 2 + 450 x − 266
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(a) Show that f ( x) = 2(3x − 5)3 − 16
(b) Sketch the graph of f ( x) for x = 0, 1, 2 and 3 only, hence estimate the value
of a zero of f ( x) .
(c) Observing that
f ( x) = 0 implies that (3 x − 5)3 = 8
find the exact value of this root.
3. By tabulating values for x from –3 to 3 in steps of 1 draw a graph of the cubic function
y = f ( x) = 2 x3 + x 2 − 5 x + 2
(a) Locate (approximately) the roots of the function from your graph.
f (−2), f (1) and f ( 12 )
4. An ashtray is made by taking the square of metal ABCD of side 12 cm (see Figure 8.7)
and removing the four squares of side x cm from each of the corners. Folding along the
lines EF, FG, GH, HE then forms an ashtray of depth x cm.
12cm
A
12cm
B
E
H
F
G
x
D
x
C
Figure 8.7: Design of ashtray
Prove that the volume V of the ashtray is given by
V = 4 x(6 − x) 2
Draw the graph of V against x for values of x between 0 and 7. From your graph
(i) state the maximum possible volume for the ashtray and the fraction of
metal removed in making the ashtray in this case.
(ii) State the possible values for the length of the ashtray when the volume is
90 cm3.
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8.6 Rational functions
•
Rational functions have the form
y = f ( x) =
polynomial p ( x)
polynomial q ( x)
For example
f ( x) =
x3 − 2 x 2 − 3x + 5
x2 + 5x + 6
is a rational function with the polynomial p ( x) being a cubic and the polynomial
•
Sketching the graphs of rational functions gives rise to the concept of an asymptote;
which is a line or a curve that a graph approaches but does not reach.
•
To illustrate consider the graph of the rational function
y = f ( x) =
x
,
1− x
0 ≤ x &lt;1
We see that as x gets closer and closer to the value 1 the values of y get larger and
larger. This is illustrated in the graph of y = f ( x) shown in Figure 8.8.
x
1− x
The line x = 1 is called a vertical asymptote to the curve, and we note that the
graph of f ( x) approaches this asymptote as x approaches 1. In general the graphs
of rational functions will have vertical asymptotes where the denominator is zero.
They may have other asymptotes.
Figure 8.8: Graph of y =
•
Consider the rational function
y = f ( x) =
x
,
x +1
x&gt;0
Expressing
x
x +1
as
( x +1) – 1
x +1
= 1–
1
x +1
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1
x
gets smaller and smaller, so that
x +1
x +1
approaches closer and closer to the value 1. This is illustrated in the graph of y = f ( x)
shown in Figure 8.9. The line y = 1 is called a horizontal asymptote to the curve, and
we note that the graph of f ( x) approaches this asymptote as x becomes large.
we see that as x gets larger and larger
Figure 8.9: Graph of y =
•
x
x +1
We can often simplify problems by expressing a rational function in terms of its
constituent fractions. This is illustrated in the following example.
Example 8.8
Sketch the graph of the rational function
y = f ( x) =
x2 − x − 6
x2 − 1
We can re-write y as
( x 2 − 1) − x − 5
x2 − 1
x+5
= 1− 2
x −1
x+5
= 1−
( x − 1)( x + 1)
2
3
= 1+
−
x +1 x −1
y=
From our two illustrative examples earlier, this shows that f ( x) has vertical asymptotes
at x = –1 and x = 1. Also where x becomes large and positive y is just less than 1, and
where x is large and negative y is just less than 1.So the line y = 1 is a horizontal
asymptote. A sketch of the graph of f ( x) is illustrated in Figure 8.10
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Figure 8.10: Graph of y =
•
x2 − x − 6
x2 − 1
An important skill is obtaining the partial fractions of rational functions, which involves
expressing it in terms of its constituent fractions
Example 8.9
Express
2x
in its partial fraction
( x + 1)( x + 2)
Write
2x
A
B
as
+
, where A and B are numbers to be found.
( x + 1)( x + 2)
x +1 x + 2
Then
2x
A
B
=
+
( x + 1)( x + 2) x + 1 x + 2
which implies
2 x = A( x + 2) + B( x + 1)
Choosing x = –1 gives
2(–1) = A(–1+2) so that A = –2
Choosing x = –2 gives
2(–2) = B (–2+1) so that B = 4
(Note: We choose values of x which makes the calculations easy! The values,
here, were chosen to make (x + 1) zero and then (x + 2) zero.)
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Thus in partial fractions:
2x
4
2
=
−
( x + 1)( x + 2) x + 2 x + 1
Check:
4
2
4( x + 1) − 2( x + 2)
2x
−
=
=
x + 2 x +1
( x + 1)( x + 2)
( x + 1)( x + 2)
Practice Questions 8.6
1. Express in partial fractions
(a)
1
( x − 2)( x + 3)
(b)
1
x( x + 2)
x
x −1
(e) 2
x − 4x − 5
x +x−2
2
x + x +1
in the form
2. (a) Express 2
x − 2x − 3
(d)
2
(c)
3x + 1
( x + 1)( x − 2)
(f)
1
( x + 1)( x + 2)( x + 3)
Ax + B
( x + 1)( x − 3)
and then in the form
1+
1+
C
D
+
x +1 x − 3
(b) Use the result obtained in (a) to sketch the graph of the rational function
x2 + x + 1
y = f ( x) = 2
x − 2x − 3
3. (a) Sketch the graph of the rational function f ( x) defined by
50 x
, x ≠ 110
x − 110
f ( x) =
(b) Using (a) as a guide, sketch the graph of
g ( x) =
px
, x≠q
x−q
where p and q are positive constants.
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(c) Sketch the graph of
h( x ) =
− px
, x≠q
x−q
(d) The cost of eliminating a large part of pollutants from a lake is relatively small.
However, to remove the remainder results in a significant increase in cost. If x
represents the percentage of pollutants removed, and C(x) the cost (in thousands of
pounds) for its removal, which of g ( x) and h( x) would you choose to model C? If
it costs &pound;5000 to remove 10%, and &pound;60,000 to remove 60% of the pollutants, use
your model to predict the cost for complete purification
8.7 Inverse functions
•
The inverse function of a function f ( x) is a function that reverses the operations carried
out by f ( x) . It is denoted by f −1 ( x).
•
We therefore have
x = f −1 ( y ), where y = f ( x)
(8.4)
that is, the independent variable x for f ( x) acts as the dependent variable for f −1 ( y ) ,
and correspondingly the dependent variable y for f ( x) becomes the independent
variable for f −1 ( y ) .
•
Since it is usual to denote the independent variable of a function by x and the dependent
variable by y, we interchange the variables x and y in (8.4) and define the inverse
function by
if y = f −1 ( x) then x = f ( y )
(8.5)
Example 8.10
Obtain the inverse function of
y = f ( x) = 2 x + 5
Here the formula for the inverse function can be found algebraically. First
rearranging
y = f ( x) = 2 x + 5
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to express x in terms of y gives
1
x = f −1 ( y ) = ( y − 5)
2
Interchanging the variables x and y then gives
1
y = f −1 ( x) = ( x − 5)
2
as the inverse function of
y = f ( x) = 2 x + 5
Check:
As a check, we have
f (2) = 2(2) + 5 = 9
while
f −1 (9) = 1 (9 − 5) = 2
2
•
The graph of the inverse function y = f −1 ( x) is the mirror image of the graph of
y = f ( x) in the line y = x. These graphs for the function in Example 8.10 are illustrated in
Figure 8.11. Note that the scales on both axes are the same.
Figure 8.11: Graph of function and inverse
Practice Questions 8.7
1. Find the inverse function of
(a) f ( x) = 2 x − 3
(b) f ( x) =
2x − 3
, x ≠ −4
x+4
Notice f ( f −1 ( x)) = x . For example, in (a)
1

f ( f −1 ( x)) = 2  ( x + 3)  − 3 = x
2

(c) Verify for (b)
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2. Obtain the inverse function of the function
y = f ( x) = 1 (4 x − 3)
5
Plot graphs of both the function and its inverse on the same set of axes.
3. Obtain the inverse function of the function
2
y=
,
x≠ 2
3
3x − 2
Plot graphs of both the function and its inverse
8.8 Trigonometric (or Circular) functions
•
In units 5 and 6 we dealt with the trigonometric ratios
sinθ, cosθ and tanθ
and saw how they could be used to solve problems involving triangles. We also saw how
they could be defined for all values, both positive and negative, of the angle θ, measured in
•
Consequently we may regard θ as being the independent variable x of the functions
y = sin x
y = cos x
y = tan x
which we call the trigonometric functions (or in modern usage circular functions, since the
independent variable x measures rotation).
(Note: We can also define functions corresponding to the other three trigonometric ratios
sec, cosec and cot).
•
Whilst x may be measured in degrees or radians it is common practice to measure it in
radians when dealing with functions. Radian measure is always used in problems, which use
calculus in the solution method.
•
To determine the value of a trigonometric function at particular vales of x we can make
reference to Figure 8.12, in which P is a point on a circle of unit radius.
Figure 8.12
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If (c, s) denote the co-ordinates of P, when OP is inclined at an angle x radians to the
horizontal axis, then
sin x = s,
cos x = c,
tan x =
s
, c≠0
c
(Note: In Figure 8.12 x measures the angle and not the x co-ordinate of the point)
We note that sin x and cos x are defined for all values of x. However, tan x is undefined at
values of x where cos x=0. Since this occurs when the point P lies on the vertical axis it
follows that tan x is undefined for
x = nπ +
π
2
where n is an integer.
•
To obtain the graph of sin x, we need to read off the values of s as P moves around the
circle. Note that as we continue around the circle for a second revolution (that is, x goes
from 2π to 4π) the values (and graph) produced is the replica of that produced as x goes
from 0 to 2π, the same being true for subsequent intervals of 2π. By allowing P to rotate
clockwise around the circle, we see that sin(-x) = -sin x, so the graph of sin x can be
extended to negative values of x. The resulting graph of y = sin x is illustrated in Figure 8.13
(Reminder: Anticlockwise rotation denotes positive angles)
Figure 8.13: Graph of y = sin x
Since the graph replicates itself for every interval of 2π
sin( x + 2π k ) = sin x, k = 0, &plusmn;1, &plusmn;2....
(8.6)
and the function sin x is said to be periodic with period 2π.
•
To obtain the graph of cos x, we need to read off the values of c as P moves around the
circle. By allowing P to rotate clockwise around the circle, we see that cos(-x) = cos(x), so
that the graph can be extended to negative values of x. The resulting graph of y = cos x is
shown in Figure 8.14
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Figure 8.14: Graph of y = cos x
Again, the function cos x is periodic with period 2π, so that
cos( x + 2π k ) = cos x, k = 0, &plusmn;1, &plusmn;2,....
•
(8.7)
Note that the graph of y = sin x is that of y = cos x moved 1 π units to the right, while that
2
1
of y = cos x is the graph of y = sin x moved π units to the left. Thus
2
sin x = cos( x − 12 π )
or
(8.8)
cos x = sin( x + π )
1
2
•
Using a similar approach the graph of y = tan x may be obtained and is illustrated in Figure
8.15
Figure 8.15: Graph of y = tan x
In this case the graph replicates itself every interval of duration π so that
tan( x + π k ) = tan x, k = 0, &plusmn;1, &plusmn;2,.....
and tan x is of period π.
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•
If k is a positive constant then the graph of the function y = f ( x + k ) is that of y = f ( x)
moved horizontally k units to the left. This is readily seen by putting x + k = 0 .
•
Similarly, by putting x − k = 0 , it is seen that the graph of y = f ( x − k ) is that of y = f ( x)
move horizontally k units to the right.
•
Using these observations graphs of functions such as
y = sin( x + α ) and y = cos( x − α )
Practice Questions 8.8
1. Sketch using the same set of axes
(a) y = 2sin x
(b) y = sin x
1
(c) y = sin x
2
for −2π ≤ x ≤ 2π . What can you observe?
2. Sketch the graph of the function
1 

y = cos  x − π 
4 

for −2π ≤ x ≤ 2π .
3. Sketch the graph of the function
3 
y = cos  x 
2 
for −2π ≤ x ≤ 2π . What is the period of the function?
4. Sketch the graph of the function
1 

y = sin  2 x + π 
3 

for 0 ≤ x ≤ 2π . What is the period of the function?
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5. Using the identity
sin( A + B ) = sin A cos B + cos A sin B
express the function
π

y = 5sin  3 x + 
3

in the form
C sin 3 x + D cos 3 x
where C and D are constants.
8.9 Inverse trigonometric functions
•
Considering the inverse of the trigonometric functions, it follows from the definition
given in (8.5) that the inverse sine function sin-1x (also sometimes denoted by arcsin x)
is such that
if y = sin −1 x then x = sin y
Here x should not be interpreted as an angle – rather sin-1x represents the angle
whose sine is x.
1
(Note: You must NOT confuse sin-1x with (sin x)-1 =
)
sin x
•
We came across the inverse problem in Units 5 and 6, when using the trigonometric
ratios to determine angles, rather than lengths of sides of triangles. We saw that
1
more than one solution was possible. For example, if sin θ = , then there is an
2
infinite number of solutions for θ :
 1  π 5π 13π
θ = sin −1   = , ,
,......
 2  6 6, 6
When interpreting sin-1x as a function, we overcome this problem by restricting the
values of x to −2π ≤ x ≤ π2 . Then
sin −1 x = the angle between −
π
2
and
π
2
whose sine is x
1 π
so, for example, sin −1   = . Thus we define the inverse function by
2 6
if y = sin −1 x then x = sin y, where – 12 π ≤ y ≤ 12 π and − 1 ≤ x ≤ 1
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Glyn James and John Searl, Modern Engineering Mathematics, 4th Edition, Refresher Units in Mathematics
The corresponding graph of the function is illustrated in Figure 8.16
Figure 8.16: Graph of y = sin −1 x
•
Similarly, in order to define the inverse cosine function cos-1x (also sometimes
denoted by arcos x), we have to restrict the values of x and define the inverse
function as
if y = cos −1 x then x = cos y, where 0 ≤ y ≤ π and − 1 ≤ x ≤ 1
(8.11)
π 1
1 π
For example cos −1   =
(since cos = )
3 2
2 3
The corresponding graph of the function is illustrated in Figure 8.17.
Figure 8.17: Graph of y = cos −1 x
•
Likewise we define the inverse tangent function tan-1x (also sometimes denoted by
arctan x) as
1
1
if y = tan −1 x then x = tan y, where − π &lt; y &lt; π and x any real number (8.13)
2
2
For example, tan −1 1 =
π
4
(since tan
π
4
= 1)
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The corresponding graph of the function is illustrated in Figure 8.18
Figure 8.18: Graph of y = tan −1 x
Practice Questions 8.8
1. Evaluate the following:
 1
(a) sin −1  − 
 2
(d) sin −1
(b) cos −1 (−1)
3
2
(e) cos −1
3
2
(c) tan −1 (−1)
(f) tan −1 3
2. If tan −1 x = α and tan −1 y = β , show that
tan(α + β ) =
x+ y
1 − xy
8.10 Exponential functions
•
Functions of the form
y = f ( x) = a x
where a is a positive constant are called exponential functions. The standard
exponential function that is used is
y = e x or y = exp x
where e is a special number approximately equal to 2.7183. A graph of the function
is shown in Figure 8.19.
Figure 8.19: Graphs of y = e x and y = e − x
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•
We note that the following properties are satisfied by the exponential function.
e x1 e x2 = e x1 + x2
e x1
= e x1 − x2
x2
e
e kx = (e k ) x = a x , where a = e k
8.11 Logarithm function
•
Given any two positive numbers a (≠1) and x, there is a unique number y such that
ay = x
We call y the logarithm to base a of x and write the function as
y = log a x
so we have that
a y = x gives y=log a x
•
(8.13)
Three values of the base a frequently occur:
a = 10 giving y = log10x and called common logarithms
a = 2 giving y = log2x and called binary logarithms
a = e giving y = logex and called natural logarithms
Usually logex is written ln x and log10x is written log x
•
The following properties of the logarithm function follow directly from the
definition (8.13).
log a 1 = 0
log a a = 1
log a ( x1 x2 ) = log a x1 + log a x2
x 
log a  1  = log a x1 − log a x2
 x2 
log a x n = n log a x
x = a log a x
y x = a x loga y
log a x
log b x =
log a b
•
this is the change of base formula
The most commonly used logarithm function is the natural logarithm
y = log e x = ln x
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From its definition this gives
x = ey
which implies that
y = ln x is the inverse function of y = e x
Thus the graph of y = ln x is the mirror image of the graph of y = e x in the line y = x
and this is illustrated in Figure 8.20.
Figure 8.20 Graph of y = ln x
Example 8.11
(a) Find the logarithm log 2 32
 10 x 
(b) Express log10  2  in terms of log10 x and log10 y
 y 
(c) Express as a single logarithm
ln 20 – ln 18 + 2ln 6
(a) ln 2 32 = ln 2 25 = 5ln 2 2 = 5
(b)
 10 x 
log10  2  =log10 10 x − log10 y 2
 y 
= 1 log10 (10 x) − 2 log10 y
2
= 1 log10 10 + 12 log10 x − 2 log10 y
2
1
= + 1 log10 x − 2 log10 y
2 2
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(c) ln 20 – ln 18 + 2ln 6 = ln 20 – ln 18 + ln62
= ln
20 &times; 36
18
= ln 40
Example 8.12
Use the substitution u = e x to reduce the equation
e 2 x − 8e x + 12 = 0
to a quadratic equation in u. Factorise this quadratic and hence find the value(s) of u which
satisfy it. What are the value(s) of x, correct to two decimal places, which satisfy the
original equation?
e 2 x − 8e x + 12 = (e x ) 2 − 8e x + 12 = 0
Substituting
u = e x reduces the equation to
u 2 − 8u + 12 = 0
which, on factorising, gives
(u − 6)(u − 2) = 0
so that
u = 6 or u = 2
Since u = e x implies x = ln u the corresponding values of x are
x = ln6 = 1.79 or x = ln2 = 0.69
correct to two dp
Practice Questions 8.9
1. Find the following logarithms without using a calculator:
(a) log 2 8
(b) log 2
(d) log 3 81
(e) log 9 3
1
4
(c) log 2
1
2
(f) log 4 0.5
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2. Express in terms of ln x and ln y:
(a) ln( x 2 y )
 x5 
(c) ln  2 
y 
(b) ln xy
3. Express as a single logarithm:
(a)
1
(b) 4 ln 2 − ln 25
2
ln14 − ln 21 + ln 6
2
8
(d) 2 ln   − ln  
3
9
(c) 1.5ln 9 − 2 ln 6
4. Use the change of base formula to simplify:
(a)
log a 32
log a 2
(b)
log 3 x
log 9 x
5. Simplify the following:
(a)
e7 x
e2 x
 1 1 − x  
(c) exp  ln 

 2 1 + x  
(b) e 2 ln x
(d) (e − x ) 2
6. Use the substitution u = e x to reduce the equation
e 2 x − 3e x + 2 = 0
to a quadratic equation in u. Factorise this quadratic and hence find the value(s) of u
which satisfy it. What are the value(s) of x, correct to two decimal places, which satisfy
the original equation?
7. A liquid which has been heated was allowed to cool down in a room where the ambient
temperature of the air remained at 10.40C. The law of cooling yields a relationship of the
form
T = b + aeλt
where T 0C is the temperature of the liquid at time t (minutes) and a, b and λ are
parameters.
Why must b = 10.4? What can you deduce about the values (relative to zero) of a and λ?
Values of T are given for particular values of t in the following table:
t
T
18.5
33
19.5
32
25
28
39
22
67
16
88
13
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Plot a graph of T against t. Let
y = ln(T − b)
and draw up a new table of y against the values of t given above. Plot a second graph, of
y against t, and, by eye, draw the ‘best’ fitting straight line through your points. Estimate
the values of a and λ, and so deduce the initial temperature of the liquid.
(Hint: ln(t − b) = ln(aeλt ) = ln a + λ t )
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Answers to Practice Questions Unit 8
Practice Questions 8.1
1. (a) 5
(b) –1
(c) –1
2. 1300, 1400, 1500, 1100+100x
3. y = 25000 x − 100 x 2 − 1020000
Practice Questions 8.2
1.
0
1
2
3
4
5
6
7
8
9
10
1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100
x
y
y
2000
1000
2
4
6
8
10
100
48
125
54.25
x
2.
25
50
x
y/10000 –45.75 –2
75
29.25
Figure 8.A1
To maximise profit produce 125 items
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150
48
175
29.25
200
–2
Glyn James and John Searl, Modern Engineering Mathematics, 4th Edition, Refresher Units in Mathematics
Practice Questions 8.3
1. (a) y = 5 ( x − 32)
9
(b) y = 4 x + 16
15
3
2. (a)
S(x)
C,S
C(x)
300
180
60
40
20
60
80
x
Since points lie on a straight line (blue line), a linear model
C = ax + b
is satisfied.
b = intersection on C axis = 180
a = slope =
34
= 1.7
20
b represents the basic overhead costs before any articles produced and a represents
the additional cost per article to manufacture.
(b) Red line S = 5.30x
(i) number articles = 50
(ii) number articles = 70
3. (a)
Figure 8.A2
Equilibrium state indicated on graph
Articles produced = 18
Selling price = &pound;54
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(b) (i) Plotted in above Figure as St
(ii) Equilibrium when 16 items produced, selling price = &pound;60 per item
(iii) Consumer pays &pound;6, manufacturer &pound;4. Thus 60% passed on to consumer.
Practice Questions 8. 4
1. (a) cup, cuts x-axis
(b) cup, cuts x-axis
(d) cap, does not cut x-axis
(c) cup, does not cut x-axis
(e) cap, cuts x-axis
2. Min, x = 11 , y = − 25
4
8
3.
Figure 8A3
Figure 8A4
Horizontal distance = 12 km
4. (b)
Figure 8A5
(c) 5, &pound;270
(d) 40
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(f) cup, touches x-axis
Glyn James and John Searl, Modern Engineering Mathematics, 4th Edition, Refresher Units in Mathematics
Practice Questions 8.5
1 (a)
–4 –3 –2 –1 0 1
2 3 4 5
–72 –28 –6 0 –4 –12 –18 –16 0 36
x
y
6
98
Figure 8A6
(b)
x
y
–4
–3 –2 –1 0 1 2 3 4
–367 –166 –53 –4 5 –2 –1 32 121
Figure 8A7
2. (b)
Figure 8A8
Estimate between 2 and 2.5
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(c) 7
3
3.
x
2x3
x2
–5x
2
f(x)
–3
–54
9
15
2
–28
–2
–16
4
10
2
0
–1
–2
1
5
2
6
0
0
0
0
2
2
1
2
1
–5
2
0
2
16
4
–10
2
12
3
54
9
–15
2
50
Figure 8A 9
Root are –2, 1, 1
2
4.
x
V
0 0.5 1 1.5
2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7
0 60.5 100 121.5 128 122.5 108 82.5 64 40.5 20 5.5 0 6.5 28
Figure 8A10
(i) max 128cm3 (x = 2) , fraction 1
9
(ii) Possible x values 0.85, 3.35
corresponding lengths 10.3 cm , 5.3 cm
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Practice Questions 8.6
1
1
−
1. (a) 5 + 5
x−2 x+3
1
2
(d) 3 + 3
x +1 x − 5
2. 1 +
3x + 4
,
( x + 1)( x − 3)
1
1
−
(b) 2 + 2
x x+2
2
7
(c) 3 + 3
x +1 x − 2
2
1
(e) 3 + 3
x + 2 x −1
1
1
1
(f) 2 −
+ 2
x +1 x + 2 x + 3
−
1
13
1+ 4 + 4
x +1 x − 3
−
Figure 8A.11
3. (a)
Figure 8A.12
(b)
Figure 8A.13
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(c)
Figure 8A 14
(d) Choose h(x) since x ∈ [0,100] and C ( x) ≥ 0
&pound;500,000
Practice Questions 8.7
1 ( x + 3)
2
1.
(a)
2.
1
(5 x + 3)
4
(b)
4x + 3
,x ≠ 2
2− x
y
y= f-1(x)
y= f(x)
1
1
3.
x
2 x +1
3 x
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Figure 8A 15
Practice Questions 8.8
1.
Figure 8A16
2.
Figure 8A17
3
Figure 8A18
Period = 4 π
3
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4.
Figure 8A19
Period = π
5.
C = 5,D = 5 3
2
2
Practice Questions 8.9
1. (a) −
π
(b) π
(c) −
(b) –2
(c) −
6
π
4
(d)
π
(e)
3
π
6
(f)
π
3
Practice Questions 8.10
1. (a) 3
1
2
(d) 4
(e)
1
2
(f) −
1
2
2. (a) 2 ln x + ln y
1
1
ln x + ln y
2
2
(c) 5ln x − 2 ln y
(b)
3. (a) ln 4
(b) ln
4. (a) 4
5. (a) e5 x
16
5
(c) ln
3
4
(d) – ln 2
(b) 2
(b) x 2
(c)
1− x
1+ x
(d) e −2 x
6. 0, ln2 = 0.69
7. The liquid eventually cools to ambient temperature so T tends to 10.4 as t gets
bigger and bigger.
This suggests that λ &lt; 0 , and since the liquid is cooling and is initially at a
temperature above 10.4, a &gt; 0
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Figure 8A.20
18.5
t
23
T
22.6
T-b
ln(T-b) 3.118
19.5
25
39
67
88
32
28
22
16
13
21.6 17.6 11.6 5.6
2.6
3.073 2.863 2.451 1.723 0.956
Figure 8A.21
ln a 3.7 so a 40
λ is the slope so λ −0.03
initial temperature = b +a = 50.40C
End of Refresher Unit 8
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1.6
0.470
```