Mathematical Methods for Engineers 1 (MATH 1063)
Calculus 1 (MATH 1054)
Week 1 Lecture Contents:
Functions, Models and Graphs
1. Functions and Mathematical Modelling

Edwards and Penney, §1.1

2. Graphs

Edwards and Penney, §1.2

3. Vertical Line Test

Edwards and Penney, §1.2

Functions and Mathematical Modelling
Functions are relationships between one variable and other(s). Some simple familiar functions are
• The volume of a sphere in terms of its radius r is
V = 4 πr 3 .
3
• The volume of a cylinder in terms of its radius r and height h is
V = πr 2 h.
• The distance travelled by a falling body dropping from rest after t seconds is s = 1 gt2 .
2
Here s will be in metres if the gravitational acceleration is g ≈ 9.8 m/s2 .
Edwards and Penney give the formal definition of a function as
Definition. A real-valued function f defined on a set D of real numbers is a rule that assigns to each number x in D exactly one real number, denoted by f (x).
Definition. The set D of all numbers for which f (x) is defined is called the domain of the function f and the set of all values y = f (x) is called the range of f .
We can represent this pictorially as
Domain

Range

f (x)

x

f (u)

u

f (v)

v

1

We call x the independent variable and y the dependent variable because the value of y depends, through f , on the choice of x.
Example 1. Consider y = f (x) = 2x2 − 6x + 5.
(a) What values can x and y take?
(b) Where does the straight line y = 3x − 2 intersect this parabola?
(c) What other straight line, parallel to y = 3x − 2, touches the parabola? y 15

y = 2x2 − 6x + 5
10

y = 3x − 2

5

−1

1

2

3

4

x

−5

Solution.

(a) The discriminant of the quadratic y = ax2 + bx + c is Δ = b2 − 4ac.

Firstly, note that the discriminant here is Δ = (−6)2 − 4 × 2 × 5 = −4 < 0, and hence there are no real zeros of f .
Complete the square: f (x) = 2(x2 − 3x) + 5
= 2[(x − 1.5)2 − 2.25] + 5
= 2(x − 1.5)2 + 0.5
Hence the minimum value of f (x) is 0.5 when x = 1.5.
The function is defined for all real values of x. We can say that the domain is
(−∞, ∞) or R.
The values that y can take are y ≥ 0.5. We can say that the range is [0.5, ∞).
(b) 2x2 − 6x + 5 = 3x − 2
2x2 − 9x + 7 = 0
(x − 1)(2x − 7) = 0 ⇒ x = 1, 3.5
They intersect at (1, 1) and (3.5, 8.5).
(c) Let the straight line be y = 3x + b. This intersects y = 2x2 − 6x + 5 whenever
2x2 − 6x + 5 = 3x + b
2x2 − 9x + (5 − b) = 0.
2

To “touch” we must have a repeated root which is the equivalent to the discriminant being zero
81 − 8(5 − b) = 0 ⇒ b = − 41
8
.·. y = 3x − 41 touches the parabola y = 2x2 − 6x + 5.
8
Exercise. Find the coordinates of the point where they touch.
(Answer: x = 2 1 , y = 1 5 )
4
8
Example 2. The Animal Pen Problem
The problem is to build a rectangular animal pen using an existing wall as one side, as shown in the diagram below. The fencing material costs $5 per metre and the wall needs painting at a cost of $1 per metre. If there is $180 available, what is the maximum area that can be enclosed. x $5/m

y $5/m

$5/m y
$1/m

x

Wall

Solution. The area of the pen is a function of the variables, the length x and the width y:
A = f (x, y) = xy.
The cost of constructing the pen is
C = 180 = x + 5y + 5x + 5y = 6x + 10y.
Hence we can find y as a function of x, namely
1
y = g(x) = 10 (180 − 6x) = 3 (30 − x).
5

Using this, we can eliminate y from the area equation to obtain
A(x) = 3 (30x − x2 ).
5
If x = 0 we have a degenerate rectangle of base zero, height 18 and zero area. If x = 30 we have a degenerate rectangle of base 30, height 0 and zero area. Thus the complete definition of the area function is
A(x) = 3 (30x − x2 ),
5

0 ≤ x ≤ 30.

We can tabulate some area values x A(x)

0 5 10 15 20 25 30
0 75 120 135 120 75 0

or plot the graph of the quadratic as shown below.
3

y
150

100

50

0

10

20

30 x

It would appear from the table, or the symmetry of the quadratic about its maximum, that the maximum area is 135 m2 when x = 15. This can be shown by completing the square on the quadratic.
A(x) = 3 (30x − x2 )
5
= − 3 [(x − 15)2 − 225]
5
= 135 − 3 (x − 15)2 .
5

Hence the maximum area is 135 m2 when x = 15 m and y = 9 m.
√
Example 3. Find the domain and range of the function y = x + 1.
We are dealing with real functions, so square roots of negative numbers are not permitted.
Hence x ≥ −1, and the domain is [−1, ∞) and the range is [0, ∞).
1
. x+1 Additionally to the previous example, division by zero is not permitted. Hence x > −1, and the domain is (−1, ∞) and the range is (0, ∞).

Example 4. Find the domain and range of the function y = √

Intervals
In denoting domains and ranges, sometimes round brackets have been used and sometimes square brackets. An open interval, such as (1, 3), is one where the endpoints ( 1 and 3) are not included; a closed interval, such as [−1, 2], is one where the endpoints ( −1 and
2) are included. Half-open intervals such as [−3, 5) or (−5, −3] are possible. Unbounded intervals [a, ∞) or (−∞, b) are always open at the ∞ or −∞ end.
Straight Line
Refer to the diagram below. y2 − y1
Δy
=
Slope: m =
Δx
x2 − x1
Equation: y − y1 = m(x − x1 ) or y = mx + b where b is the y-intercept.
4

If the line is horizontal, then m = 0, and if the line is vertical, then m is undefined.
1
If a line of slope m1 is perpendicular to another line of slope m2 , then m2 = − . m1 If θ is the angle of inclination to the positive x-axis, then tan θ = m. y (x2 , y2)
Δy = y2 − y1

y = mx + b θ Δx = x2 − x1

(x1 , y1) b x
The Absolute Value Function for x ≥ 0, for x < 0.

x,
−x,

y = |x| =

|x|
3

2

1

−3

−2

−1

y = |2x − 3| =

1
⎧
⎪
⎪
⎨
⎪
⎪
⎩

2

2x − 3,

for x ≥ 3 ,
2

−(2x − 3),

for x < 3 .
2

5

3

x

|2x − 3|
4

y = −2x + 3

y = 2x − 3

3

2

1

−3

−2

−1

1

2

3

4

x

Note that the function can be constructed from the two individual functions, or by translating the function y = 2|x| 1.5 units to the right, i.e. to the point where |2x − 3| is zero. The Floor and Ceiling Functions
The floor of x, denoted x , is the greatest integer less than or equal to x. The ceiling of x, denoted x , is the least integer greater than or equal to x.
For example
7.4 = 7,
−8.7 = −9,
5 = 5,

7.4 = 8
−8.7 = −8
5 =5

The graph of the floor function is shown below. y [

2

[

−2

−1

[

[

[

) −1

)

−2

6

)

)
1

1

2

)

3

x

x+4
3
=1+
.
x+1 x+1 We can use the ezplot command in Matlab to view the graph of the function. Typing ezplot(’(x+4)/(x+1)’) gives

Example 5. The function y =

(x+4)/(x+1)
6
5
4
3
2
1
0
−1
−2
−3
−4
−6

−4

−2

0 x 2

4

6

However, as Jensen says in the preface to “Using Matlab with Calculus”, “one can plot the graph of a function f with the symbolic toolbox’s ezplot command without any understanding of what the graph of a function is”. Calculating some (x, y) values for the function gives x −6 y 0.4

−5 −4
0.25 0

−3 −2
−0.5 −2

−1
−

0 1
4 2.5

2
3
4
2 1.75 1.6

It is apparent that there is a vertical asymptote at x = −1, and there is also a horizontal asymptote at y = 1. This is better seen if we calculate x as a function of y. x+4 x+1 y(x + 1) = x + 4
4−y
x = y−1 3
−1
= y−1 y =

If we wished to plot for a y range from −5 to 5, then we would need to plot the graph for −6 ≤ x ≤ −1.5 and −0.25 ≤ x ≤ 4. The following Matlab m-file will produce the graph shown below.
7

% M-file to plot y = (x+4)/(x+1) clear all x1 = -6:0.01:-1.5;
% or x1 = linspace(-6,-1.5,451); x2 = -0.25:0.01:4; y1 = (x1+4)./(x1+1); y2 = (x2+4)./(x2+1); plot(x1,y1,’b’,x2,y2,’b’) hold on
% Plot the asymptotes as dashed lines x3 = [-1 -1]; y3 = [-5 5]; x4 = [-6 4]; y4 = [1 1]; plot(x3,y3,’--’) plot(x4,y4,’--’) xlabel(’x’) ylabel(’y’) title(’y = (x+4)/(x+1)’) y = (x+4)/(x+1)
5

4

3

2

y

1

0

−1

−2

−3

−4

−5
−6

−5

−4

−3

−2

−1 x 8

0

1

2

3

4

Graphs
We have seen the graphs of many types of functions so far. Edwards and Penney give the formal definition of the graph of a function as
The graph of the function f is the graph of the equation y = f (x).
This is a specific form of the more general definition of the graph of an equation, which is
The graph of an equation in two variables x and y is the set of all points (x, y) in the plane that satisfy the equation.
For example, the Pythagorean theorem implies the distance formula
(x2 − x1)2 + (y2 − y1)2,

d= as shown in the figure below. y P2 (x2 , y2 ) y2 − y1

d
P1 (x1 , y1 )x2 − x1

x
This distance formula tells us that the graph of the equation x2 + y 2 = r 2 is a circle of radius r and centre at the origin (0, 0). The more general equation
(x − h)2 + (y − k)2 = r 2 is a circle of radius r and centre at the point (a, b), as shown in the figure below. y centre (h,k)

(x, y) r (h, k)

x
The circle is not the graph of a function. This is supported by the Vertical Line Test which is described below.
9

The Vertical Line Test
Each vertical line through a point in the domain of a function meets its graph in exactly one point.
Note that the top half of the circle x2 + y 2 = r 2 has the equation
√
y = r 2 − x2 , −r ≤ x ≤ r, and is a function.
Example 6. The equation x2 + y 2 − 6x − 8y − 75 = 0 can have the square completed in the x and y terms to obtain
(x − 3)2 + (y − 4)2 = 100, which is a circle of radius 10 and centre at (3, 4).
Example 7. More exotic graphs can be formed from equations in x and y. For example, see Figure 1.2.5 on page 13 of Edwards and Penney. Another example is the cardioid whose equation is
(x2 + y 2 − x)2 = x2 + y 2 , and whose graph appears below. y 1

0

1

−1

10

2

x