SMA 2101: CALCULUS I
c
Francis O. Ochieng
francokech@gmail.com
Department of Pure and Applied Mathematics
Jomo Kenyatta University of Agriculture and Technology
Course content
• Functions: definition, domain, range, codomain, composition (or composite), inverse.
• Limits, continuity and differentiability of a function.
• Differentiation by first principle and by rule for xn (integral and fractional n).
• Other techniques of differentiation, i.e., sums, products, quotients, chain rule; their applications
to algebraic, trigonometric, logarithmic, exponential, and inverse trigonometric functions all of
a single variable.
• Implicit and parametric differentiation.
• Applications of differentiation to: rates of change, small changes, stationary points, equations of
tangents and normal lines, kinematics, and economics and financial models (cost, revenue and
profit).
• Introduction to integration and its applications to area and volume.
References
[1] Calculus: Early Transcendentals (8th Edition) by James Stewart
[2] Calculus with Analytic Geometry by Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards;
5th edition
[3] Calculus and Analytical Geometry (9th edition) by George B. Thomas and Ross L. Finney
[4] Advanced Engineering Mathematics (10th ed.) by Erwin Kreyszig
[5] Calculus by Larson Hostellem
Lecture 1
1 Functions
To understand the word function, we consider the following scenario and definitions. For example,
the growth of a sidling is an instance of a functional relation, since the growth may be affected by
variations in temperature, moisture, sunlight, etc. If all these factors remain constant, then the growth
is a function of time. Francis
c Oketch
Definition 1.1 (Variables). A variable is an object, event, time period, or any other type of category
you are trying
Consider to measure.
the formula used for calculating the volume of a sphere of radius r.
4
V = πr3 (1)
3
Then,
1
i) V and r vary with different spheres. Hence, they are called variables.
4
ii) π and are constants, irrespective of the size of the sphere.
3
,is a function of time. Francis
c Oketch
Definition 1.1 (Variables). A variable is an object, event, time period, or any other type of category
you are trying
Consider to measure.
the formula used for calculating the volume of a sphere of radius r.
4
V = πr3 (1)
3
Then,
1
i) V and r vary with different spheres. Hence, they are called variables.
4
ii) π and are constants, irrespective of the size of the sphere.
3
There are two types of variables, i.e., independent and dependent variables.
Definition 1.2 (Independent and dependent variables). Independent variable refers to the input value
while dependent variable refers to the output value.
For example from formula (1), the volume, V , depends on the value of the radius, r, of the sphere.
In this case, r is called the independent variable while V is called the dependent variable since it is
affected by the variation of r. Similarly, for the function y = ax2 + bx + c, a, b and c are constants, x
is the independent variable and y is the dependent variable.
Definition 1.3 (Function). A function is a rule that assigns/associates each element in the
independent set, say X, to a unique element in the dependent set, say Y .
Examples of functions are
i) Linear functions e.g., y = x + 5
ii) Quadratic functions e.g., y = x2 − 2x + 5
iii) Cubic functions e.g., y = x3 − 1
iv) Quartic functions e.g., y = 2x4 + x3 − 1
v) Trigonometric functions e.g., y = sin(2x + 5)
vi) Logarithmic functions (log to base 10) e.g., y = log(3x + 1)
vii) Natural logarithmic functions (log to base e ≈ 2.71828) e.g., y = ln(5x + 1)
viii) Inverse of trigonometric functions e.g., y = tan−1 (2x + 1)
ix) Exponential functions e.g., y = e2x+1
x) Absolute value functions e.g., y = |x|. This function is defined as
{
−x, if x < 0
y = |x| =
x, if x ≥ 0
→ Note: in the above examples the variable y depends on the variable x. Thus, we say that the
dependent variable y is a function of the independent variable x. Using function notation, we write
y = f (x), where f is a function. The function f (x) is read as f of x, meaning that f depends on x.
1.1 Domain, Range and Codomain c
Francis Oketch
1.1 Domain, Range and Codomain
Definition 1.4 (Domain). A domain consists of all the elements in the independent set (i.e., the set
of inputs), X, for which the function is defined.
Definition 1.5 (Range). A range refers to a set of all the images of the elements in the domain.
2
Definition 1.6 (Codomain). A codomain consists of all the elements in the dependent set (i.e., the
set of outputs), Y .
For example, consider the diagram below
,y = f (x), where f is a function. The function f (x) is read as f of x, meaning that f depends on x.
1.1 Domain, Range and Codomain c
Francis Oketch
1.1 Domain, Range and Codomain
Definition 1.4 (Domain). A domain consists of all the elements in the independent set (i.e., the set
of inputs), X, for which the function is defined.
Definition 1.5 (Range). A range refers to a set of all the images of the elements in the domain.
2
Definition 1.6 (Codomain). A codomain consists of all the elements in the dependent set (i.e., the
set of outputs), Y .
For example, consider the diagram below
Example(s):
1. Find the domain and range of the following functions.
(a) f (x) = (x − 4)2 + 5
Solution
Since f (x) is defined (or is a real number) for any real number x, the domain of f is
the interval (−∞, ∞).
√
Let y = (x − 4)2 + 5. Making x the subject, we have x = 4 ± y − 5. This function is
defined if y − 5 ≥ 0 or y ≥ 5. Therefore, the range is the interval [5, ∞).
(b) f (x) = 2x2 − 5x + 1
Solution
Since f (x) is defined (or is a real number) for any real number x, the domain of f is
the interval (−∞, ∞).
Let y = 2x2 − 5x + 1 or 2x√ 2
− 5x + (1 − y) = 0. Making x the subject (use quadratic
5 ± 25 − 8(1 − y )
formula), we have x = . This function is defined if 25− 8(1 − y) ≥ 0
4
17 [ )
or y ≥ − . Therefore, the range is the interval − 17 8
,∞ .
8
4
(c) f (x) = 2
x − 5x + 6
Solution
→ Note: 4/0 = ∞ (infinity), vvvv large value, undefined, indeterminate.
The function f (x) is defined when the denominator is nonzero, i.e., if x2 − 5x + 6 = 0.
Solving yields x = 2 and x = 3. Therefore, the domain of f includes all the real numbers
of x except x = 2 and x = 3, i.e., the set (−∞, ∞)\{2, 3} or (−∞, 2) ∪ (2, 3) ∪ (3, ∞).
4 ( )
Let y = 2 or x2 − 5x + 6 − 4y = 0. Making x the subject (use quadratic
x − 5x + 6
1.2 Evaluationformula), we have
of functions √ ( ) c
Francis Oketch
5 ± 25 − 4 6 − 4y
√ x=
(d) f (x) = x − 1 )
2
(
This function is defined if 25 − 4 6 − 4y ≥ 0 or y ≥ −16. Therefore, the range is the
Solution
interval [−16, ∞).
Since f (x) is defined (or is a real number) if x − 1 ≥ 0 or x ≥ 1, the domain of f is the
interval [1, ∞). 3
√
Let y = x − 1. Making x the subject, we have x = y 2 + 1. This function is defined
for any real number y. Therefore, the range is the interval (−∞, ∞).
(e) f (x) = 2|x − 3| + 4
, x 5x + 6
1.2 Evaluationformula), we have
of functions √ ( ) c
Francis Oketch
4
5± 25 − 4 6 − y
√ x=
(d) f (x) = x − 1 )
2
(
This function is defined if 25 − 4 6 − 4y ≥ 0 or y ≥ −16. Therefore, the range is the
Solution
interval [−16, ∞).
Since f (x) is defined (or is a real number) if x − 1 ≥ 0 or x ≥ 1, the domain of f is the
interval [1, ∞). 3
√
Let y = x − 1. Making x the subject, we have x = y 2 + 1. This function is defined
for any real number y. Therefore, the range is the interval (−∞, ∞).
(e) f (x) = 2|x − 3| + 4
Solution
Since f (x) is defined for all real numbers, the domain of f is the interval (−∞, ∞).
Since for all |x − 3| ≥ 0, the function f (x) = 2|x − 3| + 4 ≥ 4. Therefore, the range is
all the values of y for which y ≥ 4 or the interval [4, ∞).
Exercise:
1. Find the domain and range of the following functions.
(a) f (x) = 6 − x2 . [ans: domain (−∞, ∞), range (−∞, 6]]
6 + 3x
(b) f (x) = . [ans: domain (−∞, 0.5) ∪ (0.5, ∞), range (−∞, 1.5) ∪ (1.5, ∞)]
1 − 2x
x+5
(c) f (x) = . [ans: domain (−∞, 2) ∪ (2, ∞), range (−∞, 1) ∪ (1, ∞)]
x−2
√
(d) f (x) = 4 − 2x + 5. [ans: domain (−∞, 2], range (−∞, ∞)]
√
x2 − 16 [ ]
(e) f (x) = .[ans: domain (−∞, −4) ∪ [4, 6) ∪ (6, ∞), range − √23 , √23 \{−1, 1}]
x2 − 2x − 24
1.2 Evaluation of functions
This involves replacing x in the function by the suggested value and retaining the rule of the function.
Example(s):
f (x + h) − f (x)
1. Given f (x) = 2x + 1. Find: (i) f (0), (ii) f (1), (iii) f (x + 2), and (iv) for h = 0.
h
Solution
i) f (0) = 2(0) + 1 = 0 + 1 = 1
ii) f (1) = 2(1) + 1 = 2 + 1 = 3
iii) f (x + 2) = 2(x + 2) + 1 = 2x + 4 + 1 = 2x + 5
f (x + h) − f (x) [2(x + h) + 1] − [2x + 1] 2x + 2h + 1 − 2x − 1 2h
iv) = = = = 2.
h h h h
f (x + h) − f (x)
2. Given f (x) = 3x2 − 2x + 4. Find: (i) f (0), (ii) f (−1), (iii) f (x + 2), and (iv)
h
for h = 0.
Solution
1.3 Composite functions c
Francis Oketch
i) f (0) = 3(0)2 − 2(0) + 4 = 0 + 0 + 4 = 4
ii) f (−1) = 3(−1)2 − 2(−1) + 4 = 3 + 2 + 4 = 9
iv)
iii) f (x + 2) = 3(x + 2)2 − 2(x + 2) + 4 = 3(x2 + 4x + 4) − 2x − 4 + 4 = 3x2 + 10x + 12
3(x + h)2 − 2(x + h) + 4 − 3x2 − 2x + 4
[ ] [ ]
f (x + h) − f (x)
=
h ( 2 h
3x + 6hx + 3h2 − 2x − 2h + 4 − 3x2 − 2x + 4 6hx + 3h2 − 2h
) ( )
= 4 =
h h
= 6x + 3h − 2
3. Given f (x) = x2 − 4x + 3. Find: (i) f (1), (ii) f (2), (iii) f (a), and (iv) f (a + h).
