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1 - function notes
Industrial Engineering (ERGO1)
University of the Assumption
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OVERVIEW
This unit will serve as an introduction in the study of Calculus, where the prerequisite knowledge in developing its first concepts will be described here. We start with the treatment of a certain type of a relation which we will call as a function. Using the concept of functions as a basis, we will describe next all of the possible methods to evaluate the limit of any function. The understanding of functions and limits will be very vital in the formulation of the concept of the derivative in the next unit.
Overall, the study of Calculus is attributed to two mathematicians, namely Isaac Newton and Gottfried Wilhelm von Leibniz , wherein they work independently of each other. Surprisingly enough, they discovered a single concept, but off course with different notations. All those possible notations will be tackled as we go along with our discussion of Calculus.
LEARNING OUTCOMES: After the completion of this unit, the students are expected to:
- Determine whether a given relation is a function or not.
- Evaluate the value of a given function.
- Sketch the graphs of basic functions (linear, quadratic, piecewise function)
- Define and discuss the concept of the limit of a function.
- Familiarize and apply correctly each of the limit theorems to evaluate the limit of basic functions.
- Evaluate the limit of a function based on its graph.
- Describe the condition for a given limit to exist using the definition of one-sided limits.
- Evaluate limits of trigonometric functions using certain limit theorems.
- Differentiate the concept of evaluating infinite limits and limits at infinity.
UNIT I: FUNCTIONS AND LIMITS
andymath/limits-on-graphs-
Most people think of the possible answers on the question < Why study calculus, if it is not that useful in our everyday lives? =. While it is true that applications of calculus is not that evident and cannot be seen by our naked eyes, the study of calculus can help develop our critical and analytical thinking skills , which is a need for every engineer or science major. Also, the understanding of concepts of calculus will serve as a very important component in the thorough study of higher mathematics, chemistry, physics and specialized subjects in any engineering fields. Therefore, the study of calculus is very vital for an engineering student, a math or a science major.
To introduce ourselves with the study of this broad field of mathematics, we start with some introduction to the study of calculus. That is, our first concern must be on the type of expressions where all the definitions and concepts of calculus is based upon. By principle, the concepts of calculus are based on a certain group of expression or relation called a function.
For us to explore all the possible properties of a function to formulate its formal definition, let us have first a preliminary description of a function.
To illustrate and explain this preliminary description, suppose that we are given with a square having a side that measures x units. Recall that its area A can be computed using the formula:
ý xA 2. (1)
Given this formula, note that it is only possible to compute for the area A of the square if we are given with the measure of its side x. That is, for every value of x , there corresponds a single value of A. The description <single value= is synonymous to the word <unique value= meaning, there must be ONLY ONE value of A for a specific value of x. It is not possible given this expression that there will be more than one value of A for a value of x. This part of the concept falls under the 2nd part of our preliminary description of a function, that is, the number A is unique for a specific value of x.
Also, note from the expression that the value of the area of the square A depends on the measure of its side x. From this reason, we can also call A as our dependent variable , and x as our independent variable. The terms are defined from the reason that the value of A is <dependent= on the value of x.
To fully describe the properties of functions, let’s denote the dependent variable to be equal to y in general and the independent variable as x in (1). Doing this, we have:
ý xy 2 (2)
1. FUNCTIONS AND THEIR GRAPHS
Function – can be thought of as a correspondence from a set X or real numbers x to a set Y of real numbers y , where the number y is unique for a specific value of x.
answer as undefined because it will give us a very large positive or negative number, depending on the sign of the numerator. That is, for example:
ý 0
8 ; ý 0
8
For the values of x that we set, we obtained the resulting values in the figure by using the expression ý xy 2. Note that for every value of x that we set, the resulting values of y is always either a positive number or 0. This is so because any number (positive, negative or 0) raised to the second power will always yield a positive number or 0. To describe all these possible values
of y in interval notation we write it as ,0[ ù, which means that the values range from 0 up to
the largest possible number ( ). The open bracket notation before 0 signifies that 0 is included in the possible values of y. If ever we have an instance that the lower or upper boundary must not be included, we must place an open or close parenthesis. In terms of a formal term, we call all resulting values of y given the possible values of x as the range of that function. We will illustrate further on how to find the domain and range of a function in the examples that we will encounter later in this section, either if we are given with just the equation itself or given its graph.
To further discover some other properties of functions; let us summarize the values of x and the corresponding values of y that we obtain, in tabular form which we call as the table of values. We will use the pairs of x and y values in the table of values to sketch the graph of the function ý xy 2.
x -2 32 -1 0 1 32 2 y 4 94 1 0 1 94 4
y
x
As stated in the concept of a function, every value of x must give only one value of y. This is evident in the ordered pairs formed in the table of values. That is, given (-2, 4), , (-1, 1), (0, 0), (1, 1), , and (2, 4) notice that no two distinct ordered pairs have the same first number. The first number described corresponds to the x - coordinate of each ordered pair. The moment we are involved with two ordered pairs having the same x -coordinate, automatically those ordered pairs will not resemble a function.
The consequence of this concept can be seen if these ordered pairs are plotted in the rectangular coordinate system to form the graph of the function. Since no two x - coordinates are the same, then notice from the graph that there are no two points in the graph that are vertically collinear with each other. That is, if we are going to plot a vertical line anywhere in the graph, there will be no
such line that will intersect two or more points in the graph. This concept, commonly known as the vertical line test determines if the graph of a certain equation resembles a function. Some illustrations of the vertical line test are shown below.
Based on all of the concepts illustrated, we are now ready to state the formal definition of a function. Note that this formal definition is just a summary of all of the concepts that we talked about in this section.
A straight-line graph is a function, since any vertical line plotted will give only one point of intersection. Since the equation of this is a 1 st degree equation , all first degree equations are functions.
Equations having graphs like this are functions, since again any vertical line plotted anywhere along its graph has only one point of intersection. In general, this graph is an example of the graph of a polynomial function , therefore all polynomial equations are functions. (Recall: polynomial equations must satisfy the definition of a polynomial from algebra)
A vertical line plotted in a part of the graph has two points of intersection. Therefore this graph does not resemble a function. In general, all graphs of conic sections (except a parabola opening upward / downward) are not functions.
The graph illustrated is not a graph of a function since a vertical line intersects the graph at more than one point.
Definition : Function
A function is a set of ordered pairs ( x , y ) in which no two distinct ordered pairs have the same first number. The set of all possible values of x is called the domain of the function, and the set of all resulting values of y is called the range of the function.
e. Evaluate ø ù xf and ø ù hf first, then combine to obtain ø ù ø ù hfxf
ø ù 2 xxxf ý 43 ; ø ù 2 hhhf ý 43
ø ù ø ù ø ù ø 22 hhxxhfxf ýü 4343 ù ø ù ø ù 22 hhxxhfxf ý 833
Note in items (c), (d) and (e) that the final expressions for function values are in terms of h and x , where in (d), we applied the concept of square of a binomial from algebra to
simplify ø ù hx 2. As a strong warning, WE MUST NOT DISTRIBUTE THE <2=
EXPONENT to x and h , since it will violate concepts of algebra. The expansion of
ø ù hx 2 must have a middle term. Our solution must be in conformity with the existing or
known concepts of algebra, trigonometry or geometry, whenever applicable.
Also, note that from our final answers for (d) and (e) that ø ù ø ù ø ù hfxfhxf. There is
no such concept that < f = can be distributed to x and h to form ø ù ø ù hfxf. These warnings
must be kept in mind to avoid possible errors in evaluating function values.
Example 2 : Given (a) ø ù 2 xxxf ý 754 ; (b) ø ù xxf ý 1 , evaluate ø ù ø ù
h
xfhxf ; h 0
This example is also an illustration of evaluating function values, where we are required
to find an expression for
ø ù ø ù
h
xfhxf , given two different functions in (a) and (b). By
concept, the value of h in the expression cannot be equal to 0, because division by zero is not defined. Note that this condition for the value of h is stated in the given. From now on, we must be critical about the given expressions that we will encounter, that is we must take note if there are restricted values in our involved variables.
As a strategy to evaluate ø ù ø ù
h
xfhxf , we apply the technique that we employ in
Example 1 (e), that is we find first expressions for ø ù hxf and ø ù xf. Then we substitute
and simplify. Executing this in the two functions involved:
a. ø ù 2 xxxf ý 754
ø ù hxf : ø ù ø ù ø ù 2 hxhxhxf ý 754
ø 22 ù ø ù hxhxhx ý 7524
ø ù 22 hxhxhxhxf ý 755484
ø ù xf : ø ù 2 xxxf ý 754
Since ø ù 22 hxhxhxhxf ý 755484 and ø ù 2 xxxf ý 754 ;
ø ù ø ù ø ù ø ù
h
xxhxhxhx h
xfhxf ý 222 754755484
h
ý 222 xxhxhxhx 754755484
h
ý 2 548 hhxh ;
ø ù
h
ý hxh 548
ü ø ù ø ù hx ý 548
h
xfhxf
b. ø ù xxf ý 1
ø ù hxf : ø ù ø ù hxhxf ý 1 ø ù xf : ø ù xxf ý 1
ø ù hxhxf ý 1
ø ù ø ù
h
xhx h
xfhxf ý 11
It seems that no further simplification process can be applied to our expression for
ø ù ø ù
h
xfhxf because from algebra, radicals in our expression is not in the
denominator. But in certain parts of our study of calculus, it is more advisable to transfer the radicals in the denominator. For us to do this, we are to multiply both the numerator and denominator by the conjugate of the numerator. Recall that the conjugate of a binomial of the form ba is ba , and vice-versa. That is, we just change the sign of the second term to form the conjugate of a binomial.
Then in our expression so far, the conjugate of xhx 11 must be xhx 11 , and we multiply this in the numerator and denominator. Executing this and simplifying:
factor out h in numerator to cancel common factors in the numerator and denominator to simplify further
we will think first of the possible values of x , that is if there are any value of x that are not allowed in the function. Often times, these restrictions for the value of x can occur if we are given with a fraction (denominator cannot be zero) or a radical expression (negative radicands are not allowed for even indices). After determining those possible values of x , we substitute some of them in the function to determine all of the possible values of y or
ø ù xf.
a. Given ø ù xxf ý 1 , we first think if there are any restrictions for the possible values of
x. Since we are just given with a linear function , all possible values of x are allowed starting from the smallest possible number ( ) up to the largest possible number
( ). Therefore, the domain of ø ù xf consists of the set of real numbers.
After determining the domain, we substitute those possible values of x from to
to determine the range. Note that the corresponding values of ø ù xf upon
substitution ranges from the smallest possible number ( ) up to the largest
possible number ( ). Therefore, the range of ø ù xf also consists of the set of
real numbers. Writing these results in interval notation and denoting the domain as set D and the range as set R, we have:
D: ø , ù or ; R: ø , ù or
NOTE : Given any polynomial function, all values of x are possible to be substituted with
no restrictions; therefore its domain in general is always at the interval ø , ù. The
previous example of a linear function is an example of a polynomial function. As an
illustration, the domain of the functions ø ù 2 xxxf ý 85 , ø ù xxxxf 234 ý 154 and
ø ù xxxf 26 ý 97 are all ø , ù since they are all polynomial functions having non-
negative and non-fractional exponents. The range of these functions is in general not
ø , ù, as will be illustrated in (b).
b. For the function ø ù xxf 2 ý 2 , note that it is a polynomial function because the
exponents are not negative or a fraction. Then its domain is automatically
ø , ù or the set of all real numbers.
To think of all the corresponding values of ø ù xf , notice that the variable x is raised to
the second power. Recall that any number raised to an even exponent will always give us a positive number or zero (if the base is 0). Therefore we have an assurance that the term x 2 will always give us a positive (non-negative) number or zero. Zero so far is the lowest possible answer for x 2. Then considering the second term which is 2, the lowest possible answer for x 2 2 will give us 2 ( 0 + 2 = 2 ). We conclude
therefore that the range of ø ù xf starts from 2 (where 2 is included) up to the
largest possible number ( ). In interval notation:
D: ø , ù or ; R: ,2[ ù
Note that since 2 is a possible value (included in the range), we place an open bracket before it. By notation, an open or close parenthesis must be placed before or after or since it is not a symbol for a real number.
c. For the possible values of x of the function ø ù xxf ý 3 , we must see to it that the
radicand x – 3 is not negative (square root of a negative number will yield an imaginary number). From this, note that the lowest possible allowable value of x is 3. Any value of x less than 3 will make the radical imaginary. Therefore, the domain of
ø ù xf starts from 3 up to the largest possible number ( ).
Given that 3 is the lowest possible value of x , note that substituting 3 to ø ù xxf ý 3
will yield an answer of 0 for ø ù xf. Therefore we can consider 0 as the lowest
possible value of ø ù xf , up to the largest possible number ( ). To write the
domain and range in interval notation:
D: ,3[ ù ; R: ,0[ ù
NOTE : By definition, finding the square root of a number must result to a positive and a negative root. But considering both square roots will violate the definition of a
function. For example, substituting x = 4 to ø ù xxf ý 3 will give us ø ù yxf ýý 1.
Forming the corresponding ordered pairs, we have (4, 1) and (4, -1). But recall that for an expression to be a function, there must be no two ordered pairs that has the same first number or x -coordinates. For this reason, we just take the answer for the square root of a number to be equal to its positive root or zero (this is called the principal
square root ). This explains the justification on the range of ø ù xxf ý 3 that starts
from 0 up to a very large positive number.
d. Same as (c), we must see to it that given ø ù xxf 2 ý 9 , the answer for the radical
must not be imaginary. Setting this condition for the possible value of x , note that the value of x 2 must not be less than 9. From this, one possible group for the value of x
is that x must be greater than or equal to 3 and in interval form, we have û ,3 ù.
But taking note again that x 2 always yields a positive number, negative values of x can also be considered as long as the answer for x 2 is not less than 9. Therefore, an another possible group for allowable values of x is that x can also be less than or
ø ù
3
2372
ý x
xf xx ; x 3
ø ùø ù
3
123
ý x
xx
ø ù xxf ýü 12 ; x 3
The condition that x 3 must STILL be applied to our simplified expression for ø ù xf
which is ø ù xxf ý 12 since this function just came from the original function having
x 3 as a preliminary condition. Notice that our simplified expression is a linear function , where the technique of graphing is discussed from junior or senior high school algebra. By concept, the graph of a linear function must be a straight line. In particular, we will apply the method of transforming the equation to its slope-intercept form.
Recalling that ø ù xf can be replaced by y and the slope-intercept form is given by
ý bmxy ( m = slope of the line and b = y-intercept):
ø ù xxf ý 12 ; x 3
xy ý 12 ; m ý 2 and b ý 1
Since b = 1, we obtained one point in the graph of ø ù xf , which is (0, 1). Since the graph
is just a straight line, we just need to find an another point in the graph and just connect them and extend to form its graph.
Now, recall that x cannot take the value of -3. We must also take this into consideration
in the graph of ø ù xf , that is, we must also find the restricted value of y by substitution to
ø ù xxf ý 12 :
*if x 3 ; y ø ù 132 y 16
y 5 ; restricted point is at (-3, -5)
From the word <restricted=, the point (-3, -5) must not be a part of the graph of the
original function ø ù
3
2372
ý x
xf xx. To emphasize this, we plot this point as a hollow
point (hole) in the graph.
Now, we have two points, the point (0, 1) obtained from the y -intercept b = 1, and the restricted point (-3, -5). Plotting these points on the Cartesian plane noting that (-3, -5) must be plotted as a hole and joining these points to form a line, we obtain the graph in the next page.
For the domain and range of the function, recall that x cannot take a value of -3 and this is the single value of x that serves as its restricted value. Then in this case, the domain of the function can be written as the combination of the
intervals ø 3, ù and ø ,3 ù or
ø ù ø ,33, ù.
Upon substituting -3 to ø ù xf , recall that
we obtained -5 as the restricted value of y. Therefore, the range of the function can be written as the combination of
the intervals ø 5, ù and ø ,5 ù or
ø ù ø ,55, ù.
D: ø ù ø ,33, ù or except -3 ; R: ø ù ø ,55, ù or except -
Example 5 : Sketch the graph of the function þ
ý
ü ý
ý 34
)( 239
x if
xf x if x and determine its
domain and range.
The function that we are going to graph is an example of a piecewise function. From the root word <piece=, a piecewise function is a type of a function that is subdivided into
two or more parts. In the example, the function is composed of two parts such that ø ù xf
is 9 – x 2 for values of x 3 , and ø ù xf is equal to 4 if x ý 3.
To sketch the graph of a piecewise function, we will graph these two parts in a single Cartesian plane, taking note of the values of x that each part represents.
Starting with ø ù 9 ý xxf 2 at values of x 3 , note that the restricted value for x is
already given, which is -3. Later, we can find the corresponding restricted value for y by
substituting -3 to x in ø ù 9 ý xxf 2. Recalling that the graph of ø ù xf here is a parabola
( ø ù xf is a second-degree equation), we transform first the equation to its standard form
for us to find the coordinates of its vertex upon replacing ø ù xf by y. Doing this:
(i) for ø ù 9 ý xxf 2 ; x 3
9 ý xy 2 ; Recall (standard form of a parabola): ø ù 24 ø ùý kyahx
2 9 ý yx 2 yx ý 9 2 ø ù yx ý 9
the piecewise function is composed of the set of all possible numbers or ø , ù
For the range of the piecewise function, we just combine the range of its 1 st and 2nd
parts. Given ø ù 9 ý xxf 2 , its range is at ø , 9 ý since x 2 will always yield a positive
number and in most instances, the difference between 9 and x 2 is negative. The highest possible value of 9 x 2 is 9 and it occurs if x = 0. For the second part, automatically
ø ù yxf ýý 4. Since this value of 4 already lies at the interval ø , 9 ý, we conclude that
the range of the piecewise function is ø , 9 ý.
D: ø , ù or ; R: ø , 9 ý
As an alternative, we can also directly determine from a mere look of the graph that we constructed the expressions for the domain and range of the function.
Example 6 : Sketch the graph of the function ÿþ
ÿ ý
ü
þ
ý
ü ý 312
35
31
)(
x if x
x if
x if x xf and determine its
domain and range.
The piecewise function involved consists of three parts. Just like our previous example, we consider and graph each part one by one. For each portion, we must take note and understand very carefully on what values of x each portion will be applicable.
The first part reads ø ù xxf ý 1 that is applicable for values of x that are less than 3.
Understanding very carefully the meaning of x ü 3 , it turns out that the graph is only
applicable for values of x that is less than 3. The graph of ø ù xxf ý 1 then is only
applicable for the left side of x = 3 in the Cartesian plane , where 3 is NOT included. The value x = 3 can be considered here as the restricted value of x. To determine on what point the graph will end or terminate, we must find also the corresponding restricted
value of y, and plot the point obtained as a hole. Noting that the graph of ø ù xxf ý 1 is a
straight line , we find first its slope and y-intercept:
(i) for ø ù xxf ý 1 ; x ü 3
xy ý 1 ; m = 1 and b = -
Again, for us to sketch a line, we need to determine any two points that passes through
the graph of ø ù xxf ý 1. To see if we can use the y -intercept, we note that the value of x
there is 0. Since 0 < 3, then the y-intercept satisfies the condition x < 3 and we can use (0, -1) as our first point.
For the second point, we find the point where the first part of the graph ends, by using the restricted value of x. Solving for the restricted value of y:
*if x 3 ; y 13
y 2 ;
restricted point is at (3, 2)
The restricted point must be plotted as a hollow point or a hole. We note that since the part of the graph is only applicable for values of x that are less than 3, the graph must be at the right side of x = 3 and the point (3, 2) will serve as the terminating point. This portion of the graph is shown in the right.
Note that we are not yet through with the graph of the piecewise function itself. We still have two portions of the piecewise function that we will plot on the same coordinates. To
proceed with the second portion, we have ø ù xf ý 5 applicable if x ý 3. Replacing ø ù xf by
y, it turns out that y ý 5 if x ý 3. Therefore the second portion corresponds to a point , located at (3, 5). In terms of a formal solution:
(ii) for ø ù xf ý 5 ; x ý 3
y ý 5 ; point: (3, 5)
For the third part of the function, we are given that ø ù xxf ý 12 is applicable for values
of x that are greater than 3. If x > 3, then it means that the graph of ø ù xf here must be
located at the right side of x = 3. Unlike the first part where the restricted point is our terminating point, here the restricted point will serve as on where the graph will start. The
same, the restricted point is found by substituting x = 3 to ø ù xxf ý 12.
To help us sketch the graph of the third part, we note that the function is linear and we need to determine two points in its graph. Transforming to slope-intercept form:
(iii) for ø ù xxf ý 12 ; x ü 3
xy ý 12 ; m = 2 and b = 1
In the first part of the piecewise function, we recall that the y -intercept is included in the graph since it satisfies the condition x > 3. For this third part, note that it is only applicable for values of x that is greater than 3. The y -intercept has an x -coordinate of 0
For the range, we will just combine the range of the three parts of the piecewise function.
Recall in the first part that if x = 3, y = 2 for ø ù xxf ý 1. The value of y = 2 will serve as
an upper boundary for the range of the first part , taking note that 2 is NOT included
because it is a restricted value. Then the range for this part is ø 2, ù. For the second
part, the only value of y here is 5. Since it is not included in our previous range of
ø 2, ù, we must consider the value of 5 separately, by writing it as û ý5,5. This notation
of û ý5,5 means that only the number 5 is considered. Lastly for the 3rd part, recall that
the graph starts at the point (3, 7), where 7 here is not included and then proceeding up
to the largest possible number ( ). Therefore, the range for this part is ø,7 ù.
Combining the range from the three portions, we have ø ù û ý ø,75,52, ù.
D: ø , ù or ; R: ø ù û ý ø,75,52, ù
Example 7 : Sketch the graph of the function ÿ þ
ÿ ý
ü
ü
ü ý x if
x if
x if xf 23
221
24
)( and determine its
domain and range.
This is an another example of a piecewise function. Given three portions of the graph of
ø ù xf , we plot them one by one on a single coordinate system.
Starting with the first part, we have ø ù xf ý 4. Since it is applicable for x ü 2 , the
function is applicable to the left side of x = -2. Replacing ø ù xf by y , we obtain y ý 4
where its graph is a horizontal line extending from the smallest possible number ( ) up to -2 (but -2 is NOT included). The point (-2, -4) serves as a restricted point. In terms of a formal solution:
(i) for ø ù xf ý 4 ; x ü 2
y ý 4 ; restricted point: (-2, -4)
The graph of the second part is still a horizontal line since the form of the function is the same. The difference is that the line involved is applicable for x 22 which means
that the horizontal line defined by ø ù xf ý 1 is only applicable between x = -2 and x = 2
where both -2 and 2 are included (no restricted point).
(ii) for ø ù xf ý 1 ; x 22
y ý 1
For the third portion, still the graph of
ø ù xf ý 3 is a horizontal line , where it is
applicable starting with x = 2 (NOT included) up to the largest possible number ( ). The point (2, 3) serves as a restriction point and also, a starting point.
(i) for ø ù xf ý 3 ; 2 ü x or x þ 2
y ý 3 ; restricted point: (2, 3)
The graph of the function is shown on the right. Since the function is defined for all values of x , its domain is the set of real
numbers or ø , ù.
The range of this function is just the right sides of ø ù xf since these are the only values of
y where the graph has corresponding points. We can just emphasize and enumerate
them by enclosing them in a pair of braces. Therefore the range of ø ù xf is given by
3,1,4 .
D: ø , ù or ; R: 3,1,4 or û ý û ý û ý 3,31,14,
Test Your Understanding : Sketch the graphs of the following functions and determine their domain and range.
1. ø ù
5
225
ý x
xf x 4. þ
ý
ü ý
ý 20
212
)(
x if
x if x xf
2. ø ù
1
234
ý x
xf xx 5. þ
ý
ü ý
ý 32
)( 234
x if
xf x if x
3.
þ
ý
ü þ
ý 32
32
)(
x if
x if xf 6. þ
ý
ü
ý ü 013
)( 201
x if x
xf x if x
1 - function notes
Course: Industrial Engineering (ERGO1)
University: University of the Assumption
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