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MODULE 1 - FUNCTIONS
OVERVIEW
Often in practical applications the value of one quantity depends on the value of another. A person’s salary may depend on the number of machines used; the distance traveled by an object may depend on the number of machines used; the resistance of an electrical cable of fixed length depends on its diameter; and so forth. A relationship between such quantities is often given by means of a function. In calculus, the quantities in the relationship are real numbers. This module serves as a review of the basic concepts of functions. This particular topic is needed because they are fundamental in calculus and serve as a unifying concept throughout the text.
Often in practical applications the value of one quantity depends on the value of another. A person’s salary may depend on the number of machines used; the distance traveled by an object may depend on the number of machines used; the resistance of an electrical cable of fixed length depends on its diameter; and so forth. A relationship between such quantities is often given by means of a function. In calculus, the quantities in the relationship are real numbers. This module serves as a review of the basic concepts of functions. This particular topic is needed because they are fundamental in calculus and serve as a unifying concept throughout the text.
LEARNING OBJECTIVES:
At the end of this module, you should be able to:1. Determine the domain and range of a function including piecewise functions;2. Evaluate operations of functions; and3. Represents real life situations using functions.
At the end of this module, you should be able to:1. Determine the domain and range of a function including piecewise functions;2. Evaluate operations of functions; and3. Represents real life situations using functions.
DEFINITION OF A FUNCTION
From the image on the left, it can be perceived that a function is a relation for which each value from the set of first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair. In simple terms, "a function is an equation for which any x that can be plugged into the equation will yield exactly one y out of the equation." A function is normally written as: f(x) and read as "function f of x," though other letters can be used such as g(x) or y(x). Meanwhile, in solving real problems, we use meaningful letters like:• P(t) for power at time t,• F(t) for force at time t,• h(x) for height of an object, x horizontal units from a fixed point.
From the image on the left, it can be perceived that a function is a relation for which each value from the set of first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair. In simple terms, "a function is an equation for which any x that can be plugged into the equation will yield exactly one y out of the equation." A function is normally written as: f(x) and read as "function f of x," though other letters can be used such as g(x) or y(x). Meanwhile, in solving real problems, we use meaningful letters like:• P(t) for power at time t,• F(t) for force at time t,• h(x) for height of an object, x horizontal units from a fixed point.
DOMAIN AND RANGE OF A FUNCTION
The domain and range of function is the set of all possible inputs and outputs of a function respectively. In general, we determine the domain of each function by looking for those values of the independent variable (usually x) which we are allowed to use. (Usually, we have to avoid 0 on the denominator of a fraction, or negative values under the square root sign). The range of a function is the complete set of all possible resulting values of the dependent variable (usually y), after we have substituted the domain. The range is the resulting y-values we get after the substituting all the possible x-values.
The domain and range of function is the set of all possible inputs and outputs of a function respectively. In general, we determine the domain of each function by looking for those values of the independent variable (usually x) which we are allowed to use. (Usually, we have to avoid 0 on the denominator of a fraction, or negative values under the square root sign). The range of a function is the complete set of all possible resulting values of the dependent variable (usually y), after we have substituted the domain. The range is the resulting y-values we get after the substituting all the possible x-values.
Here are the general formulas used to find the domain of different types of functions [where R is the set of all real numbers].
- Domain of any polynomial (linear, quadratic, cubic, etc) function is R.
- Domain of a square root function √x is x≥0.
- Domain of an exponential function is R.
- Domain of logarithmic function is x>0.
- To find the domain of a rational function y = f(x), set the denominator ≠ 0.
Also, here are the general formulas used to find the range of different types of functions [where R is the set of all real numbers here].
- Range of a linear function is R.
- Range of a quadratic function y = a(x-h)2 + k is:
- Range of a square root function is y≥0.
- Range of an exponential function is y>0.
- Range of logarithmic function is R.
- To find the range of a rational function y = f(x), solve it for x and set the denominator ≠ 0.
This example from Quiz 1 shows finding the domain and range of a rational function. As can be seen from the first picture, I got the domain by simply setting the denominator equal to 0. Meanwhile, in finding the range, I used the domain of the inverse function by replacing every x with y and every y with x. Then, by algebraic manipulation, I solved for the new y. That new y will be the inverse function and its domain will be used to find the range of the original function. For instance, the domain of the inverse function is x ≠ 0, therefore, the range of the original function is y ≠ 0.
PIECEWISE FUNCTION
A piecewise function is a function that is defined by different formulas or functions for each given interval. It is also in the name: piece. The function is defined by pieces of functions for each part of the domain. On the right is an example of a piecewise function taken from my assignment no. 2. Graphing a piecewise function is simply plugging in x values to each function given while considering the domain stated to get the y values. Then simply plot the points on the graph while taking in consideration which point is included and which is not. Then connect the dots accordingly.
A piecewise function is a function that is defined by different formulas or functions for each given interval. It is also in the name: piece. The function is defined by pieces of functions for each part of the domain. On the right is an example of a piecewise function taken from my assignment no. 2. Graphing a piecewise function is simply plugging in x values to each function given while considering the domain stated to get the y values. Then simply plot the points on the graph while taking in consideration which point is included and which is not. Then connect the dots accordingly.
OPERATIONS ON FUNCTIONS
Functions with overlapping domains can be added, subtracted, multiplied and divided. If f(x) and g(x) are two functions, then for all x in the domain of both functions the sum, difference, product and quotient are defined in the image on the left.
Functions with overlapping domains can be added, subtracted, multiplied and divided. If f(x) and g(x) are two functions, then for all x in the domain of both functions the sum, difference, product and quotient are defined in the image on the left.
COMPOSITION OF FUNCTIONS
The composition of two functions f and g is the new function h, where h(x) = f(g(x)), for all x in the domain of g such that g(x) is in the domain of f. The notation for function composition is h = f ◦ g or h(x) = (f ◦ g)(x) and is read as 'f of g of x'. The procedure is called composition because the new function is composed of the two given functions f and g, where one function is substituted into the other. The new function is called a composite function as defined on the picture in the right.
The composition of two functions f and g is the new function h, where h(x) = f(g(x)), for all x in the domain of g such that g(x) is in the domain of f. The notation for function composition is h = f ◦ g or h(x) = (f ◦ g)(x) and is read as 'f of g of x'. The procedure is called composition because the new function is composed of the two given functions f and g, where one function is substituted into the other. The new function is called a composite function as defined on the picture in the right.
FUNCTIONS AS MATHEMATICAL MODELS
In applications of calculus, we need to express a real-world situation in terms of a functional relationship, called mathematical model of the situation. For instance, see the example on the left taken from my Quiz 1. To solve the word problem, the first thing I did is read and understand the problem thoroughly. Then, I had drawn a box and assigned its dimensions with variables according to what is stated on the problem. Then, I wrote an equation representing the volume of the box. By using algebraic manipulation, I arrived to the value of y or the height in terms of x. Consequently, I got another equation which expresses the surface area of the box while considering the given dimensions. It was stated in the problem that what's need to be done is to express the surface area as a function of the width x. Thereby, it is required to transform the surface area equation into a function of x. In other words, only x will be the lone variable. To do that, it is needed to eliminate the variable y by substituting the value of y in terms of x which was gained in the previous equation. After substituting, just simplify the function by algebraic manipulation.
In applications of calculus, we need to express a real-world situation in terms of a functional relationship, called mathematical model of the situation. For instance, see the example on the left taken from my Quiz 1. To solve the word problem, the first thing I did is read and understand the problem thoroughly. Then, I had drawn a box and assigned its dimensions with variables according to what is stated on the problem. Then, I wrote an equation representing the volume of the box. By using algebraic manipulation, I arrived to the value of y or the height in terms of x. Consequently, I got another equation which expresses the surface area of the box while considering the given dimensions. It was stated in the problem that what's need to be done is to express the surface area as a function of the width x. Thereby, it is required to transform the surface area equation into a function of x. In other words, only x will be the lone variable. To do that, it is needed to eliminate the variable y by substituting the value of y in terms of x which was gained in the previous equation. After substituting, just simplify the function by algebraic manipulation.
MODULE 2 - LIMITS
DEFINITIONS
A limit is a value at which a function approaches the output for the given input values in mathematical perspective. Limits are used to define integrals, derivatives, and continuity in calculus and mathematical analysis. It is employed in the analysis process and is always concerned with the function's behavior at a specific point.
A limit is a value at which a function approaches the output for the given input values in mathematical perspective. Limits are used to define integrals, derivatives, and continuity in calculus and mathematical analysis. It is employed in the analysis process and is always concerned with the function's behavior at a specific point.
In the same manner, if we require that x be greater than a, we get the “right-hand limit of f(x) as x approaches a is equal to L” which is written in the same manner only that the notation x → a- will be changed into x → a+ which means that we consider only x greater than a. By comparing the definition of limit and the definition of one-sided limits, it can be inferred that "limit only exists if and only if left-hand limit is equal to right-hand limit" and "limit does not exist (DNE) if the left and right limits are different."
LIMIT LAWS
Limit laws are simply the properties of limit which are used to calculate the limit of a function. It allows us to evaluate limits of functions without having to go through the step-by-step processes every time we are asked to get the limit of a function. There are 11 limit laws found in the table on the right.
DIRECT SUBSTITUTION PROPERTY
Aside from utilizing the limit laws on the right, there is another easy way of finding limits of functions and that is through direct substitution. The substitution rule for calculating limits is a method of determining limits by simply replacing the value of x with the point at which the limit should be calculated. Consider a function f(x), the purpose of which is to discover the function's limit at x = a. In this manner, in the expression for the function f(x), x is simply replaced with "a." Functions with the direct substitution property are called continuous at a. However, NOT all limits can be evaluated by direct substitution. See the example below.
Limit laws are simply the properties of limit which are used to calculate the limit of a function. It allows us to evaluate limits of functions without having to go through the step-by-step processes every time we are asked to get the limit of a function. There are 11 limit laws found in the table on the right.
DIRECT SUBSTITUTION PROPERTY
Aside from utilizing the limit laws on the right, there is another easy way of finding limits of functions and that is through direct substitution. The substitution rule for calculating limits is a method of determining limits by simply replacing the value of x with the point at which the limit should be calculated. Consider a function f(x), the purpose of which is to discover the function's limit at x = a. In this manner, in the expression for the function f(x), x is simply replaced with "a." Functions with the direct substitution property are called continuous at a. However, NOT all limits can be evaluated by direct substitution. See the example below.
I got this example from my Quiz 1. As we can see, it is impossible to find the limit of the given function through direct substitution because by substituting x = 4 in the denominator, we will get 0 which will make the function undefined. Even the quotient law cannot be used since the limit of the denominator is 0. So, to find the limit, I did some preliminary algebra by first factoring the denominator. Then by observing the factored form of the function, we can say that there is a common factor which can be cancelled. By cancelling the common factor x - 4, we are now left with 1/(x + 3). At this point, we can now apply the direct substitution, thus, the limit is 1/7.
I got here another example from my Quiz 1 again. As can be observed, both direct substitution and quotient law cannot be applied to find the limit of the given function since it will make the function undefined given that the limit of the denominator is zero (0). The same as the previous example, I did some preliminary algebra not by factoring but by rationalizing the numerator. The rationalizing technique is simply multiplying the numerator or denominator by the conjugate of the numerator or denominator. In the given example, the conjugate of the numerator √(x2 + 4) - 2 is √(x2 + 4) + 2. Basically, conjugate can be derived by changing the sign of the second term in case of binomials. After multiplying the conjugate, I simplified the expression by cancellations. Then, I used the direct substitution in the latter part to get the limit which is 1/4.
FINDING LIMITS AT INFINITY
Limits at infinity are used to describe the behavior of functions as the independent variable increases or decreases without bound. In the example I got from my Quiz 1, it can be noticed that as x becomes large, both numerator and denominator become large, so it is not obvious what happens to their ratio. To evaluate the limit at infinity of the given rational function, it is advised to divide both the numerator and denominator by the highest exponent of x that occurs in the denominator, which is in this case x3. After dividing, we need to get the limit of each of the term in the numerator and denominator while considering the limit laws and other properties discussed above. How did I arrive in 4/5? The limit of 4 is 4 and the limit of 5 is 5 because the limit of a constant function is simply the constant itself. Moreover, the limit at infinity of 1/x or any fraction with a constant as the numerator and x as the denominator of any power is zero (0). Therefore, the limit at infinity of the given function is indeed 4/5.
Limits at infinity are used to describe the behavior of functions as the independent variable increases or decreases without bound. In the example I got from my Quiz 1, it can be noticed that as x becomes large, both numerator and denominator become large, so it is not obvious what happens to their ratio. To evaluate the limit at infinity of the given rational function, it is advised to divide both the numerator and denominator by the highest exponent of x that occurs in the denominator, which is in this case x3. After dividing, we need to get the limit of each of the term in the numerator and denominator while considering the limit laws and other properties discussed above. How did I arrive in 4/5? The limit of 4 is 4 and the limit of 5 is 5 because the limit of a constant function is simply the constant itself. Moreover, the limit at infinity of 1/x or any fraction with a constant as the numerator and x as the denominator of any power is zero (0). Therefore, the limit at infinity of the given function is indeed 4/5.
Salazar (BSCE 1B) - Assignment No. 1.pdfASSIGNMENT NO. 1
Salazar - (BSCE 1B) - Quiz 1 Solutions.pdfPROBLEM SET 1 (20/20)
Salazar (BSCE 1B) - Problem Set 1.pdfASSIGNMENT NO. 2
Salazar (BSCE 1B) - Assignment No. 2.pdfQUIZ 1 SOLUTIONS (35/35)
REFLECTION
In general, Calculus 1 helped me to review some concepts and at the same time taught me new lessons that I believe would be valuable in future computations in the civil engineering course. The topics contained in the first and second learning outcomes (CLO1 and CLO2) were already discussed to us during my Senior High School through the subject Basic Calculus. Thanks to Calculus 1 because I was able to review some useful topics such as functions in Module 1, limits in Module 2, derivatives in Module 3, implicit differentiation and higher order derivatives in Module 4, and derivatives of trigonometric functions in Module 5. Relearning the aforementioned topics gave me satisfaction in this subject since I had already forgotten some of them. The subject somewhat allowed me to reminisce my last days in school because I remember I was already reviewing my notes back then in Basic Calculus in preparation of the final exam, but unfortunately, it was cancelled due to the start of the nationwide lockdown. Going back, I found new topics in this subject such as the derivatives of inverse trigonometric functions of Module 6; derivatives of logarithmic and exponential functions in module 7; tangents and normals, extrema and first derivative test (FDT), concavity, points of inflection, second derivative test, increasing/decreasing intervals, and ultimately graphing functions. Furthermore, I found the last topics to be a bit challenging especially when it comes to graphing of functions that I encountered some confusions. Yet, I am grateful to our instructor for addressing my concerns regarding the topics. Nevertheless, I still enjoyed the whole subject because it is aligned on my interest of solving math problems. I just cannot explain the feeling of satisfaction after finally arriving to every correct answer of the problem sets, assignments, and quizzes given to us. A big thank you to our instructor for discussing the course in a manner that everybody could understand. Probably, I wouldn't enjoy the subject without the pure efforts and dedication of our instructor.
In general, Calculus 1 helped me to review some concepts and at the same time taught me new lessons that I believe would be valuable in future computations in the civil engineering course. The topics contained in the first and second learning outcomes (CLO1 and CLO2) were already discussed to us during my Senior High School through the subject Basic Calculus. Thanks to Calculus 1 because I was able to review some useful topics such as functions in Module 1, limits in Module 2, derivatives in Module 3, implicit differentiation and higher order derivatives in Module 4, and derivatives of trigonometric functions in Module 5. Relearning the aforementioned topics gave me satisfaction in this subject since I had already forgotten some of them. The subject somewhat allowed me to reminisce my last days in school because I remember I was already reviewing my notes back then in Basic Calculus in preparation of the final exam, but unfortunately, it was cancelled due to the start of the nationwide lockdown. Going back, I found new topics in this subject such as the derivatives of inverse trigonometric functions of Module 6; derivatives of logarithmic and exponential functions in module 7; tangents and normals, extrema and first derivative test (FDT), concavity, points of inflection, second derivative test, increasing/decreasing intervals, and ultimately graphing functions. Furthermore, I found the last topics to be a bit challenging especially when it comes to graphing of functions that I encountered some confusions. Yet, I am grateful to our instructor for addressing my concerns regarding the topics. Nevertheless, I still enjoyed the whole subject because it is aligned on my interest of solving math problems. I just cannot explain the feeling of satisfaction after finally arriving to every correct answer of the problem sets, assignments, and quizzes given to us. A big thank you to our instructor for discussing the course in a manner that everybody could understand. Probably, I wouldn't enjoy the subject without the pure efforts and dedication of our instructor.
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