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CHAPTER 1: FUNCTIONS AS MODELS
LESSON 1.1: FUNCTIONS AS MODELS
FUNCTIONS
A function "f" is a type of relation between two sets of elements in which for each element in the first set there corresponds one and only one element in the second set. The first set, which is usually the set of x − values , is called the domain , and the set of all corresponding elements in the second set, which is usually the set of y − values , is called the range.
RELATIONS
A relation “r” is a rule of correspondence between two sets of elements in which for each element of the first set ( domain ) there corresponds two or more elements of the second set ( co − domain ). A relation is also a set of ordered pairs (x, y).
FUNCTIONS VS. RELATIONS TABLE
The following table shows how relations and functions are the same and how they are different. RELATIONS FUNCTIONS One − to − one Many − to − one One − to − many Many − to − many
One − to − one Many − to − one
All functions are relation but not all relations are function.
LESSON 1.2: FUNCTIONS AND RELATIONS AS A
TABLE OF VALUES
Example 1.1. Which of the following relations are functions? p = {(2,3), (3,3), (4,6), (5,6)} o Function: Many to one q = {(2,4), (2,5), (3,6), (3,7), (4,8)} o NF: One to many r = {(1,4), (2,8), (3,12), (4,16), ... , (n, 4n)} o Function: One to one
Example 1.2. Relations and functions can be represented by mapping diagrams where the elements of the domain are mapped to the elements of the co-domain using arrows.
Example 1.3. Which of the following represents a function? a.) y = 2x + 1 = Function (Linear F.) b.) y = x^2 − 2x + 2 = Function (Quadratic F.) c.) x^2 + y^2 = 1 = Not a Function (Unit Circle) d.) y = √𝑥 + 1 = Function (Square root Function) e.) y = 2𝑥+ 𝑥−1 = Function (Reciprocal Function) f.) y = ⌊𝑥⌋ = Function
if even nos. exponent nung y = NF if odd nos. exponent nung y = F x – domain y – range ⌊^ ⌋^ floor variable y = f(x)
LESSON 1.3: FUNCTIONS AS A GRAPH IN THE
CARTESIAN PLANE
Recall : A relation between two sets of numbers can be illustrated by a graph in the Cartesian plane, and that a function passes the vertical line test. Vertical line test – is a vertical line can be drawn anywhere and intersect the graph in at most one point.
VERTICAL LINE TEST
A graph represents a function if and only if each vertical line intersects the graph at most once. If a vertical line x = a intersects a graph twice, say at ( a, b ) and at ( a, c ), then the graph cannot represent a function because two different y − values correspond to x = a. Example 1.4 Which of the following graphs can be graphs of functions?
y – values = Vertical Line Test x – value = Horizontal Line test Vertical: I Horizontal: _____
LESSON 1.4: DIFFERENT TYPES OF FUNCTIONS
AND THEIR DOMAIN AND RANGE
These are the basic functions which are often encounter.
1. Constant Function f(x) = k D: {𝑥|𝑥 ∈ ℝ} or ℝ or set of all real numbers or (-∞,∞) R: {𝑦|𝑦 ∈ 𝑘}
f(x) = x
2. Identity Function f(x) = x D: {𝑥|𝑥 ∈ ℝ} or ℝ or set of all real numbers or (-∞,∞) R: {𝑦|𝑦 ∈ ℝ} or ℝ or set of all real numbers or (-∞,∞) 3. Absolute Value h(x) = |𝑥| D: {𝑥|𝑥 ∈ ℝ}^ or ℝ or set of all real numbers or (-∞,∞) R: {𝑦|𝑦 ≥ 0}^ or [0, ∞) or y ≥ 0 ( ) = not included [ ] = included 4. Quadratic Function f(x) = x^2 D: {𝑥|𝑥 ∈ ℝ}^ or ℝ or set of all real numbers or (-∞,∞) R: {𝑦|𝑦 ≥ 0}^ – set notation or [0, ∞) – set interval 5. Cubic Function f(x) = x^3 D: {𝑥|𝑥 ∈ ℝ}^ or ℝ or set of all real numbers or (-∞,∞) R: {𝑦|𝑦 ∈ ℝ} or ℝ or set of all real numbers or (-∞,∞) 6. Square Root Function g(x) = (^) √𝑥 D: {𝑥|𝑥 ≥ 0} or [0, ∞) R: {𝑦|𝑦 ≥ 0} 7. Cube Root Function f(x) = √𝑥 3 D: {𝑥|𝑥 ∈ ℝ} or ℝ or set of all real numbers or (-∞,∞) R: {𝑦|𝑦 ∈ ℝ} or ℝ or set of all real numbers or (-∞,∞) 8. Reciprocal Function f(x) = 1 𝑥 D: {𝑥|𝑥 ≠ 0} or (-∞,0)∪(0, ∞) or Set of all ℝ except 0 R: {𝑦|𝑦 ≠ 0} Asymptote: a line that a curve approaches, as it heads towards infinity. 9. Reciprocal Squared Function f(x) = 1 𝑥^2 D: {𝑥|𝑥 ≠ 0}^ or (-∞,0)∪(0, ∞) or Set of all ℝ except 0 R: {𝑦|𝑦 > 0}
LESSON 1.5: FUNCTIONS AS
REPRESENTATIONS OF REAL-LIFE SITUATION
USING PIECEWISE FUNCTION.
Some situations can only be described by more than one formula, depending on the value of the independent variable. a. A user is charge ₱300.00 monthly for a particular mobile plan, which includes 100 free text messages. Messages in excess of 100 are charged ₱1.00 each. Represent the amount a consumer pays each month as a function of the number of messages “m” sent in a month. Solution : Let t(m) represent the amount paid by the consumer each month. It can be expressed by the piecewise function.
t(m) = {
F F NF
F: Many – to – one F: Many – to – one NF: One – to – many
x
y
F: Many – to – one F: One – to – one NF: Many – to – many
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b. A jeepney ride costs ₱9.00 for the first 4 kilometers, and each additional integer kilometer adds ₱1.50 to the fare. Use a piecewise function to represent the jeepney fare in terms of the distance “d” in kilometers. Solution : The input value is distance and the output is the cost of the jeepney fare. if F(d) represents the fare as a function of distance, the function can be represented as follows:
F(d) = {
⌊ ⌋ - floor varia (↓) ⌈ ⌉ - ceiling value (↑) 𝑡(𝑚)−> input value
c. A videoke machine can be rented for ₱1,000 for three days, but for the fourth day onwards, an additional costs ₱400 per day is added. Represent the cost of renting a videoke machine as a piecewise function of the number of days it is rented. Solution: Let C(d) represent the cost of rented videoke machine per day.
C(d) = {
A piecewise function is a function that is defined by two or more equations where each equations applies to a certain interval over a specified domain.
LESSON 2: EVALUATION OF FUNCTIONS
With a prior knowledge of what the graph of a function looks like, one way to sketch the graph of a function is to determine a set of few points satisfying the function. For example, in the function f(x) = 2x – 1 (Linear), the value of x can be any real number since there is no restriction in its domain. We now assign value for x then substitute it to the given function to identify the corresponding value for y. This process is called evaluation of functions. x - 2 - 1 0 1 2 f(x) or y - 5 - 3 - 1 1 3
f(x) = 2(-2) -1 = -4-1 = - f(x) = 2(-1) -1 = -2-1 = - f(x) = 2(0) -1 = 0-1 = - f(x) = 2(1) -1 = 2-1 = 1 f(x) = 2(2) -1 = 4-1 = 3 -5 + 2 = -
Hence, some points on the graph of f(x) = 2x − 1 are (−2, −5), (−1, −3), (0, −1), (1,1) and (2,3)
Example 2.1. Evaluate the following functions at x = 2.5: a. f(x) = 2x + 1 2(2.5)+1 = 6 b. q(x) = x^2 − 2x + 2 (2.5)^2 – 2(2.5) + 2 = 3.
c. r(x) = √𝑥 + 1 (^) √2.5 + 1 = (^) √3.
d. g(x) = 2𝑥+ 𝑥− 2(2.5)+ 2.5 −
e. t(x) = ⌊x⌋ ⌊2.5⌋ = 2
Example 2.2. Find r(−3) and g(1) where r and g are as defined in the previous example. This is not possible because −3 is not in the domain of r(x), and if you evaluate it √−3 is an imaginary number. While 1 is not in the domain of g(x) because when you substitute 1 into the denominator of g(x), zero (0) will be the denominator which is undefined.
Example 2.2. Evaluate the following functions, where f and q are as defined in Example 1.2.1. a.) f(2x − 1) b.) q(3x + 2)
DPMA: Distributive Property of Multiplication over Addition Square of a Binomial: (a + b)^2 = a^2 + 2ab + b^2 (a - b)^2 = a^2 - 2ab + b^2
a^2 + b^2 = (a + b)^2 – 2ab
LESSON 3: OPERATIONS ON FUNCTIONS
RECALL: ADDITION AND SUBTRACTION
a) Find the least common denominator (LCD) of both fractions. To find the LCD, multiply the two denominators together and divide that number by the greatest common factor. For example, for ⅜ and 5/12, the greatest common factor is 4, and the two denominators multiplied are 96. 96 divided by four is 24, which is the LCD. b) Rewrite the fractions as equivalent fractions with the same LCD. c) The LCD is the denominator of the resulting fractions. d) The sum or difference of the numerators is the numerator of the resulting fractions.
Example 3.1. Find the sum of 3 4 and^
2 5 (fraction) Solution: The LCD of the two fractions is 20. 3 4
2 5
15 20
8 20
= 23 20 or 1 3 20
Example 3.2. Find the sum of 1 2𝑥− and 2 𝑥− (rational) Solution: The LCD of the two rational expressions is (2x − 3)(x − 5) or 2x^2 − 13x + 15. So, better to choose the LCD that may easy to use in performing functions. 1 2𝑥−
2 𝑥− = 𝑥− (2𝑥−3)(𝑥−5)
2(2𝑥−3) (2𝑥−3)(𝑥−5)
= 𝑥− (2𝑥−3)(𝑥−5)
4𝑥− (2𝑥−3)(𝑥−5)
= 5𝑥− (2𝑥−3)(𝑥−5) =^
5𝑥− 2𝑥^2 − 13x + 15
RECALL: MULTIPLICATION
a) Rewrite the numerator and denominator in terms of its prime factors. b) Common factors in the numerator and denominator can be simplified as (this is often called cancelling) c) Multiply the numerators together to get the new numerator. d) Multiply the denominators together to get the new denominator.
Example 3.3. Find the product of 12 24 and^
16 9.^ Use^ cancellation^ of^ factors^ when convenient. Solution: 12 24
16 9
= 8 9
Example 3.4. Find the product of 𝑥^2 −4𝑥− 𝑥^2 −3𝑥+2 and^
𝑥^2 −5𝑥+ 𝑥^2 −3𝑥−10. Solution: Express the numerators and denominators of the two rational expressions into their prime factors. Multiply and cancel common factors in the numerator and the denominator to reduce the final answer into lowest terms. Note the similarity in the process between this example and the previous.
RECALL: DIVISION
To divide two fractions or rational expressions, multiply the dividend with the reciprocal of the divisor.
Example 3.5. Find the quotient of x + 2 and x^2 − 4.
- Apply the rules.
- Multiply numerator to numerator and denominator to denominator.
- Sum and difference of x^2 − 4 is (x + 2)(x − 2), then apply cancellation.
SUM AND DIFFERENCE OF FUNCTIONS
If f and g are functions respectively, their sum is the function defined as (f + g)(x) = f(x) + g(x), and their difference is the function defined as (f − g)(x) = f(x) − g(x).
Example 3.6. Let f(x) = 2x^3 − 5x + 4 and g(x) = 3x^2 + 2x − 6. Find (f + g)(x) and (f - g)(x).
15 is from 20÷4 × 3 8 is from 20÷5 × 2
x-5 is from (2x-3)(x-5)÷2x-3 × 1 2(2x-3) is from(2x-3)(x-5)÷x-5 × 2
(2x-3)(x-5) 2x^2 -10x-3x+ 2x^2 -13x+
4 2 3 3
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LESSON 8: SOLVING RATIONAL INEQUALITIES
RECALL: INTERVAL AND SET NOTATION
An inequality may have infinitely many solutions. The set of all solutions can be expressed using set notation or interval notation. These notations are presented in the table below.
To solve Rational Inequalities : a.) Rewrite the inequality as a single fraction on one side of the inequality symbol and 0 on the other side. b.) Determine the x-values for which the rational expression is zero or undefined (factoring the numerator and denominator is a useful strategy). Plot on a number line and use shaded circle to indicate that the value is included in the solution set, and a hollow circle to indicated that the value is excluded. c.) Construct a table of signs to determine the sign of the function in each interval determined by the x-values. d.) Summarize the intervals satisfying the inequality. Plot the intervals containing the solutions on the number line.
Example 8.1. Solve the inequality 2𝑥 𝑥+1 ≥ 1
Rewrite the inequality as a single fraction with zero on one side and simplify. Determine the x-values. Plot on a number line. The value x = 1 is included (shaded) in the solution since it makes the fraction equal to zero, while x = − is not included (hallow) since it makes the fraction undefined. Construct a table of signs to determine the sign of the function in each interval determined by 1 and -1. Since we are looking for the intervals where the fraction is positive or zero, we determine the solution intervals to be x < −1 and x ≥ 1 (greater than or equal 1 since it will equal to zero and it satisfy the inequality). Summarize the intervals satisfying the inequality. Plot these intervals on the number line. The solution set is {x ∈ R| x < −1 or x ≥ 1. It can also be written using interval notation (−∞, −1) ∪ [1, ∞).
Example 8.2. Solve the inequality 3 𝑥−
1 𝑥
Example 8.3. Solve the inequality 𝑥− 4 −^
𝑥 2 > 1
If ang symbol ng inequality ay walang "EQUAL TO" meaning lahat sila Hollow Circle kahit nasa numerator pa yan. Magkakaroon lang ng SHADED CIRCLE yung may zeroes ng numerator (x-values) kapag MAYROONG "EQUAL TO". If the inequality symbol ay walang "equal to" meaning po yung x-values ng numerator is also not included sa solution, Kaya magiging hollow circle po lahat.
LESSON 9: SOLVING RATIONAL EQUATION AND
RATIONAL INEQUALITY INVOLVING WORD
PROBLEMS.
Example 9.1 In an inter-barangay basketball league, the team from Barangay San Juan has won out of 25 games, a winning percentage of 48%. How many games should they win in a row to improve their win percentage to 60%? Solution: Let x represent the number of games that they need to win to raise their percentage to 60%. The team has already won 12 out of 25 games. If they win x games in a row to increase their percentage to 60%, then they would have played 12 + x games out of 25 + x games. The equation is 12 + 𝑥 25 + 𝑥
The equation is a rational equation. Solve the equation using the techniques previously discussed. Since 25 + x is the only denominator, we multiply it to both sides of the equation. We then solve the resulting equation
Since x represents the number of games, this number should be an integer. Therefore, Barangay San Juan needs to win 8 games in a row to raise their winning average percentage to 60%.
LESSON 10: GRAPHING RATIONAL FUNCTION
In sketching the graph of a rational function, we will make use of the x and y- intercepts of the function. In addition, we will observe how the function behaves near the values where the function is undefined as well as in the extreme values of x.
Recall of important concepts: a. Domain – is the set of all values that the variable x can take. b. Range – is the set of all values that f(x) can take. c. Zeroes – are the values of x which make the function zero. The real- numbered zeroes are also x − intercepts of the graph of the function. (is the function value when y = 0) d. y − intercepts – is the function value when x = 0. e. Vertical Asymptotes – find the values of x where the denominator of the function reduced rational function is zero. f. Horizontal Asymptotes – Use the degree of the polynomial in the numerator and denominator. (As indicated below in the procedure in sketching graph of rational function).
DRASIG METHOD Sketching the graph of rational function using DRASIG Method (Domain, Range, Asymptote, Symmetry, Intercept, and Graph).
DOMAIN Observe where the function is undefined and is not part of the domain (restriction form the domain). X+1 = {𝑥|𝑥 ≠ −1}
RANGE Observe the horizontal asymptote whether it is part of the solution or not (if the line will go through the HA). You may see it after you sketch the graph of rational function. (tingnan yung H.asymptote if natatamaan all real numbers, if hindi, ayun yung exception ≠)
ASYMPTOTE We have two kinds of asymptote. a. Vertical Asymptote (restriction from the domain) b. Horizontal Asymptote: if n < m, then HA is y = 0. if n = m, then HA is y = 𝑎 𝑏 , where a is the leading coefficient of the numerator and b is the leading coefficient of the denominator. Ex. 1/ if n > m, then there is no HA.
SYMMETRY a. Restriction forms the domain. (Where the function undefined) b. Zeroes function. (The values of x which make the function zero) sa numerator ex. X+1 so it is = -
Line of symmetry will determine the behavior of the graph whether it is above or below the x − axis. Construction of table of signs of values of x will be used to determine the intervals as boundaries of the graph.
INTERCEPT a. x − intercept (from zeroes function and where the value of y = 0) (2,0)
- ex. numerator x-2 so ang x intercept ay (2,0) b. y − intercept (in terms of f(x), and where all the values of x = 0) (0,-1)
- papalitan mo lang lahat ng x sa given ng 0, then solve.
GRAPH a. Draw a Cartesian plane and plot the Vertical and Horizontal Asymptote in a broken line (VA and HA). b. Plot the x − intercept and y − intercept. c. Follow the line of symmetry where the behavior of graphs will pass through using the intervals as boundaries in the table of sign.
Note: If the sign is positive (+) means it is above the x − axis but if the sign is negative (−) means it is below the x – axis.
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Example 10.1. Sketch the graph of f(x) = 𝑥− 𝑥+
Domain: {x ∈ R|x ≠ −2} (restriction from the domain) Range: {y ∈ R|y ≠ 1} (since horizontal asymptote also not part of the solution) Asymptote: VA: −2, and HA: 1 (if n = m, then HA is y = 𝑎 𝑏
Symmetry: a. −2 (restriction from the domain); b. 2 (zeroes function in the numerator x − 2 = 0)
Intercept : x-intercept: (2,0) zeroes function y − intercept: (0, −1) in terms of f(x) and when all x is
Graph:
Example 10.2. Sketch the graph of f(x) = 3𝑥^2 −8𝑥− 2𝑥^2 +7𝑥−
CHAPTER 3: ONE-TO-ONE AND INVERSE
FUNCTION - HANDOUTS
LESSON 11: ONE-TO-ONE FUNCTION
The function f is one-to-one if for any x 1 , x 2 in the domain of f, then f(x 1 ) ≠ f(x 2 ). That is, the same y − value is never paired with two different x – values. There are no two distinct inputs that correspond to the same output. If the highest degree of your function is odd number then it is a one to one function
LESSON 1.1: REPRESENTATION OF REAL-LIFE
SITUATIONS USING ONE-TO-ONE FUNCTION
a. The relation pairing an SSS member to his or her SSS number. / b. The relation pairing a real number to its square. Not one to one 4=-2^2 /2^2 c. The relation pairing an airport to its airport code. Airport code are three letter codes used to uniquely identify airports around the world and prominently displayed on checked-in bags to denote the destination of these bags. / d. The relation pairing a person to his or her citizenship. Relation is not a function because a person may have a two citizenship meaning there’s a two output in one input. e. The relation pairing a distance travelled along a given jeepney route to the jeepney fare travelling those distances. Not one to one
LESSON 11.2. GRAPHS OF A ONE-TO-ONE
FUNCTION
HORIZONTAL LINE TEST
A function is one-to-one if each horizontal line does not intersect the graph at more than one point. All functions satisfy the Vertical Line Test. All one-to-one functions satisfy both Vertical Line Test and Horizontal Line Test.
LESSON 12: INVERSE OF ONE-TO-ONE
FUNCTION
A relation reversing the process performed by any function f(x) is called inverse of f(x). This means that the domain of the inverse is the range of the original function and that the range of the inverse is the domain of the original function.
Consider the table of values for the function given by the equation: y = 2x − 1 given below x - 4 - 3 - 2 - 1 0 1 2 3 4 y - 9 - 7 - 5 - 3 - 1 1 3 5 7
Verify that it is one-to-one function by showing that no two y-values share the same x − value. Let us invert the values of x and y. x - 9 - 7 - 5 - 3 - 1 1 3 5 7 y - 4 - 3 - 2 - 1 0 1 2 3 4 Still represent a function and a one-to-one function
INVERTING FUNCTIONS:
a. If the x – and y – values of a one-to-one function are interchanged, the result is a function, but b. If the x – and y – values of a function that is not one-to-one are inverted the results is no longer a function.
DEFINITION 12.1. INVERSE ONE-TO-ONE FUNCTION
Let f be a one-to-one function with domain A and range B. Then the inverse of f, denoted f-1, is a function with domain B and range A defined by f-1(y) = x and if and only if f(x) = y for any y in B. A function has an inverse if and only if it’s one-to-one. Inverting the x- and y- values of a function results in a function if and only if the original function is one-to-one.
TO FIND THE INVERSE OF ONE-TO-ONE FUNCTION:
a. Write the function in the form y = f(x). b. Interchange the x and y variables. c. Solve for y in terms of x.
Example 12.1: Find the inverse of f(x) = 3x + 1. Solution: The equation of the function is y = 3x + 1. Interchange the x and y variables: x = 3y + 1. Solve for y in terms of x.
Therefore, the inverse of f(x) = 3x + 1 is 𝒇−𝟏^ (𝒙) = 𝒙−𝟏 𝟑
Example 12.2 : Find the inverse of g(x) = x^3 − 2.
Example 12.3 : Find the inverse of the rational function f(x) = 2𝑥+ 3𝑥−
Example 12.4. Find the inverse of f(x) = x^2 + 4x – 2, if it exists. This is quadratic function with a graph in the shape of a parabola that opens upwards. It is not a one-to-one function as it fails the horizontal line test.
Example 12.5. Find the inverse of f(x) = |3x|. The graph of f(x) = |3x|, is shaped like a “V” whose vertex is located at the origin. This function fails the horizontal line test and therefore has no inverse
