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This document provides an overview of functions and their key concepts. It defines relations, domains, ranges, and functions. It discusses different types of functions including constant, linear, quadratic, cubic, and others. It also covers evaluating functions, performing operations on functions, and piecewise functions. The document is intended to help understand functions and how they can represent real-life situations. It provides examples of evaluating functions at different inputs, adding and subtracting functions, and finding function values for piecewise functions.

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GENERAL MATHEMATICS ANGELO S. REYES SAMUEL CHRISTIAN COLLEGE
UNDERSTANDING FUNCTIONS
UNDERSTANDING FUNCTIONS •REVISITING FUNCTIONS •EVALUATING FUNCTIONS •PERFORMING OPERATIONS ON FUNCTIONS •COMPOSITION OF FUNCTIONS •SOLVING PROBLEMS INVOLVING FUNCTIONS
Why functions are important in representing real-life situations? Essential Questions:
REVISITING FUNCTIONS
AT THE END OF THE LESSON •I can represent real-life situations using functions, including piecewise functions. •I can differentiate functions from mere relations •I can be representing rational functions through its (a) table of values (b) graph (c) equations •I can find the domain and range of rational functions; and •I can describe and graph different types of functions.
DEFINITION: RELATIONS ∥ DOMAIN ∥ RANGE ∥ FUNCTION
WHAT IS RELATION, DOMAIN, RANGE? •A 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 is a set of ordered pairs. •The 𝑑𝑜𝑚𝑎𝑖𝑛 of a relation is the set of the first coordinates. •The 𝑟𝑎𝑛𝑔𝑒 is the set of the second coordinates. • 𝑅𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠 are often defined by equations with no domain stated. •If the 𝑑𝑜𝑚𝑎𝑖𝑛 is not stated, we agree that the domain consists of all real numbers that, when used to substitute the independent variable, produce real numbers for the dependent variable.
EXAMPLE:
EXAMPLE:
WHAT IS FUNCTION? •A 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 is a relation that assigns to each member of the domain exactly one member of the range. •It is a 𝑠𝑒𝑡 𝑜𝑓 𝑜𝑟𝑑𝑒𝑟𝑒𝑑 𝑝𝑎𝑖𝑟𝑠 (𝑥, 𝑦) in which no two distinct ordered pairs have the same first member. •The set of all permissible 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.
WHAT IS FUNCTION? •We can also define a function as a rule that assigns to each element x in set A exactly one element, called 𝒇(𝒙), in set B. •We usually represent a function by the equation 𝒚 = 𝒇(𝒙) where 𝒚 is the dependent variable and 𝒙 is the independent variable. •Take note: that for every input (domain), there corresponds a unique output (range).
WAYS ON HOW TO DETERMINE IF A RELATION IS A FUNCTION OR NOT
FUNCTIONS DEFINED BY GRAPH VERTICAL LINE TEST
VERTICAL LINE TEST •The graph of a function can be intersected by a vertical line in at most one point, since in a function; there is a unique value of to each value of in the domain of the function. •But if the vertical line intersects the graph at more than one point, the graph is not that of a function (it is only a relation). •We usually represent a function by the equation
Graph the relation 𝒇 𝒙 = 𝒙𝟐 + 𝟐𝒙 − 𝟑 Using Vertical Line Test 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕? Answer: 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏
Graph the relation 𝒙𝟐 + 𝒚𝟐 = 𝟏𝟔 Using Vertical Line Test 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕? Answer: 𝑵𝒐𝒕
Using Vertical Line Test 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕? Answer: 𝑵𝒐𝒕
TYPES OF FUNCTION
CONSTANT FUNCTION LINEAR FUNCTION QUADRATIC FUNCTION
CUBIC FUNCTION ABSOLUTE FUNCTION SQUARE ROOT FUNCTION
EXPONENTIAL FUNCTION RATIONAL FUNCTION LOGARITHMIC FUNCTION
INVERSE FUNCTION FLOORING FUNCTION CEILING FUNCTION
SIGNUM FUNCTION OSCILLATING FUNCTION
PIECEWISE FUNCTION
PIECEWISE FUNCTION •Is a function defined by two or more formulas on different parts of its domain.
Example A merchant sells rice at 45 pesos a kilo. If, however, a buyer buys more than 150 kilos of rice. The merchant gives a 10% discount. •If x represents the number of kilos of rice purchased, express the amount to be paid by f(x). •How much should be paid for 100 kilos of rice? •How much should be paid for 200 kilos of rice?
Step 1: Express the situation, using function If the number of kilos of purchased rice exceeds 150. That’s 𝑥 > 150. Then the number of kilos in excess of 150 kilos is given a 10% discount If the number of kilos of purchased rice does not exceed 150. That is 0 ≤ 𝑥 ≤ 150 then 𝑓 𝑥 = 45𝑥
Step 1: Express the situation, using function 𝑓 𝑥 = 45𝑥 𝑖𝑓 0 ≤ 𝑥 ≤ 150 40.50 + 675𝑥 𝑖𝑓 𝑥 > 150 So, the function that describes the situation is: Amount of the first 150 kilos of rice 45 150 = 6 750 Amount in excess of 150 kilos 𝟒𝟓 𝟎. 𝟗𝟎 𝒙 − 𝟏𝟓𝟎 = 𝟒𝟎. 𝟓𝟎(𝒙 − 𝟏𝟓𝟎) Total Amount 𝟔 𝟕𝟓𝟎 + 𝟒𝟎. 𝟓𝟎 𝒙 − 𝟏𝟓𝟎 = 𝟒𝟎. 𝟓𝟎𝒙 + 𝟔𝟕𝟓
Step 2: Find the amount to be paid for 100 kilos of rice. 𝑓 𝑥 = 45𝑥 𝑓 100 = 45 100 = 4 500 Find f(100) by substituting x = 100 in the first piece or chunk of the function because 100 < 150. So, the amount to be paid for 100 kilos of rice is Php 4,500
Step 2: Find the amount to be paid for 200 kilos of rice. 𝑓 𝑥 = 40.50𝑥 + 675 𝑓 200 = 40.50 200 + 675 𝑓 200 = 8 775 Find f(200) by substituting x=200 in the second piece or chunk of the function because 200>150. So, the amount to be paid for 200 kilos of rice is Php 8 775.
GRAPHING PIECEWISE FUNCTION
Graph: 𝒇 𝒙 = 𝑥 − 2 , 𝑥 ≤ −2 3𝑥2 − 5 , 𝑥 > −2 The function is described by two intervals. Take note that when x=-2, f(x)=x-2 is defined. That is why the point (-2,-4) is shaded.
Graph: 𝒇 𝒙 = 𝑥 − 2 , 𝑥 ≤ −2 3𝑥2 − 5 , 𝑥 > −2 In 𝑓 𝑥 = 3𝑥2 − 5 when x=-2 the point (-2,7) is not shaded because it is not defined. Clearly, by vertical line test, the piecewise is indeed a function.
EVALUATING FUNCTION
Example: If 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the following: 𝑓 5 𝑎𝑛𝑑 𝑓 0 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 5 = 4 5 2 − 5 5 + 3 𝑓 5 = 100 − 25 + 3 𝑓 5 = 78 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 0 = 4 0 2 − 5 0 + 3 𝑓 0 = 0 − 0 + 3 𝑓 0 = 3
Example: If 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the following: 𝑓 1 4 𝑎𝑛𝑑 𝑓 2 2 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 1 4 = 4 1 4 2 − 5 1 4 + 3 𝑓 1 4 = 1 4 − 5 4 + 3 𝑓 1 4 = 2 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 2 2 = 4 2 2 2 − 5 2 2 + 3 𝑓 2 2 = 4 1 2 − 5 2 4 + 3 𝑓 2 2 = − 5 2 4 + 5
Example: If 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the following: 𝑓(𝑥 + 1) 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 𝑥 + 1 = 4 𝑥 + 1 2 − 5 𝑥 + 1 + 3 𝑓 𝑥 + 1 = 4 𝑥2 + 2𝑥 + 1 − 5𝑥 − 5 + 3 𝑓 𝑥 + 1 = 4𝑥2 + 8𝑥 + 4 − 5 − 5𝑥 − 2 𝑓 𝑥 + 1 = 4𝑥2 + 3𝑥 − 3
Example: If 𝑓 𝑥 = 3𝑥 + 2 evaluate the following: 𝑓 𝑥+ℎ −𝑓(𝑥) ℎ 𝑓 𝑥 = 3𝑥 + 2 𝑓 𝑥 + ℎ = 3 𝑥 + ℎ + 2 𝑓 𝑥 + ℎ = 3𝑥 + 3ℎ + 2 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 + 3ℎ + 2 − 3𝑥 + 2 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 + 3ℎ + 2 − 3𝑥 − 2 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 − 3𝑥 + 3ℎ + 2 − 2 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3ℎ 𝑓 𝑥 + ℎ − 𝑓 𝑥 ℎ = 3ℎ ℎ = 3
EVALUATING PIECEWISE FUNCTION
Example: Given the piecewise below Find the following: ℎ −10 , ℎ 1 2 , ℎ(6)
Find: ℎ(−10) Step 1: Determine the piece that should be used. Since -10 belongs to the interval x < -5. therefore, we shall use the third piece of function Step 2: Find h(-10) using the 3rd piece of the function ℎ 𝑥 = 2 𝑥 + 4 2 + 10 ℎ −10 = 2 −10 + 4 2 + 10 ℎ −10 = 2 36 + 10 ℎ −10 = 82
Find: ℎ 1 2 Step 1: Determine the piece that should be used. Since 1/2 belongs to the interval −5 ≤ 𝑥 < 6. therefore, we shall use the second piece of function Step 2: Find h(1/2) using the 2nd piece of the function ℎ 𝑥 = 3 2𝑥 ℎ 1 2 = 3 2 1 2 ℎ 1 2 = 3
Find: ℎ(6) Step 1: Determine the piece that should be used. Since x = 6 belongs to the interval 𝑥 ≥ 6. therefore, we shall use the first piece of function Step 2: Find h(6) using the 1st piece of the function ℎ 𝑥 = 𝑥3 + 2𝑥 ℎ 6 = 6 3 + 2(6) ℎ 6 = 216 + 12 ℎ 6 = 228
PERFORMING OPERATIONS ON FUNCTION
SUM, DIFFERENCE, PRODUCT, AND QUOTIENT OF FUNCTIONS Given two functions 𝑓(𝑥) and 𝑔(𝑥), we define the following. •Sum: 𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔(𝑥) •Difference: 𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔(𝑥) •Product: 𝑓 ∗ 𝑔 𝑥 = 𝑓 𝑥 ∗ 𝑔(𝑥) •Quotient: 𝑓 𝑔 𝑥 = 𝑓 𝑥 𝑔 𝑥 , 𝑔(𝑥) ≠ 0
Example: If 𝑓 𝑥 = 6𝑥2 − 12𝑥 − 19 and 𝑔 𝑥 = 𝑥3 − 𝑥2 + 8𝑥 − 1 find the following: 𝑓 + 𝑔 𝑥 𝑎𝑛𝑑 (𝑓 − 𝑔)(𝑥)
Find: (𝑓 + 𝑔)(𝑥) Step 1: Apply the definition for adding functions. 𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔 𝑥 𝑓 + 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 + (𝑥3 − 𝑥2 + 8𝑥 − 1) Step 2: Group similar terms and make sure that the terms are arranged in descending order: 𝑓 + 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 + 𝑥3 − 𝑥2 + 8𝑥 − 1 𝑓 + 𝑔 𝑥 = 𝑥3 + 6𝑥2 − 𝑥2 + −12𝑥 + 8𝑥 + (−19 − 1)
Find: (𝑓 + 𝑔)(𝑥) Step 3: Combined similar terms 𝑓 + 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 + 𝑥3 − 𝑥2 + 8𝑥 − 1 𝑓 + 𝑔 𝑥 = 𝑥3 + 6𝑥2 − 𝑥2 + −12𝑥 + 8𝑥 + −19 − 1 𝑓 + 𝑔 𝑥 = 𝒙𝟑 + 𝟓𝒙𝟐 − 𝟒𝒙 − 𝟐𝟎
Find: (𝑓 − 𝑔)(𝑥) Step 1: Apply the definition for subtracting functions. 𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔 𝑥 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − (𝑥3 − 𝑥2 + 8𝑥 − 1) Step 2: Group similar terms and make sure that the terms are arranged in descending order: 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − 𝑥3 − 𝑥2 + 8𝑥 − 1 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − 𝑥3 + 𝑥2 − 8𝑥 + 1 𝑓 − 𝑔 𝑥 = −𝑥3 + 6𝑥2 + 𝑥2 + −12𝑥 − 8𝑥 + (−19 + 1)
Find: (𝑓 + 𝑔)(𝑥) Step 3: Combined similar terms 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − 𝑥3 − 𝑥2 + 8𝑥 − 1 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − 𝑥3 + 𝑥2 − 8𝑥 + 1 𝑓 − 𝑔 𝑥 = −𝑥3 + 6𝑥2 + 𝑥2 + −12𝑥 − 8𝑥 + −19 + 1 𝑓 − 𝑔 𝑥 = −𝒙𝟑 + 𝟕𝒙𝟐 − 𝟐𝟎𝒙 − 𝟏𝟖
Example: If 𝑓 𝑥 = 3 𝑥2−1 and 𝑔 𝑥 = 𝑥 𝑥−1 find 𝑓 + 𝑔 𝑥 .
Find: (𝑓 + 𝑔)(𝑥) Step 1: Apply the definition for adding functions. 𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔 𝑥 𝑓 + 𝑔 𝑥 = 3 𝑥2 − 1 + 𝑥 𝑥 − 1 Step 2: Find the LCD of 𝑥2 − 1 𝑎𝑛𝑑 𝑥 − 1, the 𝐿𝐶𝐷 𝑖𝑠 𝑥 + 1 𝑥 − 1 𝑓 + 𝑔 𝑥 = 3 𝑥2 − 1 + 𝑥 𝑥 − 1 𝑓 + 𝑔 𝑥 = 3 + 𝑥(𝑥 + 1) (𝑥 + 1)(𝑥 − 1)
Find: (𝑓 + 𝑔)(𝑥) Step 3: Simplify the expression: 𝑓 + 𝑔 𝑥 = 3 𝑥2 − 1 + 𝑥 𝑥 − 1 𝑓 + 𝑔 𝑥 = 3 + 𝑥(𝑥 + 1) (𝑥 + 1)(𝑥 − 1) 𝑓 + 𝑔 𝑥 = 𝒙𝟐 + 𝒙 + 𝟑 (𝒙 + 𝟏)(𝒙 − 𝟏)
Example: If 𝑓 𝑥 = 𝑥−3 𝑥+4 and 𝑔 𝑥 = 𝑥+1 𝑥+4 find 𝑓 − 𝑔 𝑥 .
Find: (𝑓 − 𝑔)(𝑥) Step 1: Apply the definition for subtracting functions. 𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔 𝑥 𝑓 − 𝑔 𝑥 = 𝑥 − 3 𝑥 + 4 − 𝑥 + 1 𝑥 + 4 Step 2: Simplify 𝑓 − 𝑔 𝑥 = 𝑥 − 3 𝑥 + 4 − 𝑥 + 1 𝑥 + 4 𝑓 − 𝑔 𝑥 = 𝑥 − 3 − (𝑥 + 1) 𝑥 + 4 = 𝑥 − 3 − 𝑥 − 1 𝑥 + 4 = −𝟒 𝒙 + 𝟒
Example: If 𝑓 𝑥 = 𝑥3 − 27, 𝑔 𝑥 = 𝑥2 + 3𝑥 + 9 and ℎ 𝑥 = 4𝑥 − 1. . Find the following: (𝑓 ∗ ℎ)(𝑥), 𝑓 𝑔 (𝑥), 𝑓 ℎ (𝑥)
Find: (𝑓 ∗ ℎ)(𝑥) Step 1: Apply the definition for multiplying functions. 𝑓 ∗ ℎ 𝑥 = 𝑓 𝑥 ∗ ℎ 𝑥 𝑓 ∗ ℎ 𝑥 = 𝑥3 − 27 (4𝑥 − 1) Step 2: Find the product of polynomials using the distributive law. 𝑓 ∗ ℎ 𝑥 = 𝑥3 − 27 4𝑥 − 1 𝑓 ∗ ℎ 𝑥 = 𝟒𝒙𝟒 − 𝒙𝟑 − 𝟏𝟎𝟖𝒙 + 𝟐𝟕
Find: 𝑓 𝑔 (𝑥) Step 1: Apply the definition for dividing functions. 𝑓 𝑔 𝑥 = 𝑓 𝑥 𝑔 𝑥 Step 2: Factor the numerator 𝑓 𝑔 𝑥 = 𝑥3 − 27 𝑥2 + 3𝑥 + 9 𝑓 𝑔 𝑥 = 𝑥3 − 27 𝑥2 + 3𝑥 + 9 𝑓 𝑔 𝑥 = (𝑥 − 3)(𝑥2 + 3𝑥 + 9) 𝑥2 + 3𝑥 + 9 Step 3: Cancel the common factors and simplify 𝑓 𝑔 𝑥 = (𝑥 − 3)(𝑥2 + 3𝑥 + 9) 𝑥2 + 3𝑥 + 9 𝑓 𝑔 𝑥 = 𝒙 − 𝟑
Find: 𝑓 ℎ (𝑥) Step 1: Apply the definition for dividing functions. 𝑓 ℎ 𝑥 = 𝑓 𝑥 𝑔 𝑥 Step 2: Since there are no common factors in the numerator and the denominator, this is already the simplest form 𝑓 ℎ 𝑥 = 4𝑥 − 1 𝑥3 − 27 𝑓 ℎ 𝑥 = 𝟒𝒙 − 𝟏 𝒙𝟑 − 𝟐𝟕
Example: If 𝑓 𝑥 = 𝑥2−3𝑥−28 𝑥2−8𝑥+16 , 𝑔 𝑥 = 𝑥2−5𝑥+4 𝑥2−49 . Find the following: 𝑓 ∗ 𝑔 (𝑥) and 𝑓 𝑔 (𝑥)
Find: (𝑓 ∗ 𝑔)(𝑥) Step 1: Apply the definition for multiplying functions. 𝑓 ∗ 𝑔 𝑥 = 𝑓 𝑥 ∗ 𝑔 𝑥 𝑓 ∗ 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 5𝑥 + 4 𝑥2 − 49 Step 2: Factor the numerator and denominator of both rational function. 𝑓 ∗ 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 5𝑥 + 4 𝑥2 − 49 𝑓 ∗ 𝑔 𝑥 = (𝑥 − 7)(𝑥 + 4) (𝑥 − 4)(𝑥 − 4) ∗ (𝑥 − 1)(𝑥 − 4) (𝑥 − 7)(𝑥 + 7)
Find: (𝑓 ∗ 𝑔)(𝑥) Step 3: Cancel common factors and simplify. 𝑓 ∗ 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 5𝑥 + 4 𝑥2 − 49 𝑓 ∗ 𝑔 𝑥 = (𝑥 − 7)(𝑥 + 4) (𝑥 − 4)(𝑥 − 4) ∗ (𝑥 − 1)(𝑥 − 4) (𝑥 − 7)(𝑥 + 7) 𝑓 ∗ 𝑔 𝑥 = (𝒙 + 𝟒)(𝒙 − 𝟏) (𝒙 − 𝟒)(𝒙 + 𝟕)
Find: 𝑓 𝑔 (𝑥) Step 1: Apply the definition for dividing functions. 𝑓 𝑔 𝑥 = 𝑓 𝑥 𝑔 𝑥 → Step 2: Apply the rule in dividing rational expressions. 𝑓 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 49 𝑥2 − 5𝑥 + 4 𝑓 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ÷ 𝑥2 − 5𝑥 + 4 𝑥2 − 49 𝑓 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ÷ 𝑥2 − 5𝑥 + 4 𝑥2 − 49
Find: (𝑓 ∗ 𝑔)(𝑥) Step 3: Factor the numerator and denominator of both rational functions. 𝑓 𝑔 𝑥 = (𝑥 − 7)(𝑥 + 4) (𝑥 − 4)(𝑥 − 4) ∗ (𝑥 − 7)(𝑥 + 7) (𝑥 − 4)(𝑥 − 1) 𝑓 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 49 𝑥2 − 5𝑥 + 4
Find: (𝑓 ∗ 𝑔)(𝑥) Step 4: There are no common factors to cancel. So we just simplify 𝑓 𝑔 𝑥 = (𝑥 − 7)(𝑥 + 4) (𝑥 − 4)(𝑥 − 4) ∗ (𝑥 − 7)(𝑥 + 7) (𝑥 − 4)(𝑥 − 1) 𝑓 𝑔 𝑥 = 𝒙 − 𝟕 𝟐 (𝒙 + 𝟕)(𝒙 + 𝟒) 𝒙 − 𝟒 𝟑(𝒙 − 𝟏)
COMPOSITION OF FUNCTION
Definition of Composition of Function and the domain of (𝑓 ∘ 𝑔)(𝑥) is the set of all numbers x in the domain of 𝑔(𝑥), such that 𝑔(𝑥) is the domain of 𝑓(𝑥). Given two function 𝑓(𝑥) and 𝑔(𝑥), the composition of function is denoted by (𝑓 ∘ 𝑔)(𝑥) is defined by: 𝒇 ∘ 𝒈 𝒙 = 𝒇(𝒈(𝒙)) Similarly, we define (𝑔 ∘ 𝑓)(𝑥) as follows: 𝒈 ∘ 𝒇 𝒙 = 𝒈(𝒇(𝒙))
Example: If 𝑓 𝑥 = 𝑥2 − 6 and 𝑔 𝑥 = 3𝑥 + 2. Find the following: (𝑓 ∘ 𝑔)(𝑥), (𝑔 ∘ 𝑓)(𝑥), (𝑔 ∘ 𝑔)(𝑥), (𝑓 ∘ 𝑓)(𝑥)
Find:(𝑓 ∘ 𝑔)(𝑥) Step 1: Apply the definition of composition of functions 𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥 → (𝑓 ∘ 𝑔)(𝑥) = 𝑓(3𝑥 + 2) Step 2: Apply the definition of 𝑓 𝑓 ∘ 𝑔 𝑥 = 𝑓 3𝑥 + 2 → 𝑓 ∘ 𝑔 𝑥 = 3𝑥 + 2 2 − 6 Step 3: Simplify the results. 𝑓 ∘ 𝑔 𝑥 = 3𝑥 + 2 2 − 6 𝑓 ∘ 𝑔 𝑥 = 9𝑥2 + 12𝑥 + 4 − 6 𝑓 ∘ 𝑔 𝑥 = 𝟗𝒙𝟐 + 𝟏𝟐𝒙 − 𝟐
Find:(𝑔 ∘ 𝑓)(𝑥) Step 1: Apply the definition of composition of functions 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 → (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑥2 − 6) Step 2: Apply the definition of 𝑓 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑥2 − 6 → 𝑔 ∘ 𝑓 𝑥 = 3 𝑥2 − 6 + 2 Step 3: Simplify the results. 𝑔 ∘ 𝑓 𝑥 = 3𝑥2 − 18 + 2 𝑔 ∘ 𝑓 𝑥 = 𝟑𝒙𝟐 − 𝟏𝟔 𝑔 ∘ 𝑓 𝑥 = 3 𝑥2 − 6 + 2
Find:(𝑔 ∘ 𝑔)(𝑥) Step 1: Apply the definition of composition of functions 𝑔 ∘ 𝑔 𝑥 = 𝑔 𝑔 𝑥 → (𝑔 ∘ 𝑔)(𝑥) = 𝑔(3𝑥 + 2) Step 2: Apply the definition of 𝑓 𝑔 ∘ 𝑔 𝑥 = 𝑔 3𝑥 + 2 → 𝑔 ∘ 𝑔 𝑥 = 3 3𝑥 + 2 + 2 Step 3: Simplify the results. 𝑔 ∘ 𝑔 𝑥 = 9𝑥 + 6 + 2 𝑔 ∘ 𝑔 𝑥 = 𝟑𝒙𝟐 + 𝟖 𝑔 ∘ 𝑔 𝑥 = 3 3𝑥 + 2 + 2
Find:(𝑓 ∘ 𝑓)(𝑥) Step 1: Apply the definition of composition of functions 𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑓 𝑥 → (𝑓 ∘ 𝑓)(𝑥) = 𝑓(𝑥2 − 6) Step 2: Apply the definition of 𝑓 𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑥2 − 6 → 𝑓 ∘ 𝑓 𝑥 = 𝑥2 − 6 2 − 6 Step 3: Simplify the results. 𝑔 ∘ 𝑓 𝑥 = 𝑥4 − 12𝑥2 + 36 − 6 𝑔 ∘ 𝑓 𝑥 = 𝒙𝟒 − 𝟏𝟐𝒙𝟐 + 𝟑𝟎 𝑓 ∘ 𝑓 𝑥 = 𝑥2 − 6 2 − 6
Example: Given the following functions 𝑓 𝑥 = 𝑥3, 𝑔 𝑥 = 𝑥+2 3𝑥−1 ,and ℎ 𝑥 = 6𝑥 − 7 Find the following: 𝑓 ∘ 𝑓 𝑥 𝑎𝑛𝑑 (𝑓 ∘ 𝑓)(2) , 𝑔 ∘ ℎ 𝑥 𝑎𝑛𝑑 (𝑔 ∘ ℎ)(−1)
Find:(𝑓 ∘ 𝑓)(𝑥) Step 1: Apply the definition of composition of functions 𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑓 𝑥 → (𝑓 ∘ 𝑓)(𝑥) = 𝑓(𝑥3 ) Step 2: Apply the definition of 𝑓 𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑥3 → 𝑓 ∘ 𝑓 𝑥 = 𝑥3 3 Step 3: Simplify the results. 𝑔 ∘ 𝑓 𝑥 = 𝒙𝟗 𝑓 ∘ 𝑓 𝑥 = 𝑥3 3 𝑔 ∘ 𝑓 2 = (𝟐)𝟗 = 𝟓𝟏𝟐
Find:(𝑔 ∘ ℎ)(𝑥) Step 1: Apply the definition of composition of functions 𝑔 ∘ ℎ 𝑥 = 𝑔 ℎ 𝑥 → (𝑔 ∘ ℎ)(𝑥) = 𝑔 6𝑥 + 7 Step 2: Apply the definition of 𝑓 𝑔 ∘ ℎ 𝑥 = 𝑔 6𝑥 + 7 𝑔 ∘ ℎ 𝑥 = 6𝑥 − 7 + 2 3 6𝑥 − 7 − 1 Step 3: Simplify the results. 𝑔 ∘ ℎ 𝑥 = 6𝑥 − 7 + 2 3 6𝑥 − 7 − 1 𝑔 ∘ ℎ 𝑥 = 𝟔𝒙 − 𝟓 𝟏𝟖𝒙 − 𝟐𝟐 𝑔 ∘ ℎ −1 = 6(−1) − 5 18(−1) − 22 = − 𝟏𝟏 𝟒
SOLVING PROBLEMS INVOLVING FUNCTION
The following steps will be helpful in solving problems involving functions. • Read and comprehend the problem carefully. • Identify the function to be used (linear, quadratic, cubic, etc.) in representing the given problem. • Write the function that describes the problem. • Use the function to solve the problem. • Use the function to solve the problem. • Check the answer.
EXAMPLE: A Science major student earned 6 units after enrolling for a semester. Then, he took an academic leave for one year. After his leave, he enrolled again, ever more desirous this time to finish all the 150 units that he needs. In how many semesters does he need to enroll if he takes 9 units of Science subjects every semester?
Step:1 Represent the given and unknown quantities in the problem. Let: 𝑦 = total number of units to be taken in x semesters. 𝑥 = number of semesters. Step:2 Determine the function or equation that relates the quantities or variables in the problem. The problem can be described best by a linear function which relates the number of semesters (x) to the total number of units (y). The function that describes the relationship between x and y is: 𝑦 = 9𝑥 + 16
Step:3 Solve the problem. We want to know the number of semesters that the student needs to take in order to finish all 150 units. 𝑦 = 150 The student needs 16 semesters to complete the program. 150 = 9𝑥 + 6 150 − 6 = 9𝑥 144 = 9𝑥 𝑥 = 16
EXAMPLE: The area of a rectangle is 420 𝑐𝑚2. The length of the rectangle is one less than thrice its width. What are the dimensions of the given rectangle?
Step:1 Represent the given and unknown quantities in the problem. Express the width and the length of the rectangle in terms of one variable. Let: 𝑦 = area of the rectangle 𝑥 = width 3𝑥 − 1 = Length Step:2 Use function to relate the quantities or variables in the problem. The area of the rectangle is given by: 𝐴 = 𝑙𝑤; where 𝑙 = length and 𝑤 = width Using the representation in Step 1, we have: 𝑦 = 𝑥 3𝑥 − 1 𝑦 = 3𝑥2 − 𝑥 The function that describes the relationship is 𝑦 = 3𝑥2 − 𝑥
Step:3 Solve the problem. The area of rectangle is 𝑦 = 420 Substitute the value in 𝑦 = 3𝑥2 − 𝑥 3𝑥2 − 𝑥 − 420 = 0 Find the length and the width by substituting the value of x in the representation found in Step 1. Reject the negative value of x because the measurement of width cannot be negative. 3𝑥 + 35 𝑥 − 12 = 0 420 = 3𝑥2 − 𝑥 3𝑥 + 35 = 0 If 𝑥 = 12, then: width = 𝑥 = 12 Length = 3𝑥 − 1 = 3 12 − 1 = 35 Thus the dimensions of the given rectangle is 12 and 35. 3𝑥 = −35 𝒙 = − 𝟑𝟓 𝟑 𝑥 − 12 = 0 𝒙 = 𝟏𝟐 Step:4

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Gen Math topic 1.pptx

  • 1.
    GENERAL MATHEMATICS ANGELO S.REYES SAMUEL CHRISTIAN COLLEGE
  • 2.
  • 3.
    UNDERSTANDING FUNCTIONS •REVISITING FUNCTIONS •EVALUATINGFUNCTIONS •PERFORMING OPERATIONS ON FUNCTIONS •COMPOSITION OF FUNCTIONS •SOLVING PROBLEMS INVOLVING FUNCTIONS
  • 7.
    Why functions areimportant in representing real-life situations? Essential Questions:
  • 8.
  • 9.
    AT THE ENDOF THE LESSON •I can represent real-life situations using functions, including piecewise functions. •I can differentiate functions from mere relations •I can be representing rational functions through its (a) table of values (b) graph (c) equations •I can find the domain and range of rational functions; and •I can describe and graph different types of functions.
  • 10.
    DEFINITION: RELATIONS ∥ DOMAIN∥ RANGE ∥ FUNCTION
  • 11.
    WHAT IS RELATION,DOMAIN, RANGE? •A 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 is a set of ordered pairs. •The 𝑑𝑜𝑚𝑎𝑖𝑛 of a relation is the set of the first coordinates. •The 𝑟𝑎𝑛𝑔𝑒 is the set of the second coordinates. • 𝑅𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠 are often defined by equations with no domain stated. •If the 𝑑𝑜𝑚𝑎𝑖𝑛 is not stated, we agree that the domain consists of all real numbers that, when used to substitute the independent variable, produce real numbers for the dependent variable.
  • 12.
  • 13.
  • 14.
    WHAT IS FUNCTION? •A𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 is a relation that assigns to each member of the domain exactly one member of the range. •It is a 𝑠𝑒𝑡 𝑜𝑓 𝑜𝑟𝑑𝑒𝑟𝑒𝑑 𝑝𝑎𝑖𝑟𝑠 (𝑥, 𝑦) in which no two distinct ordered pairs have the same first member. •The set of all permissible 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.
  • 15.
    WHAT IS FUNCTION? •Wecan also define a function as a rule that assigns to each element x in set A exactly one element, called 𝒇(𝒙), in set B. •We usually represent a function by the equation 𝒚 = 𝒇(𝒙) where 𝒚 is the dependent variable and 𝒙 is the independent variable. •Take note: that for every input (domain), there corresponds a unique output (range).
  • 17.
    WAYS ON HOWTO DETERMINE IF A RELATION IS A FUNCTION OR NOT
  • 24.
  • 25.
    VERTICAL LINE TEST •Thegraph of a function can be intersected by a vertical line in at most one point, since in a function; there is a unique value of to each value of in the domain of the function. •But if the vertical line intersects the graph at more than one point, the graph is not that of a function (it is only a relation). •We usually represent a function by the equation
  • 26.
    Graph the relation 𝒇𝒙 = 𝒙𝟐 + 𝟐𝒙 − 𝟑 Using Vertical Line Test 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕? Answer: 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏
  • 27.
    Graph the relation 𝒙𝟐 +𝒚𝟐 = 𝟏𝟔 Using Vertical Line Test 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕? Answer: 𝑵𝒐𝒕
  • 28.
    Using Vertical LineTest 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕? Answer: 𝑵𝒐𝒕
  • 29.
  • 30.
  • 31.
  • 32.
  • 33.
  • 34.
  • 35.
  • 36.
    PIECEWISE FUNCTION •Is afunction defined by two or more formulas on different parts of its domain.
  • 37.
    Example A merchant sellsrice at 45 pesos a kilo. If, however, a buyer buys more than 150 kilos of rice. The merchant gives a 10% discount. •If x represents the number of kilos of rice purchased, express the amount to be paid by f(x). •How much should be paid for 100 kilos of rice? •How much should be paid for 200 kilos of rice?
  • 38.
    Step 1: Expressthe situation, using function If the number of kilos of purchased rice exceeds 150. That’s 𝑥 > 150. Then the number of kilos in excess of 150 kilos is given a 10% discount If the number of kilos of purchased rice does not exceed 150. That is 0 ≤ 𝑥 ≤ 150 then 𝑓 𝑥 = 45𝑥
  • 39.
    Step 1: Expressthe situation, using function 𝑓 𝑥 = 45𝑥 𝑖𝑓 0 ≤ 𝑥 ≤ 150 40.50 + 675𝑥 𝑖𝑓 𝑥 > 150 So, the function that describes the situation is: Amount of the first 150 kilos of rice 45 150 = 6 750 Amount in excess of 150 kilos 𝟒𝟓 𝟎. 𝟗𝟎 𝒙 − 𝟏𝟓𝟎 = 𝟒𝟎. 𝟓𝟎(𝒙 − 𝟏𝟓𝟎) Total Amount 𝟔 𝟕𝟓𝟎 + 𝟒𝟎. 𝟓𝟎 𝒙 − 𝟏𝟓𝟎 = 𝟒𝟎. 𝟓𝟎𝒙 + 𝟔𝟕𝟓
  • 40.
    Step 2: Findthe amount to be paid for 100 kilos of rice. 𝑓 𝑥 = 45𝑥 𝑓 100 = 45 100 = 4 500 Find f(100) by substituting x = 100 in the first piece or chunk of the function because 100 < 150. So, the amount to be paid for 100 kilos of rice is Php 4,500
  • 41.
    Step 2: Findthe amount to be paid for 200 kilos of rice. 𝑓 𝑥 = 40.50𝑥 + 675 𝑓 200 = 40.50 200 + 675 𝑓 200 = 8 775 Find f(200) by substituting x=200 in the second piece or chunk of the function because 200>150. So, the amount to be paid for 200 kilos of rice is Php 8 775.
  • 42.
  • 43.
    Graph: 𝒇 𝒙= 𝑥 − 2 , 𝑥 ≤ −2 3𝑥2 − 5 , 𝑥 > −2 The function is described by two intervals. Take note that when x=-2, f(x)=x-2 is defined. That is why the point (-2,-4) is shaded.
  • 44.
    Graph: 𝒇 𝒙= 𝑥 − 2 , 𝑥 ≤ −2 3𝑥2 − 5 , 𝑥 > −2 In 𝑓 𝑥 = 3𝑥2 − 5 when x=-2 the point (-2,7) is not shaded because it is not defined. Clearly, by vertical line test, the piecewise is indeed a function.
  • 45.
  • 46.
    Example: If 𝑓𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the following: 𝑓 5 𝑎𝑛𝑑 𝑓 0 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 5 = 4 5 2 − 5 5 + 3 𝑓 5 = 100 − 25 + 3 𝑓 5 = 78 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 0 = 4 0 2 − 5 0 + 3 𝑓 0 = 0 − 0 + 3 𝑓 0 = 3
  • 47.
    Example: If 𝑓𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the following: 𝑓 1 4 𝑎𝑛𝑑 𝑓 2 2 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 1 4 = 4 1 4 2 − 5 1 4 + 3 𝑓 1 4 = 1 4 − 5 4 + 3 𝑓 1 4 = 2 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 2 2 = 4 2 2 2 − 5 2 2 + 3 𝑓 2 2 = 4 1 2 − 5 2 4 + 3 𝑓 2 2 = − 5 2 4 + 5
  • 48.
    Example: If 𝑓𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the following: 𝑓(𝑥 + 1) 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 𝑓 𝑥 + 1 = 4 𝑥 + 1 2 − 5 𝑥 + 1 + 3 𝑓 𝑥 + 1 = 4 𝑥2 + 2𝑥 + 1 − 5𝑥 − 5 + 3 𝑓 𝑥 + 1 = 4𝑥2 + 8𝑥 + 4 − 5 − 5𝑥 − 2 𝑓 𝑥 + 1 = 4𝑥2 + 3𝑥 − 3
  • 49.
    Example: If 𝑓𝑥 = 3𝑥 + 2 evaluate the following: 𝑓 𝑥+ℎ −𝑓(𝑥) ℎ 𝑓 𝑥 = 3𝑥 + 2 𝑓 𝑥 + ℎ = 3 𝑥 + ℎ + 2 𝑓 𝑥 + ℎ = 3𝑥 + 3ℎ + 2 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 + 3ℎ + 2 − 3𝑥 + 2 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 + 3ℎ + 2 − 3𝑥 − 2 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 − 3𝑥 + 3ℎ + 2 − 2 𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3ℎ 𝑓 𝑥 + ℎ − 𝑓 𝑥 ℎ = 3ℎ ℎ = 3
  • 50.
  • 51.
    Example: Given thepiecewise below Find the following: ℎ −10 , ℎ 1 2 , ℎ(6)
  • 52.
    Find: ℎ(−10) Step 1:Determine the piece that should be used. Since -10 belongs to the interval x < -5. therefore, we shall use the third piece of function Step 2: Find h(-10) using the 3rd piece of the function ℎ 𝑥 = 2 𝑥 + 4 2 + 10 ℎ −10 = 2 −10 + 4 2 + 10 ℎ −10 = 2 36 + 10 ℎ −10 = 82
  • 53.
    Find: ℎ 1 2 Step 1:Determine the piece that should be used. Since 1/2 belongs to the interval −5 ≤ 𝑥 < 6. therefore, we shall use the second piece of function Step 2: Find h(1/2) using the 2nd piece of the function ℎ 𝑥 = 3 2𝑥 ℎ 1 2 = 3 2 1 2 ℎ 1 2 = 3
  • 54.
    Find: ℎ(6) Step 1:Determine the piece that should be used. Since x = 6 belongs to the interval 𝑥 ≥ 6. therefore, we shall use the first piece of function Step 2: Find h(6) using the 1st piece of the function ℎ 𝑥 = 𝑥3 + 2𝑥 ℎ 6 = 6 3 + 2(6) ℎ 6 = 216 + 12 ℎ 6 = 228
  • 55.
  • 56.
    SUM, DIFFERENCE, PRODUCT,AND QUOTIENT OF FUNCTIONS Given two functions 𝑓(𝑥) and 𝑔(𝑥), we define the following. •Sum: 𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔(𝑥) •Difference: 𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔(𝑥) •Product: 𝑓 ∗ 𝑔 𝑥 = 𝑓 𝑥 ∗ 𝑔(𝑥) •Quotient: 𝑓 𝑔 𝑥 = 𝑓 𝑥 𝑔 𝑥 , 𝑔(𝑥) ≠ 0
  • 57.
    Example: If 𝑓𝑥 = 6𝑥2 − 12𝑥 − 19 and 𝑔 𝑥 = 𝑥3 − 𝑥2 + 8𝑥 − 1 find the following: 𝑓 + 𝑔 𝑥 𝑎𝑛𝑑 (𝑓 − 𝑔)(𝑥)
  • 58.
    Find: (𝑓 +𝑔)(𝑥) Step 1: Apply the definition for adding functions. 𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔 𝑥 𝑓 + 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 + (𝑥3 − 𝑥2 + 8𝑥 − 1) Step 2: Group similar terms and make sure that the terms are arranged in descending order: 𝑓 + 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 + 𝑥3 − 𝑥2 + 8𝑥 − 1 𝑓 + 𝑔 𝑥 = 𝑥3 + 6𝑥2 − 𝑥2 + −12𝑥 + 8𝑥 + (−19 − 1)
  • 59.
    Find: (𝑓 +𝑔)(𝑥) Step 3: Combined similar terms 𝑓 + 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 + 𝑥3 − 𝑥2 + 8𝑥 − 1 𝑓 + 𝑔 𝑥 = 𝑥3 + 6𝑥2 − 𝑥2 + −12𝑥 + 8𝑥 + −19 − 1 𝑓 + 𝑔 𝑥 = 𝒙𝟑 + 𝟓𝒙𝟐 − 𝟒𝒙 − 𝟐𝟎
  • 60.
    Find: (𝑓 −𝑔)(𝑥) Step 1: Apply the definition for subtracting functions. 𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔 𝑥 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − (𝑥3 − 𝑥2 + 8𝑥 − 1) Step 2: Group similar terms and make sure that the terms are arranged in descending order: 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − 𝑥3 − 𝑥2 + 8𝑥 − 1 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − 𝑥3 + 𝑥2 − 8𝑥 + 1 𝑓 − 𝑔 𝑥 = −𝑥3 + 6𝑥2 + 𝑥2 + −12𝑥 − 8𝑥 + (−19 + 1)
  • 61.
    Find: (𝑓 +𝑔)(𝑥) Step 3: Combined similar terms 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − 𝑥3 − 𝑥2 + 8𝑥 − 1 𝑓 − 𝑔 𝑥 = 6𝑥2 − 12𝑥 − 19 − 𝑥3 + 𝑥2 − 8𝑥 + 1 𝑓 − 𝑔 𝑥 = −𝑥3 + 6𝑥2 + 𝑥2 + −12𝑥 − 8𝑥 + −19 + 1 𝑓 − 𝑔 𝑥 = −𝒙𝟑 + 𝟕𝒙𝟐 − 𝟐𝟎𝒙 − 𝟏𝟖
  • 62.
    Example: If 𝑓𝑥 = 3 𝑥2−1 and 𝑔 𝑥 = 𝑥 𝑥−1 find 𝑓 + 𝑔 𝑥 .
  • 63.
    Find: (𝑓 +𝑔)(𝑥) Step 1: Apply the definition for adding functions. 𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔 𝑥 𝑓 + 𝑔 𝑥 = 3 𝑥2 − 1 + 𝑥 𝑥 − 1 Step 2: Find the LCD of 𝑥2 − 1 𝑎𝑛𝑑 𝑥 − 1, the 𝐿𝐶𝐷 𝑖𝑠 𝑥 + 1 𝑥 − 1 𝑓 + 𝑔 𝑥 = 3 𝑥2 − 1 + 𝑥 𝑥 − 1 𝑓 + 𝑔 𝑥 = 3 + 𝑥(𝑥 + 1) (𝑥 + 1)(𝑥 − 1)
  • 64.
    Find: (𝑓 +𝑔)(𝑥) Step 3: Simplify the expression: 𝑓 + 𝑔 𝑥 = 3 𝑥2 − 1 + 𝑥 𝑥 − 1 𝑓 + 𝑔 𝑥 = 3 + 𝑥(𝑥 + 1) (𝑥 + 1)(𝑥 − 1) 𝑓 + 𝑔 𝑥 = 𝒙𝟐 + 𝒙 + 𝟑 (𝒙 + 𝟏)(𝒙 − 𝟏)
  • 65.
    Example: If 𝑓𝑥 = 𝑥−3 𝑥+4 and 𝑔 𝑥 = 𝑥+1 𝑥+4 find 𝑓 − 𝑔 𝑥 .
  • 66.
    Find: (𝑓 −𝑔)(𝑥) Step 1: Apply the definition for subtracting functions. 𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔 𝑥 𝑓 − 𝑔 𝑥 = 𝑥 − 3 𝑥 + 4 − 𝑥 + 1 𝑥 + 4 Step 2: Simplify 𝑓 − 𝑔 𝑥 = 𝑥 − 3 𝑥 + 4 − 𝑥 + 1 𝑥 + 4 𝑓 − 𝑔 𝑥 = 𝑥 − 3 − (𝑥 + 1) 𝑥 + 4 = 𝑥 − 3 − 𝑥 − 1 𝑥 + 4 = −𝟒 𝒙 + 𝟒
  • 67.
    Example: If 𝑓 𝑥= 𝑥3 − 27, 𝑔 𝑥 = 𝑥2 + 3𝑥 + 9 and ℎ 𝑥 = 4𝑥 − 1. . Find the following: (𝑓 ∗ ℎ)(𝑥), 𝑓 𝑔 (𝑥), 𝑓 ℎ (𝑥)
  • 68.
    Find: (𝑓 ∗ℎ)(𝑥) Step 1: Apply the definition for multiplying functions. 𝑓 ∗ ℎ 𝑥 = 𝑓 𝑥 ∗ ℎ 𝑥 𝑓 ∗ ℎ 𝑥 = 𝑥3 − 27 (4𝑥 − 1) Step 2: Find the product of polynomials using the distributive law. 𝑓 ∗ ℎ 𝑥 = 𝑥3 − 27 4𝑥 − 1 𝑓 ∗ ℎ 𝑥 = 𝟒𝒙𝟒 − 𝒙𝟑 − 𝟏𝟎𝟖𝒙 + 𝟐𝟕
  • 69.
    Find: 𝑓 𝑔 (𝑥) Step 1: Apply thedefinition for dividing functions. 𝑓 𝑔 𝑥 = 𝑓 𝑥 𝑔 𝑥 Step 2: Factor the numerator 𝑓 𝑔 𝑥 = 𝑥3 − 27 𝑥2 + 3𝑥 + 9 𝑓 𝑔 𝑥 = 𝑥3 − 27 𝑥2 + 3𝑥 + 9 𝑓 𝑔 𝑥 = (𝑥 − 3)(𝑥2 + 3𝑥 + 9) 𝑥2 + 3𝑥 + 9 Step 3: Cancel the common factors and simplify 𝑓 𝑔 𝑥 = (𝑥 − 3)(𝑥2 + 3𝑥 + 9) 𝑥2 + 3𝑥 + 9 𝑓 𝑔 𝑥 = 𝒙 − 𝟑
  • 70.
    Find: 𝑓 ℎ (𝑥) Step 1: Apply thedefinition for dividing functions. 𝑓 ℎ 𝑥 = 𝑓 𝑥 𝑔 𝑥 Step 2: Since there are no common factors in the numerator and the denominator, this is already the simplest form 𝑓 ℎ 𝑥 = 4𝑥 − 1 𝑥3 − 27 𝑓 ℎ 𝑥 = 𝟒𝒙 − 𝟏 𝒙𝟑 − 𝟐𝟕
  • 71.
    Example: If 𝑓 𝑥= 𝑥2−3𝑥−28 𝑥2−8𝑥+16 , 𝑔 𝑥 = 𝑥2−5𝑥+4 𝑥2−49 . Find the following: 𝑓 ∗ 𝑔 (𝑥) and 𝑓 𝑔 (𝑥)
  • 72.
    Find: (𝑓 ∗𝑔)(𝑥) Step 1: Apply the definition for multiplying functions. 𝑓 ∗ 𝑔 𝑥 = 𝑓 𝑥 ∗ 𝑔 𝑥 𝑓 ∗ 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 5𝑥 + 4 𝑥2 − 49 Step 2: Factor the numerator and denominator of both rational function. 𝑓 ∗ 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 5𝑥 + 4 𝑥2 − 49 𝑓 ∗ 𝑔 𝑥 = (𝑥 − 7)(𝑥 + 4) (𝑥 − 4)(𝑥 − 4) ∗ (𝑥 − 1)(𝑥 − 4) (𝑥 − 7)(𝑥 + 7)
  • 73.
    Find: (𝑓 ∗𝑔)(𝑥) Step 3: Cancel common factors and simplify. 𝑓 ∗ 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 5𝑥 + 4 𝑥2 − 49 𝑓 ∗ 𝑔 𝑥 = (𝑥 − 7)(𝑥 + 4) (𝑥 − 4)(𝑥 − 4) ∗ (𝑥 − 1)(𝑥 − 4) (𝑥 − 7)(𝑥 + 7) 𝑓 ∗ 𝑔 𝑥 = (𝒙 + 𝟒)(𝒙 − 𝟏) (𝒙 − 𝟒)(𝒙 + 𝟕)
  • 74.
    Find: 𝑓 𝑔 (𝑥) Step 1: Applythe definition for dividing functions. 𝑓 𝑔 𝑥 = 𝑓 𝑥 𝑔 𝑥 → Step 2: Apply the rule in dividing rational expressions. 𝑓 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 49 𝑥2 − 5𝑥 + 4 𝑓 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ÷ 𝑥2 − 5𝑥 + 4 𝑥2 − 49 𝑓 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ÷ 𝑥2 − 5𝑥 + 4 𝑥2 − 49
  • 75.
    Find: (𝑓 ∗𝑔)(𝑥) Step 3: Factor the numerator and denominator of both rational functions. 𝑓 𝑔 𝑥 = (𝑥 − 7)(𝑥 + 4) (𝑥 − 4)(𝑥 − 4) ∗ (𝑥 − 7)(𝑥 + 7) (𝑥 − 4)(𝑥 − 1) 𝑓 𝑔 𝑥 = 𝑥2 − 3𝑥 − 28 𝑥2 − 8𝑥 + 16 ∗ 𝑥2 − 49 𝑥2 − 5𝑥 + 4
  • 76.
    Find: (𝑓 ∗𝑔)(𝑥) Step 4: There are no common factors to cancel. So we just simplify 𝑓 𝑔 𝑥 = (𝑥 − 7)(𝑥 + 4) (𝑥 − 4)(𝑥 − 4) ∗ (𝑥 − 7)(𝑥 + 7) (𝑥 − 4)(𝑥 − 1) 𝑓 𝑔 𝑥 = 𝒙 − 𝟕 𝟐 (𝒙 + 𝟕)(𝒙 + 𝟒) 𝒙 − 𝟒 𝟑(𝒙 − 𝟏)
  • 77.
  • 78.
    Definition of Compositionof Function and the domain of (𝑓 ∘ 𝑔)(𝑥) is the set of all numbers x in the domain of 𝑔(𝑥), such that 𝑔(𝑥) is the domain of 𝑓(𝑥). Given two function 𝑓(𝑥) and 𝑔(𝑥), the composition of function is denoted by (𝑓 ∘ 𝑔)(𝑥) is defined by: 𝒇 ∘ 𝒈 𝒙 = 𝒇(𝒈(𝒙)) Similarly, we define (𝑔 ∘ 𝑓)(𝑥) as follows: 𝒈 ∘ 𝒇 𝒙 = 𝒈(𝒇(𝒙))
  • 79.
    Example: If 𝑓 𝑥= 𝑥2 − 6 and 𝑔 𝑥 = 3𝑥 + 2. Find the following: (𝑓 ∘ 𝑔)(𝑥), (𝑔 ∘ 𝑓)(𝑥), (𝑔 ∘ 𝑔)(𝑥), (𝑓 ∘ 𝑓)(𝑥)
  • 80.
    Find:(𝑓 ∘ 𝑔)(𝑥) Step1: Apply the definition of composition of functions 𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥 → (𝑓 ∘ 𝑔)(𝑥) = 𝑓(3𝑥 + 2) Step 2: Apply the definition of 𝑓 𝑓 ∘ 𝑔 𝑥 = 𝑓 3𝑥 + 2 → 𝑓 ∘ 𝑔 𝑥 = 3𝑥 + 2 2 − 6 Step 3: Simplify the results. 𝑓 ∘ 𝑔 𝑥 = 3𝑥 + 2 2 − 6 𝑓 ∘ 𝑔 𝑥 = 9𝑥2 + 12𝑥 + 4 − 6 𝑓 ∘ 𝑔 𝑥 = 𝟗𝒙𝟐 + 𝟏𝟐𝒙 − 𝟐
  • 81.
    Find:(𝑔 ∘ 𝑓)(𝑥) Step1: Apply the definition of composition of functions 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 → (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑥2 − 6) Step 2: Apply the definition of 𝑓 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑥2 − 6 → 𝑔 ∘ 𝑓 𝑥 = 3 𝑥2 − 6 + 2 Step 3: Simplify the results. 𝑔 ∘ 𝑓 𝑥 = 3𝑥2 − 18 + 2 𝑔 ∘ 𝑓 𝑥 = 𝟑𝒙𝟐 − 𝟏𝟔 𝑔 ∘ 𝑓 𝑥 = 3 𝑥2 − 6 + 2
  • 82.
    Find:(𝑔 ∘ 𝑔)(𝑥) Step1: Apply the definition of composition of functions 𝑔 ∘ 𝑔 𝑥 = 𝑔 𝑔 𝑥 → (𝑔 ∘ 𝑔)(𝑥) = 𝑔(3𝑥 + 2) Step 2: Apply the definition of 𝑓 𝑔 ∘ 𝑔 𝑥 = 𝑔 3𝑥 + 2 → 𝑔 ∘ 𝑔 𝑥 = 3 3𝑥 + 2 + 2 Step 3: Simplify the results. 𝑔 ∘ 𝑔 𝑥 = 9𝑥 + 6 + 2 𝑔 ∘ 𝑔 𝑥 = 𝟑𝒙𝟐 + 𝟖 𝑔 ∘ 𝑔 𝑥 = 3 3𝑥 + 2 + 2
  • 83.
    Find:(𝑓 ∘ 𝑓)(𝑥) Step1: Apply the definition of composition of functions 𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑓 𝑥 → (𝑓 ∘ 𝑓)(𝑥) = 𝑓(𝑥2 − 6) Step 2: Apply the definition of 𝑓 𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑥2 − 6 → 𝑓 ∘ 𝑓 𝑥 = 𝑥2 − 6 2 − 6 Step 3: Simplify the results. 𝑔 ∘ 𝑓 𝑥 = 𝑥4 − 12𝑥2 + 36 − 6 𝑔 ∘ 𝑓 𝑥 = 𝒙𝟒 − 𝟏𝟐𝒙𝟐 + 𝟑𝟎 𝑓 ∘ 𝑓 𝑥 = 𝑥2 − 6 2 − 6
  • 84.
    Example: Given the followingfunctions 𝑓 𝑥 = 𝑥3, 𝑔 𝑥 = 𝑥+2 3𝑥−1 ,and ℎ 𝑥 = 6𝑥 − 7 Find the following: 𝑓 ∘ 𝑓 𝑥 𝑎𝑛𝑑 (𝑓 ∘ 𝑓)(2) , 𝑔 ∘ ℎ 𝑥 𝑎𝑛𝑑 (𝑔 ∘ ℎ)(−1)
  • 85.
    Find:(𝑓 ∘ 𝑓)(𝑥) Step1: Apply the definition of composition of functions 𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑓 𝑥 → (𝑓 ∘ 𝑓)(𝑥) = 𝑓(𝑥3 ) Step 2: Apply the definition of 𝑓 𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑥3 → 𝑓 ∘ 𝑓 𝑥 = 𝑥3 3 Step 3: Simplify the results. 𝑔 ∘ 𝑓 𝑥 = 𝒙𝟗 𝑓 ∘ 𝑓 𝑥 = 𝑥3 3 𝑔 ∘ 𝑓 2 = (𝟐)𝟗 = 𝟓𝟏𝟐
  • 86.
    Find:(𝑔 ∘ ℎ)(𝑥) Step1: Apply the definition of composition of functions 𝑔 ∘ ℎ 𝑥 = 𝑔 ℎ 𝑥 → (𝑔 ∘ ℎ)(𝑥) = 𝑔 6𝑥 + 7 Step 2: Apply the definition of 𝑓 𝑔 ∘ ℎ 𝑥 = 𝑔 6𝑥 + 7 𝑔 ∘ ℎ 𝑥 = 6𝑥 − 7 + 2 3 6𝑥 − 7 − 1 Step 3: Simplify the results. 𝑔 ∘ ℎ 𝑥 = 6𝑥 − 7 + 2 3 6𝑥 − 7 − 1 𝑔 ∘ ℎ 𝑥 = 𝟔𝒙 − 𝟓 𝟏𝟖𝒙 − 𝟐𝟐 𝑔 ∘ ℎ −1 = 6(−1) − 5 18(−1) − 22 = − 𝟏𝟏 𝟒
  • 87.
  • 88.
    The following stepswill be helpful in solving problems involving functions. • Read and comprehend the problem carefully. • Identify the function to be used (linear, quadratic, cubic, etc.) in representing the given problem. • Write the function that describes the problem. • Use the function to solve the problem. • Use the function to solve the problem. • Check the answer.
  • 89.
    EXAMPLE: A Sciencemajor student earned 6 units after enrolling for a semester. Then, he took an academic leave for one year. After his leave, he enrolled again, ever more desirous this time to finish all the 150 units that he needs. In how many semesters does he need to enroll if he takes 9 units of Science subjects every semester?
  • 90.
    Step:1 Represent the givenand unknown quantities in the problem. Let: 𝑦 = total number of units to be taken in x semesters. 𝑥 = number of semesters. Step:2 Determine the function or equation that relates the quantities or variables in the problem. The problem can be described best by a linear function which relates the number of semesters (x) to the total number of units (y). The function that describes the relationship between x and y is: 𝑦 = 9𝑥 + 16
  • 91.
    Step:3 Solve the problem.We want to know the number of semesters that the student needs to take in order to finish all 150 units. 𝑦 = 150 The student needs 16 semesters to complete the program. 150 = 9𝑥 + 6 150 − 6 = 9𝑥 144 = 9𝑥 𝑥 = 16
  • 92.
    EXAMPLE: The areaof a rectangle is 420 𝑐𝑚2. The length of the rectangle is one less than thrice its width. What are the dimensions of the given rectangle?
  • 93.
    Step:1 Represent the givenand unknown quantities in the problem. Express the width and the length of the rectangle in terms of one variable. Let: 𝑦 = area of the rectangle 𝑥 = width 3𝑥 − 1 = Length Step:2 Use function to relate the quantities or variables in the problem. The area of the rectangle is given by: 𝐴 = 𝑙𝑤; where 𝑙 = length and 𝑤 = width Using the representation in Step 1, we have: 𝑦 = 𝑥 3𝑥 − 1 𝑦 = 3𝑥2 − 𝑥 The function that describes the relationship is 𝑦 = 3𝑥2 − 𝑥
  • 94.
    Step:3 Solve the problem. Thearea of rectangle is 𝑦 = 420 Substitute the value in 𝑦 = 3𝑥2 − 𝑥 3𝑥2 − 𝑥 − 420 = 0 Find the length and the width by substituting the value of x in the representation found in Step 1. Reject the negative value of x because the measurement of width cannot be negative. 3𝑥 + 35 𝑥 − 12 = 0 420 = 3𝑥2 − 𝑥 3𝑥 + 35 = 0 If 𝑥 = 12, then: width = 𝑥 = 12 Length = 3𝑥 − 1 = 3 12 − 1 = 35 Thus the dimensions of the given rectangle is 12 and 35. 3𝑥 = −35 𝒙 = − 𝟑𝟓 𝟑 𝑥 − 12 = 0 𝒙 = 𝟏𝟐 Step:4

Editor's Notes

  • #51 Observe that the function consist of two parts: the first part is a line, while the second part is a parabola. Graph them in Cartesian plane. First, graph the line f(x)=x-2 for all x<= -2. Then graph the parabola f(x)=3x^2 -5 for all x>2.
  • #52 Observe that the function consist of two parts: the first part is a line, while the second part is a parabola. Graph them in Cartesian plane. First, graph the line f(x)=x-2 for all x<= -2. Then graph the parabola f(x)=3x^2 -5 for all x>2.