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An overall examination reviewer beneficial to the Senior High School students of San Pedro College This reviewer serves as the compilation of all the lessons discussed by different teachers of the same subject, it also includes examples and short activities to sharpen one’s knowledge about the topic. If there are topics included that are not discussed, please clarify with your subject teachers. OUTLINE I. Functions A. Functions vs. Relations B. Evaluating Functions C. Piecewise Functions II. Operations on Functions A. Addition B. Subtraction C. Multiplication D. Division E. Composite Functions III.Rational Functions A. Rational Expression B. Rational Equation, Function & Inequality C. Understanding and Solving Rational Equation D. Understanding and Solving Rational Inequality IV.Inverse of a Function A. One-to-one Function B. Steps in Obtaining the Inverse of a Function f V. Exponential Functions A. Exponential Growth B. Exponential Decay C. Half-life D. Compound Interest E. Representing Exponential Functions through Tables Graphs, and Equations VI. Exponential Equations A. Terms B. Laws of Exponents C. Solving Exponential Equations VII. Review Questions VIII. Validated
I. FUNCTIONS
Function
- A relation in which each element of the domain (x, the independent variable) is paired with exactly one element of the range (y, the dependent variable). - In other words, there is one and only one output (y) with each input (x). - Usually of the form: 𝑦 = 𝑓(𝑥) Function Notation
- A way of expressing a function using variables and symbols
A. FUNCTIONS VS. RELATIONS
Relation
- It is a set of ordered pairs that are arranged in an orderly manner. Functions
- It is a special kind of relation where each input has only one output. One-to-one
- each x-value corresponds to one distinct y-value One-to-many
- one x-value corresponds to multiple y-values Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
Many-to-one
- multiple x-values correspond to the same y-value Many-to-many
- multiple x-values correspond to multiple y-values Examples: Determine whether each relation is a function or not and whether the relationship is One-to-one, Many-to-one, One-to-many, or Many-to-many.
- {(2,3), (4,3), (5,2), (3,0)} x y YES , it is a function since 2—>3 the x-values are paired to 4—>3 exactly one y-value.It is also 5—>2 an example of a Many-to- 3—>0 one-relationship since two domain values (2 and 4) are . assigned to one y value (3).
- {(4,1), (5,2), (5,3), (6,6), (1,9)} x y 4—>1 NO , it is not a function since 5—>2 one x value (5) is assigned to 5—>3 two y values (2 and 3).Given 6—>6 this, it is also an example of a 1—>9 One-to-many relationship.
- {(6,3),(2,5),(5,4),(4,2),(1,1)} x y 6—>3 YES , it is a function since 2—>5 each x value is assigned to 5—>4 only one y value.This is an 4—>2 example of a One-to-one 1—>1 relationship. Vertical Line Test
- Another way of determining whether a relation is a function or not is when the given is a graph. First, we would need to draw vertical lines on the Cartesian plane and if any vertical line passes through more than one point on the graph, then that relation is not a function. Examples:
Let's start by drawing vertical lines on the graph. We can see that no vertical line passed through more than one point on the graph. Therefore, it is a Function.
Once again, start by drawing vertical lines on the graph. Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
Solution:
1. Identify and substitute the given. 𝑓(𝑥) = 2𝑥 + 1; 𝑥 = 3 , 𝑥 = 1 𝑓(3) = 2(3) + 1 ; 𝑓(1) = 2(1) + 1 Step 2: Perform the operations. 𝑓(3) = 6 + 1 𝑓(3) = 7 ; 𝑓(1) = 2 + 1 𝑓(1) = 3; find − 4[𝑓(3) − 𝑓(1)] 𝑓(3) = 7 ; 𝑓(1) = 3 − 4[7 − 3] − 4[4] − 16 ANSWER: 𝑓(3) = 7; 𝑓(1) = 3 𝑎𝑛𝑑 − 4[𝑓(3) − 𝑓(1)] =− 16
- 𝑓(𝑥) = 𝑥 2 − 3𝑥 + 1; 𝑓𝑖𝑛𝑑 𝑓(2𝑥 − 1) Solution: 1. Substitute the given. 𝑓(𝑥) = 𝑥 2 − 3𝑥 + 1; 𝑥 = 2𝑥 − 1 𝑓(2𝑥 − 1) = (2𝑥 − 1) 2 − 3(2𝑥 − 1) + 1 NOTE : Use the FOIL method for (2𝑥 − 1) 2
- Perform the indicated operations and combine like terms.
2 − 4𝑥 + 1 − 6𝑥 + 3 + 1 𝑓(2𝑥 − 1) = 4𝑥 2 − 4𝑥 − 6𝑥 + 1 + 3 + 1 𝑓(2𝑥 − 1) = 4𝑥 2 − 10𝑥 + 5 ANSWER: 𝑓(2𝑥 − 1) = 4𝑥 2 − 10𝑥 + 5
C. PIECEWISE FUNCTIONS
- A function that is made up of separate functions each with their own intervals or the conditions in which they become valid.
- Graph and example of a piecewise function: 𝑓(𝑥) = {𝑥 − 1, 𝑖𝑓 𝑥 < − 2 = {− 3 , 𝑖𝑓 𝑥 ≥ − 2 Examples:
- A jeepney ride costs ₱8. 00for the first 4 kilometers, and each additional integer kilometers adds ₱1. 50. Use a piecewise function to represent the jeepney fare in terms of distance (𝑑)in kilometers. Solution: Let the number of kilometers be represented by (𝑑). And 𝑓(𝑑)as the cost at a given distance. Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
For the first 4 kilometers, a jeepney ride costs ₱8. 00. 𝑆𝑜, 𝑓(𝑑) = { 8 𝑖𝑓 0 < 𝑑 ≤ 4 This can be thought of as the jeepney fare at a distance that it is greater than 0 km and less than or equal to 4 kilometers. If the distance traveled exceeds 4 kilometers, you would have to pay an additional ₱1. 50 per 1km. 𝑆𝑜, 𝑓(𝑑) = { 8 + 1. 50(𝑑 − 4) 𝑖𝑓 𝑑 > 4 This can be thought of as the jeepney fare at a distance that is greater than 4. For example, for your jeepney ride, you would have to travel for 7 km. How much would you have to pay? 𝑓(7) = { 8 + 1. 50(7 − 4) 𝑠𝑖𝑛𝑐𝑒 7 > 4 𝑓(7) = { 8 + 1. 50(3) 𝑓(7) = { 8 + 4. 5 𝑓(7) = 12. 5 Therefore, the total amount that you would have to pay if the distance traveled is 7 km is ₱12. 5. FINAL ANSWER: 𝑓(𝑑) = { 8 𝑖𝑓 0 < 𝑑 ≤ 4 = { 8 + 1. 50(𝑑 − 4) 𝑖𝑓 𝑑 > 4
- A user is charged ₱300monthly for a particular mobile plan, which includes free 100 text messages. Messages in excess of 100 are charge ₱1each. Represent the amount a consumer pays each month as a function of the number of messages (𝑚)sent in a month. Solution: Let (𝑚)be the number of messages sent in a month and 𝑝(𝑚)as the amount paid depending on the number of messages. For the first 100 messages sent, a user is charged ₱. 𝑆𝑜, 𝑝(𝑚) = {300 𝑖𝑓 0 < 𝑚 ≤ 100 Thought of as the total amount paid if messages sent are greater than 0 and less than or equal to 100. For messages in excess of 100, the user is charged ₱1 each. 𝑆𝑜, 𝑝(𝑚) = {300 + (𝑚 − 100) 𝑖𝑓 𝑚 > 100 𝑂𝑅 𝑝(𝑚) = {300 + 1(𝑚 − 100) 𝑖𝑓 𝑚 > 100 Thought of as the amount paid if messages sent exceeds 100. NOTE : Both are the same since 1 does not need to be written. FINAL ANSWER: 𝑝(𝑚) = {300 𝑖𝑓 0 < 𝑚 ≤ 100 {300 + (𝑚 − 100) 𝑖𝑓 𝑚 > 100
- The cost of hiring a catering service to serve food for a party is ₱150per head for 20 persons or less, ₱130per head for 21 to 50 persons, and ₱110per head for 51 to 100 persons. For more than 100 persons, the cost is at ₱ per head. Represent the total cost as a piecewise function of the number of attendees of the party. Solution: Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
2 − 5𝑥 + 5 𝑓𝑖𝑛𝑑(𝑓 − 𝑔)(5) NOTE : For problems similar to this, solve the equation for now before substituting the value of x. Solution:
1. Plug the given values into the equation. (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) (𝑓 + 𝑔)(𝑥) = (3𝑥 − 2) + (𝑥 2 − 5𝑥 + 5) 2. Simplify, combine like terms, and arrange in descending order of degrees. (𝑓 + 𝑔)(𝑥) = 3𝑥 − 2 + 𝑥 2 − 5𝑥 + 5 (𝑓 + 𝑔)(𝑥) = 𝑥 2 + 3𝑥 − 5𝑥 − 2 + 5 (𝑓 + 𝑔)(𝑥) = 𝑥 2 − 2𝑥 + 3 Start solving for (𝑓 + 𝑔)(5) (𝑓 + 𝑔)(5) = 𝑥 2 − 2𝑥 + 3 (𝑓 + 𝑔)(5) = (5) 2 − 2(5) + 3 (𝑓 + 𝑔)(5) = 25 − 10 + 3 (𝑓 + 𝑔)(5) = 18 ANSWER/S: (𝑓 + 𝑔)(𝑥) = 𝑥 2 − 2𝑥 + 3 (𝑓 + 𝑔)(5) = 18
- 𝑓(𝑥) = 1 𝑥+3 ; 𝑔(𝑥) =^ 1 𝑥− 𝑓𝑖𝑛𝑑 (𝑓 + 𝑔)(𝑥) Solution: 1. Plug the given values into the equation. (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) (𝑓 + 𝑔)(𝑥) = 1 𝑥+3 +^ 1 𝑥− NOTE: For cases such as these, multiply both sides by their opposite fraction’s denominator.(top and bottom) 2. Simplify. (𝑓 + 𝑔)(𝑥) = (𝑥−2) (𝑥−2) •^ 1 𝑥+3 +^ 1 𝑥−2 •^ (𝑥+3) (𝑥+3) (𝑓 + 𝑔)(𝑥) = 𝑥− (𝑥+3)(𝑥−2) +^ 𝑥+ (𝑥−2)(𝑥+3) Now that they have the same denominators, add the top terms. (𝑓 + 𝑔)(𝑥) = 𝑥−2+𝑥+ (𝑥+3)(𝑥−2) (𝑓 + 𝑔)(𝑥) = 𝑥+𝑥−2+ (𝑥+3)(𝑥−2) (𝑓 + 𝑔)(𝑥) = 2𝑥+ (𝑥+3)(𝑥−2) ANSWER: (𝑓 + 𝑔)(𝑥) = 2𝑥+ (𝑥+3)(𝑥−2) OR (𝑓 + 𝑔)(𝑥) = (by FOIL) 2𝑥+ 𝑥 2 +𝑥−
B. SUBTRACTION
- The difference of two functions 𝑓(𝑥) and 𝑔(𝑥) is denoted by(𝑓 − 𝑔)(𝑥)
- The difference of these two functions is defined as (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) **Steps on Solving:
- Plug the given functions** 𝑓(𝑥) and 𝑔(𝑥) into the equation: (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
2. Simplify by combining like terms. Follow the usual format wherein terms are in descending order of degree. Also remember to distribute the (-) sign. Examples:
- 𝑓(𝑥) = 2𝑥 2 − 4𝑥 + 1 ; 𝑔(𝑥) =− 𝑥 2 − 3𝑥 − 1 Solution: 1. Plug the given values into the equation. (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥 ) (𝑓 − 𝑔)(𝑥) = (2𝑥 2 − 4𝑥 + 1) − (− 𝑥 2 − 3𝑥 − 1) 2. Distribute the (-) sign to the right side, combine like terms, and arrange in descending order of degrees. (𝑓 − 𝑔)(𝑥) = 2𝑥 2 − 4𝑥 + 1 + 𝑥 2
- 3𝑥 + 1 (𝑓 − 𝑔)(𝑥) = 2𝑥 2
- 𝑥 2 − 4𝑥 + 3𝑥 + 1 + 1 (𝑓 − 𝑔)(𝑥) = 3𝑥 2 − 𝑥 + 2 ANSWER: (𝑓 − 𝑔)(𝑥) = 3𝑥 2 − 𝑥 + 2
- 𝑓(𝑥) = 4𝑥 − 7; 𝑔(𝑥) = 3𝑥 2
- 2𝑥 + 1 𝑓𝑖𝑛𝑑 (𝑓 − 𝑔)(5) Solution: 1. Plug the given into the equation. (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) (𝑓 − 𝑔)(𝑥) = (4𝑥 − 7) − (3𝑥 2
- 2𝑥 + 1) 2. Distribute the (-) sign to the right side, combine like terms, and arrange in descending order of degrees. (𝑓 − 𝑔)(𝑥) = 4𝑥 − 7 − 3𝑥 2 − 2𝑥 − 1 (𝑓 − 𝑔)(𝑥) =− 3𝑥 2
- 4𝑥 − 2𝑥 − 1 − 7 (𝑓 − 𝑔)(𝑥) =− 3𝑥 2
- 2𝑥 − 8 Now, solve for (𝑓 − 𝑔)(5) (𝑓 − 𝑔)(𝑥) =− 3𝑥 2
- 2𝑥 − 8 (𝑓 − 𝑔)(5) =− 3(5) 2
- 2(5) − 8 (𝑓 − 𝑔)(5) =− 3(25) + 10 − 8 (𝑓 − 𝑔)(5) =− 75 + 10 − 8 (𝑓 − 𝑔)(5) =− 73 ANSWER/S: (𝑓 − 𝑔)(𝑥) =− 3𝑥 2
- 2𝑥 − 8 (𝑓 − 𝑔)(5) =− 73
- 𝑓(𝑥) = 𝑥+ 𝑥−3 ; 𝑔(𝑥) =^ 𝑥− 𝑥+ 𝑓𝑖𝑛𝑑 (𝑓 − 𝑔)(5) Solution: 1. Plug the given values into the equation. (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) (𝑓 − 𝑔)(𝑥) = 𝑥+ 𝑥−3 −^ 𝑥− 𝑥+ NOTE: For problems like these, multiply both sides by their opposite fraction’s denominator. (top and bottom) 2. Distribute the (-) sign to the right side, combine like terms, and arrange in descending order of degrees. Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
terms and arrange in decreasing order of degrees. (𝑓 • 𝑔)(𝑥) = (2𝑥)(𝑥) + (− 1)(𝑥) + (5)(2𝑥) + (5)(− 1) (𝑓 • 𝑔)(𝑥) = 2𝑥 2 − 𝑥 + 10𝑥 − 5 (𝑓 • 𝑔)(𝑥) = 2𝑥 2
- 9𝑥 − 5 Now, solve for (𝑓 • 𝑔)(3) (𝑓 • 𝑔)(𝑥) = 2𝑥 2
- 9𝑥 − 5 (𝑓 • 𝑔)(3) = 2(3) 2
- 9(3) − 5 (𝑓 • 𝑔)(3) = 18 + 27 − 5 (𝑓 • 𝑔)(3) = 40 ANSWER/S: (𝑓 • 𝑔)(𝑥) = 2𝑥 2
- 9𝑥 − 5 (𝑓 • 𝑔)(3) = 40
- 𝑓(𝑥) = 𝑥 + 5 ; 𝑔(𝑥) = 2𝑥 2 + 9𝑥 − 5 𝑓𝑖𝑛𝑑 (𝑓 • 𝑔)(2) Solution: 1. Plug the given into the equation. (𝑓 • 𝑔)(𝑥) = (𝑥 + 5)(2𝑥 2
- 9𝑥 − 5) 2. Multiply each term in the first function by every term in the second function. Combine like terms and arrange in decreasing order of degrees. (𝑓 • 𝑔)(𝑥) = (2𝑥 2 )(𝑥) + (9𝑥)(𝑥) + (− 5)(𝑥)
- (2𝑥 2 )(5) + (9𝑥)(5) + (− 5)(𝑥) (𝑓 • 𝑔)(𝑥) = 2𝑥 3
- 9𝑥 2 − 5𝑥 + 10𝑥 2
- 45𝑥 − 25 (𝑓 • 𝑔)(𝑥) = 2𝑥 3
- 9𝑥 2
- 10𝑥 2 − 5𝑥 + 45𝑥 − 25 (𝑓 • 𝑔)(𝑥) = 2𝑥 3
- 19𝑥 2
- 40𝑥 − 25 Now, solve for (𝑓 • 𝑔)(2)
3
- 19𝑥 2
- 40𝑥 − 25 (𝑓 • 𝑔)(2) = 2(2) 3
- 19(2) 2
- 40(2) − 25 (𝑓 • 𝑔)(2) = 2(8) + 19(4) + 40(2) − 25 (𝑓 • 𝑔)(2) = 16 + 76 + 80 − 25 (𝑓 • 𝑔)(2) = 147 ANSWERS/S: (𝑓 • 𝑔)(𝑥) = 2𝑥 3
- 19𝑥 2
- 40𝑥 − 25 (𝑓 • 𝑔)(2) = 147
D. DIVISION
- The quotient of two functions 𝑓(𝑥)and 𝑔(𝑥) is denoted by 𝑓
- The quotient of the two functions is defined as: 𝑓
( 𝑔)(𝑥) =^
𝑓(𝑥) 𝑔(𝑥) 𝑤ℎ𝑒𝑟𝑒 𝑔(𝑥) ≠ 0 Steps on Solving:
1. Substitute the given values to the equation in fraction form 𝑓
( 𝑔)(𝑥) =^
𝑓(𝑥) 𝑔(𝑥)
2. Try to simplify the expression by factoring and canceling. Examples:
- 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 3; 𝑔(𝑥) = 𝑥 2
Solution:
1. Substitute the given values to the equation in fraction form 𝑓
( 𝑔)(𝑥) =^
𝑓(𝑥) 𝑔(𝑥) 𝑓
( 𝑔)(𝑥) =^
𝑥 2 −2𝑥− 𝑥^2 +2𝑥+ Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
2. Try to simplify the expression by factoring and canceling. 𝑓
( 𝑔)(𝑥) =^
(𝑥−3)(𝑥+1) (𝑥+1)(𝑥+1) Cancel (𝑥 + 1) 𝑓
( 𝑔)(𝑥) =^
(𝑥−3) (𝑥+1) ANSWER: 𝑓
( 𝑔)(𝑥) =^
(𝑥−3) (𝑥+1)
- 𝑓(𝑥) = 𝑥 2 − 2𝑥 − 3; 𝑔(𝑥) = 𝑥 + 1 𝑓𝑖𝑛𝑑 𝑓
Solution:
1. Substitute the given values to the equation in fraction form 𝑓
( 𝑔)(𝑥) =^
𝑓(𝑥) 𝑔(𝑥) 𝑓
( 𝑔)(𝑥) =^
𝑥 2 −2𝑥− 𝑥+ 𝑓
( 𝑔)(𝑥) =^
(𝑥−3)(𝑥+1) 𝑥+ Cancel 𝑥 + 1 𝑓
Now, solve for 𝑓
𝑓
𝑓
𝑓
ANSWER/S: ;
𝑓
( 𝑔)(𝑥) = 𝑥 + 3^
𝑓
𝑥+ 𝑥−2 𝑔(𝑥) =^ 𝑥− 𝑥+1 𝑓𝑖𝑛𝑑^ 𝑓
Solution:
1. Substitute the given values to the equation in fraction form 𝑓
( 𝑔)(𝑥) =^
𝑓(𝑥) 𝑔(𝑥) 𝑓
𝑥+ 𝑥− 𝑥− 𝑥+
2. Try to simplify the expression by factoring and canceling. In this case nothing needs to be simplified. NOTE: For cases such as these, change the operation from division to multiplication and find the reciprocal of the denominator. From this, 𝑓
𝑥+ 𝑥− 𝑥− 𝑥+ To this: 𝑓
( 𝑔)(𝑥) =^
𝑥+ 𝑥−2 •^ 𝑥+ 𝑥− 𝑓
( 𝑔)(𝑥) =^
(𝑥+3)(𝑥+1) (𝑥−2)(𝑥−1) Nothing can be simplified at this point so this will be our final answer or you can FOIL the top and bottom terms. ANSWER: 𝑓
( 𝑔)(𝑥) =^
𝑥^2 +4𝑥+ 𝑥 2 −3𝑥+
E. COMPOSITION OF FUNCTIONS
- The composition of two functions 𝑓(𝑥) and 𝑔(𝑥) is denoted by(𝑓 ◦ 𝑔)(𝑥) Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
Polynomial Function
- A polynomial function p of degree n is a function that can be written in the form
- where 𝑎 0 , 𝑎 1 ,... , 𝑎 n (^) ∈ ℝ , 𝑎 n (^) ≠ 0is a positive integer. Each summand is a term of the polynomial function. The constants 𝑎 0 , 𝑎 1 , 𝑎 2 ,.. ., 𝑎n are the coefficients. The leading coefficient is n. The leading term is^ n 𝑎 𝑎 𝑥n^ and the constant term is 𝑎 0. Rational Function
- A function of the form 𝑓(𝑥) = 𝑝(𝑥) 𝑞(𝑥) where 𝑝(𝑥) and 𝑞(𝑥)are polynomial functions, and 𝑞(𝑥)is not zero (i.e., 𝑞(𝑥) ≠ 0 ). The domain of 𝑓(𝑥)is all values of x where 𝑞(𝑥) ≠ 0. NOTE: 𝑞(𝑥)or the denominator of a Rational Function must not be zero (0) because it will make the function undefined.
A. RATIONAL EXPRESSION
Rational Expression
- Is an expression that can be written as a ratio of two polynomals. Examples: Factors that make an expression NOT be classified as Rational Expression:
- Square roots of variables Example: 𝑥
- Variables as exponents Example: 3 x
- Trigonometric Functions Example: 𝑠𝑖𝑛(𝑥)
B. RATIONAL EQUATION, FUNCTION &
INEQUALITY
Rational Equation
- An equation involving rational expressions.
- The goal is to solve for x that will make the equation true. Examples:
2 𝑥 + 3 =^ 5 𝑥−
𝑥+ 𝑥−1 = 𝑥 3
2 𝑥 −^ 3 2𝑥 =^ 1 5 Rational Inequality
- An Inequality involving rational expressions.
- It may use symbols like <, >, ≤, 𝑜𝑟 ≥ Examples:
5 𝑥−3 ≤^ 2 𝑥
- 6𝑥 − 5 𝑥+3 > 0
𝑥+ 𝑥−3 ≥ 0 Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
- The underline below the expression means “ Equal to” so it can be read as “Greater than or equal to” and “Less than or equal to." It allows the value to be exactly the same as the value on the other side of the inequality. Rational Function
- A function of the form of 𝑓(𝑥) = 𝑝(𝑥) 𝑞(𝑥) where 𝑝(𝑥) and 𝑞(𝑥)are polynomial functions, and 𝑞(𝑥)is not zero function. Examples:
- 𝑓(𝑥) = or 𝑥^2 +2𝑥+ 𝑥+1 𝑦 =^ 𝑥^2 +2𝑥+ 𝑥+
- 𝑦 = 5𝑥 3 − 2𝑥 + 1
C. UNDERSTANDING AND SOLVING
RATIONAL EQUATION
Rational Equation
- Is an equation whose terms are rational expressions. Examples:
3 𝑥−2 = 1
𝑥− 𝑥−1 = 2 Rational Expression
- Is a fraction whose numerator and denominator are both polynomials. Examples:
3 𝑥−
𝑥− 𝑥−
𝑥^2 + 𝑥+ Least Common Denominator (LCD)
- Is the least common multiple of the denominators. Examples:
4 𝑥^2 −
𝑥− 𝑥^2 +3𝑥+ The LCD of the rational expressions is (𝑥 − 2)(𝑥 + 2)(𝑥 + 1) (𝑥 + 2) 𝑎𝑛𝑑 (𝑥 + 1) are the factored out form of 𝑥 2
- 3𝑥 + 2 Solutions or Roots
- Values (being substituted to x) that satisfy a given rational equation. Example: The value 𝑥 = 3is a solution of the rational expression 6 𝑥−1 = 3 Explanation: If we substitute 3 for the variable x, it will make the expression. And 6 4−1 = 3 by solving it, we will end up with 3 = 3. Therefore 𝑥 = 3satisfies the expression. Extraneous Solutions
- Values that arrive upon solving a rational equation but do not satisfy the given equation. Example: In solving , x will be equal to 2 and 5𝑥− 𝑥^2 −
- However, 2 is an extraneous solution since Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
Therefore, -4 is the solution or root of the given equation. ANSWER: − 4
4 𝑦+4 =^ 𝑦 3 Solution:
1. Find the Least Common Denominator (LCD). The denominators are 𝑦 + 4 𝑎𝑛𝑑 3. Both of these expressions are completely factored already. Thus, the LCD of the given equation is 3(y + 4). 2. Multiply both sides of the equation by the LCD to remove the denominators. 4 𝑦+4 =^ 𝑦 3 (3)(𝑦 + 4) ( 4 𝑦+4 ) = (^ 𝑦 3 )(3)(𝑦 + 4) Now, divide the LCD by each denominator: For ( , divide by , 4 𝑦+4 )^ (3)(𝑦 + 4)^ (𝑦 + 4) leaving 3. Multiply 3 to the numerator. For ( , divide by 3, leaving 𝑦 3 )^ (3)(𝑦 + 4) (𝑦 + 4). Multiply (𝑦 + 4)to the numerator. The result will be: 3(4) = (𝑦 + 4)(𝑦) 12 = 𝑦 2
- 4𝑦 3. Solve for the unknown variable. 12 = 𝑦 2
- 4𝑦 Transpose the values to one side. − 𝑦 2 − 4𝑦 + 12 = 0 Multiply the entire equation by -1 to make the leading coefficient (-y^2 ) positive. It will also change the sign of all terms. (− 1)(− 𝑦 2 − 4𝑦 + 12) = 0 (− 1) 𝑦 2
- 4𝑦 − 12 = 0 Factor out the equation (𝑦 + 6)(𝑦 − 2) = 0 Hence, 𝑦 + 6 = 0 𝑜𝑟 𝑦 − 2 = 0 𝑦 =− 6 𝑜𝑟 𝑦 = 2 4. Verify your answer (solution and roots) by substituting it to the variable for the original equation. Then, simplify. For y = -6, 4 𝑦+4 =^ 𝑦 3 →^ 4 −6+4 =^ − 3 →^ 4 −2 =^ − 3 − 2 =− 2 ✓ Since it satisfies the equation, 𝑦 =− 6is a solution or root of the given equation. For y = 2, 4 𝑦+4 =^ 𝑦 3 →^ 4 2+4 =^ 2 3 →^ 4 6 =^ 2 3 ✓ 2 3 =^ 2 3 Since it satisfies the equation, 𝑦 = 2is a solution or root of the given equation. Therefore, -6 and 2 are the solution or roots of the given equation. ANSWER: − 6 𝑎𝑛𝑑 2 Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
𝑥+ 𝑥−3 =^ 5 𝑥− Solution:
1. Find the Least Common Denominator (LCD). Both factors has the denominator x-3. This expression is completely factored already. Thus, the LCD of the given equation is x-. 2. Multiply both sides of the equation by the LCD to remove the denominators. 𝑥+ 𝑥−3 =^ 5 𝑥− (𝑥 − 3)( 𝑥+ 𝑥−3 ) = (^ 5 𝑥−3 )(𝑥 − 3) For ( , divide by , leaving 𝑥+ 𝑥−3 )^ (𝑥 − 3)^ (𝑥 − 3)
- Multiply 1 by the numerator (𝑥 + 2), keeping it as (𝑥 + 2). For ( , divide by , leaving 5 𝑥−3 )^ (𝑥 − 3)^ (𝑥 − 3)
- Multiply 1 by the numerator 5, keeping it as
The result will be: 𝑥 + 2 = 5 3. Solve for the unknown variable. 𝑥 + 2 = 5 Transpose 2 to the other side of the equation and change the sign to negative. 𝑥 = 5 − 2 𝑥 = 3 4. Verify your answer (solution and roots) by substituting it to the variable for the original equation. Then, simplify. 𝑥+ 𝑥−3 =^ 5 𝑥−3 →^ 3+ 3−3 =^ 5 3−3 →^ 5 0 =^ 5 0 If you substitute x = 3 back into the original equation, the denominator x-3 becomes zero, which is undefined in rational equations. Therefore x = 3 is an extraneous solution. ANSWER: 3 is an extraneous solution
D. UNDERSTANDING AND SOLVING
RATIONAL INEQUALITY
Rational Inequality
- Uses any of the symbols ≠, ≤, ﹤, ≥, ﹥, and contains at least one rational expression. Example: The expressions is a rational 𝑥− 𝑥+1 ≥ 0 inequality. Inequality
- Shows a comparison between different quantities or expressions using the symbols ≠, ≤, ﹤, ≥, or ﹥ ≠ not equal to ≤ less-than or equal to ﹤less-than ≥ greater-than or equal to ﹥greater-than Example: The expression 2𝑥 − 1 ≥ 0shows a comparison between 2𝑥 − 1 and 0. Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
3. Find the critical values of rational expression on the left-hand side of the inequality. This can be done by writing the numerator and the denominator separately, equating each of them to zero, and solving the resulting equations. For the numerator 𝑥 − 2 = 0,the critical value is 𝑥 = 2. For the denominator 𝑥 + 2 = 0,the critical value is 𝑥 =− 2. The roots of the numerator make the rational expression equal to 0. Thus, these values must be included in the final solution set. The root of the denominator, however, is not included because it will make the rational expression undefined. 4. Use the critical values as bounds to divide the set of real numbers into intervals. Remember the following guidelines: a. The symbols − ∞ and ∞always come up with parentheses () because they cannot possibly be included in any interval of real numbers. b. If the inequality involves the strict inequality symbols ﹥, ≠, or ﹤, all intervals should be enclosed in parentheses because their endpoints cannot possibly become part of the solution set. c. If the inequality involves the non-strict inequality symbols ≥ or ≤, the roots of the numerator should be included in their perspective intervals using the symbols 𝑥] or [𝑥, while the roots of the denominator should be excluded using the symbols 𝑥) or(𝑥. The number line helps you divide the set of real numbers into intervals. Use a shaded circle if the value is included in the solution set, and a hollow circle if not. (The number line below is not drawn to scale). As indicated by the previous guidelines, we use the critical values − 2 and 2 to divide the set of real numbers into intervals. Intervals: (− ∞, − 2), (− 2, 2], [2, ∞). 5. Construct a table of signs for the rational inequality. The top row of the table should contain the intervals from the previous step, while the leftmost column should contain the test point, the factors of the numerator, and the factors of the denominator. Add another row at the bottom for the entries that correspond to the entire rational expression. 6. To fill an entry on the table, choose a convenient number from the corresponding interval on top (do not choose a critical value). Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
Substitute this number into the expression on the left, then simplify. Take the sign of the answer you obtained (the actual value does not matter) and write it on the table. (Never use Critical Values as test points.) − 3 − 2 =− 5 (Negative) − 3 + 2 =− 1 (Negative) 0 − 2 =− 2 (Negative) 0 + 2 = 2 (Positive) 3 − 2 = 1 (Positive) 3 + 2 = 5 (Positive)
7. Determine the sign of the entire rational expression for each interval by multiplying the signs in each column. Write the answers in the last row of the table. Instead of substituting the test point, you can use the sign. For 𝑥− 𝑥+ (−)(−) =+ (−)(+) =− (+)(+) =+ 8. Determine the solution set by forming the union of all intervals and satisfy the inequality. (− ∞, − 2) (− 2, 2] [2, ∞)
Let’s see which of the 3 intervals can satisfy the inequality The given inequality is 𝑥− 𝑥+2 ≥ Observe the inequality symbol, we have the greater than or equal to 0. We need to look for positive integers. Let’s check which of the 3 intervals are positive and whoever are positive, they will be part of the solution to our inequality (− ∞, − 2) (− 2, 2] [2, ∞) ✔ + ✖ − ✔ + Note: if the inequality symbol is less than or equal to 0, we’ll look for 0 or negative integers, and if the inequality symbol is greater than or equal to 0, we’ll look for positive integers. Next, we need determine the solution set by forming the union of all intervals that satisfy the inequality Therefore, the solution of the inequality is ANSWER: (− ∞, − 2) U [2, ∞) Another method may be used in solving rational inequalities. We call this as the method of test values. The steps 1-4 of the abovementioned method are just the same.
- Choose a convenient test value for each interval. Substitute each of the chosen values to the given rational inequality and simplify. Disclaimer : This reviewer is only verified by one teacher in General Mathematics however, you may still use this reviewer but please do not depend fully on its contents. If ever you feel that what is written was not discussed, please do clarify with your subject teachers.
