General Mathematics Quarter 1 – Module 2: Evaluating Functions General Mathematics Alternative Delivery Mode Quarter 1 – Module 2: Evaluating Functions First Edition, 2020 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio Development Team of the Module Writer: Rey Mark R. Queaño Editors: Elizabeth B. Dizon, Anicia J. Villaruel, Roy O. Natividad Reviewers: Fritz A. Caturay, Necitas F. Constante, Dexter M. Valle, Jerome A. Chavez Illustrator: Dianne C. Jupiter Layout Artist: Noel Rey T. Estuita Management Team: Wilfredo E. Cabral, Job S. Zape Jr., Eugenio S. Adrao, Elaine T. Balaogan, Hermogenes M. Panganiiban, Babylyn M. Pambid, Josephine T. Natividad, Anicia J. Villaruel, Dexter M. Valle Printed in the Philippines by ________________________ Department of Education – Region IV-A CALABARZON Office Address: Telefax: E-mail Address: Gate 2 Karangalan Village, Barangay San Isidro Cainta, Rizal 1800 02-8682-5773/8684-4914/8647-7487 region4a@deped.gov.ph General Mathematics Quarter 1 – Module 2: Evaluating Functions Introductory Message For the facilitator: Welcome to Grade 11 General Mathematics Alternative Delivery Mode (ADM) Module on Evaluating Functions! This module was collaboratively designed, developed and reviewed by educators from public institutions to assist you, the teacher or facilitator in helping the learners meet the standards set by the K to 12 Curriculum while overcoming their personal, social, and economic constraints in schooling. This learning resource hopes to engage the learners into guided and independent learning activities at their own pace and time. Furthermore, this also aims to help learners acquire the needed 21st century skills while taking into consideration their needs and circumstances. In addition to the material in the main text, you will also see this box in the body of the module: Notes to the Teacher This contains helpful tips or strategies that will help you in guiding the learners. As a facilitator you are expected to orient the learners on how to use this module. You also need to keep track of the learners' progress while allowing them to manage their own learning. Furthermore, you are expected to encourage and assist the learners as they do the tasks included in the module. For the learner: Welcome to the General Mathematics Grade 11 Alternative Delivery Mode (ADM) Module on Evaluating Functions! The hand is one of the most symbolized part of the human body. It is often used to depict skill, action and purpose. Through our hands we may learn, create and accomplish. Hence, the hand in this learning resource signifies that you as a learner is capable and empowered to successfully achieve the relevant competencies and skills at your own pace and time. Your academic success lies in your own hands! This module was designed to provide you with fun and meaningful opportunities for guided and independent learning at your own pace and time. You will be enabled to process the contents of the learning resource while being an active learner. This module has the following parts and corresponding icons: iii What I Need to Know This will give you an idea of the skills or competencies you are expected to learn in the module. What I Know This part includes an activity that aims to check what you already know about the lesson to take. If you get all the answers correct (100%), you may decide to skip this module. What’s In This is a brief drill or review to help you link the current lesson with the previous one. What’s New In this portion, the new lesson will be introduced to you in various ways such as a story, a song, a poem, a problem opener, an activity or a situation. What is It This section provides a brief discussion of the lesson. This aims to help you discover and understand new concepts and skills. What’s More This comprises activities for independent practice to solidify your understanding and skills of the topic. You may check the answers to the exercises using the Answer Key at the end of the module. What I Have Learned This includes questions or blank sentence/paragraph to be filled in to process what you learned from the lesson. What I Can Do This section provides an activity which will help you transfer your new knowledge or skill into real life situations or concerns. Assessment This is a task which aims to evaluate your level of mastery in achieving the learning competency. Additional Activities In this portion, another activity will be given to you to enrich your knowledge or skill of the lesson learned. This also tends retention of learned concepts. Answer Key This contains answers to all activities in the module. At the end of this module you will also find: iv References This is a list of all sources used in developing this module. The following are some reminders in using this module: 1. Use the module with care. Do not put unnecessary mark/s on any part of the module. Use a separate sheet of paper in answering the exercises. 2. Don’t forget to answer What I Know before moving on to the other activities included in the module. 3. Read the instruction carefully before doing each task. 4. Observe honesty and integrity in doing the tasks and checking your answers. 5. Finish the task at hand before proceeding to the next. 6. Return this module to your teacher/facilitator once you are through with it. If you encounter any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator. Always bear in mind that you are not alone. We hope that through this material, you will experience meaningful learning and gain deep understanding of the relevant competencies. You can do it! v What I Need to Know This module was designed and written with you in mind. It is here to help you master the key concepts of functions specifically on evaluating functions. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The lessons are arranged to follow the standard sequence of the course. But the order in which you read them can be changed to correspond with the textbook you are now using. After going through this module, you are expected to: 1. recall the process of substitution; 2. identify the various types of functions; and 3. evaluate functions. What I Know Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. Which of the following is a polynomial function? a. f ( x) = 2 x 2 − 10 x + 7 3 c. p ( x) = x − 7 b. g ( x) = 4 x 2 − 3 x + 8 d. s ( x) = 2 m − 1 2. What kind of function is being illustrated by f ( x) = 2 x 3 − 3 x + 5 ? a. Rational Function c. Greatest Integer Function b. Constant Function d. Absolute Value Function 3. Find the function value given h( x) = 17 + 8 x of x = 4d . c. 17 + 32d a. 17 − 32d d. 17 + 32d 4. Which of the following shows a logarithmic function? b. 17 − 32d 2 a. f ( x) = 8 x 3 + 8 b. f ( x) = log 9 81 2 x c. f ( x) = 3 − 6 f ( x) = x − 1 − 8 d. 5. Find the function value given h( x) = 7 x − 11 , if x = 8m + 3 . a. 56m + 10 2 c. 56m + 10 b. 56m − 10 2 d. 56m − 10 1 6. Which of the following is the value of the function f ( x) = 3 x 2 − 15 x + 5 + 3 given x = 3? a. 25 b. 16 c. 19 d. 10 7. Evaluate the function h( x) = x + 31 given x = 2.5. a. 34 b. -34 c. -33 d. 33 8. Give the value of the of the function c( x) = 5 x 3 − 18 at c(3) . a. 117 b. 27 c. 153 d. 63 9. Evaluate: h( x) = 5 x 2 − 8 x + 12 given x = 5. a. 22 b. 145 c. 97 d. -3 10. Find the value of the function h( x) = 5 x − 4 if x = 6 . 2 a. 80 16 c. b. 2 19 d. 4 11. Evaluate the function f ( x) = 3x − 5 x + 2 given x = 2 x + 5 . 2 a. 12 x 2 + 50 x + 52 2 c. 12 x − 50 x + 52 b. 12 x 2 + 65 x + 77 2 d. 12 x − 65 x + 77 12. Given h( x) = 2x 2 − 5 , determine h(5). 3 a. -15 b. − c. 15 5 d. 3 5 3 13. Evaluate the function k ( x) = 5 x if x = a. 3 b. 5 c. 25 d. 14. Given g ( x) = a. 2 . 3 3 25 2 x 2 − 3x + 7 , determine g ( 2) . 3x − 4 9 2 b. − 5 8 c. 7 9 2 d. − 8 7 15. For what values of x can we not evaluate the function f ( x) = a. ±4 b. ±3 c. ±2 d. ±1 2 3x + 7 ? x2 − 4 Lesson 1 Evaluating Functions Finding the value of “x” for most of the students is what Mathematics is all about. Sometimes, it seems to be a joke for the students to evaluate an expression, like what is shown by the illustration. Find x. Here it is! X 6 8 If you want to learn how to find the value of “y”, well then, you are in the right page. WELCOME to your second module! What’s In Before we begin, let’s go back to the time when you first encounter how to evaluate expressions. Do you still remember? Given the following expressions, find its value if x = 3. 1. x − 9 2. 3x + 7 3. x 2 + 4 x − 10 4. 2 x 2 − 6 x + 26 5. 3x 2 − 6 3 We have learned that, in an algebraic expression, letters can stand for numbers. And to find the value of the expression, there are two things that you have to do. 1. Replace each letter in the expression with the assigned value. First, replace each letter in the expression with the value that has been assigned to it. To make your calculations clear and avoid mistakes, always enclose the numbers you're substituting inside parentheses. The value that's given to a variable stays the same throughout the entire problem, even if the letter occurs more than once in the expression. However, since variables "vary", the value assigned to a particular variable can change from problem to problem, just not within a single problem. 2. Perform the operations in the expression using the correct order of operations. Once you've substituted the value for the letter, do the operations to find the value of the expression. Don't forget to use the correct order of operations: first do any operations involving exponents, then do multiplication and division, and finally do addition and subtraction! If in the activity above, you do the same process in order to arrive with these answers, then, this module seems to be very easy to you. Solutions: Given the following expressions, find its value if x = 3. 1. x − 9 Since x = 3, we just replaced x by 3 in the expression, then subtract by 9. = x−9 = (3) − 9 = −6 2. 3x + 7 = 3x + 7 = 3(3) + 7 =9+7 = 16 Following the steps, we just replace x by 3, multiply it by the numerical coefficient 3, then add 7 4 3. x 2 + 4 x − 10 After replacing x by 3, we get the squared of 3 which is 9, add it to the product of 4 and 3, then lastly, we subtracted 10 from its sum. = x 2 + 4 x − 10 = (3) 2 + 4(3) − 10 = 9 + 12 − 10 = 11 4. 2 x 2 − 6 x + 26 Simply each term inside the parenthesis in order to arrive with 18 subtracted by 18 plus 26 = 2 x 2 − 6 x + 26 = 2(3) 2 − 6(3) + 26 = 18 − 18 + 26 = 26 5. 3x 3 − 6 Get the cubed of 3 which is 27, then multiply it to 3 to get 81 then subtract 6 = 3x 3 − 6 = 3(3) 3 − 6 = 3(27) − 6 = 81 − 6 = 75 Types of Functions What’s New Before you proceed to this module, try to look and analyze some of the common types of functions that you might encounter as you go on with this module. Types of Function Constant Function Identity Function Description Example A constant function is a function that has the same output value no matter what your input value is. Because of this, a constant function has the form f ( x) = b , where b is a constant (a single value that does not change). The identity function is a function which returns the same value, which was used as its argument. In other words, the y=7 5 f (2) = 2 Polynomial Function identity function is the function f ( x) = x , for all values of x. A polynomial function is defined by y = a 0 + a1 x + a 2 x 2 + ... + a n x n , where n is a 0 ✓ Linear Function ✓ Quadratic Function ✓ Cubic Function Power Function Rational Function Logarithmic Function p( x) in which numerator, p(x) and q( x) denominator, q(x) are polynomial functions of x, where q(x) ≠ 0. These are functions of the form: y = ab x , where x is in an exponent and a and b are constants. (Note that only b is raised to the power x; not a.) If the base b is greater than 1 then the result is exponential growth. Logarithmic functions are the inverses of exponential functions, and any exponential function can be expressed in logarithmic form. Logarithms are very useful in permitting us to work with very large numbers while manipulating numbers of a much more manageable size. It is written in the form y = log b x Absolute Value Function 2 A rational function is any function which can be represented by a rational fraction say, Exponential function 1 non-negative integer and a , a , a ,…, n ∈ R. The polynomial function with degree one. It is in the form y = mx + b If the degree of the polynomial function is two, then it is a quadratic function. It is expressed as y = ax 2 + bx + c , where a ≠ 0 and a, b, c are constant and x is a variable. A cubic polynomial function is a polynomial of degree three and can be denoted by f ( x) = ax 3 + bx 2 + cx + d , where a ≠ 0 and a, b, c, and d are constant & x is a variable. A power function is a function in the form y = ax b where b is any real constant number. Many of our parent functions such as linear functions and quadratic functions are in fact power functions. x 0, where b 0 and b 1 The absolute value of any number, c is represented in the form of |c|. If any 6 y = 2x + 5 y = 3x 2 + 2 x + 5 y = 5 x 3 + 3x 2 + 2 x + 5 f ( x) = 8 x 5 x 2 − 3x + 2 f ( x) = x2 − 4 y = 2x y = log 7 49 function f: R→ R is defined by f ( x) = x , it is known as absolute value function. For each non-negative value of x, f(x) = x and for each negative value of x, f(x) = -x, i.e., f(x) = {x, if x ≥ 0; – x, if x < 0. If a function f: R→ R is defined by f(x) = [x], x ∈ X. It round-off to the real number to the integer less than the number. Suppose, the given interval is in the form of (k, k+1), the value of greatest integer function is k which is an integer. Greatest Integer Function y = x−4 +2 f ( x) = x + 1 where x is the greatest integer function What is It Evaluating function is the process of determining the value of the function at the number assigned to a given variable. Just like in evaluating algebraic expressions, to evaluate function you just need to a.) replace each letter in the expression with the assigned value and b.) perform the operations in the expression using the correct order of operations. Look at these examples! Example 1: Given f ( x) = 2 x − 4 , find the value of the function if x = 3. Solution: f (3) = 2(3) − 4 f (3) = 6 − 4 f (3) = 2 Answer: Given ✓ Substitute 3 for x in the function. ✓ Simplify the expression on the right side of the equation. f ( x) = 2 x − 4 , f (3) = 2 Example 2: Given g ( x) = 3x 2 + 7 , find g (−3) . Solution: g (−3) = 3(−3) 2 + 7 g (−3) = 3(9) + 7 g (−3) = 27 + 7 g (−3) = 34 7 ✓ Substitute -3 for x in the function. ✓ Simplify the expression on the right side of the equation. Answer: Given g ( x) = 3x 2 + 7 , g (−3) = 34 Example 3: Given p( x) = 3x 2 + 5 x − 2 , find p(0) and p(−1) . Solution: p(0) = 3(0) 2 + 5(0) − 2 p(0) = 3(0) + 0 − 2 p(0) = 0 + 0 − 2 p(0) = −2 Treat each of these like two separate problems. In each case, you substitute the value in for x and simplify. Start with x = 0, then x=-1. p(0) = 3(−1) 2 + 5(−1) − 2 p(0) = 3(1) − 5 − 2 p(0) = 3 − 5 − 2 p(0) = −4 Answer: Given p( x) = 3x 2 + 5 x − 2 , Example 4: Given p(0) = −2 , p(−1) = −4 f ( x) = 5x + 1 , find f (h + 1) . ✓ Solution: ✓ f (h + 1) = 5(h + 1) + 1 f (h + 1) = 5h + 5 + 1 f (h + 1) = 5h + 6 Answer: Given This time, you substitute (h + 1) into the equation for x. Use the distributive property on the right side, and then combine like terms to simplify. f ( x) = 5x + 1 , f (h + 1) = 5h + 6 Example 5: Given g ( x) = 3x − 2 , find g (9) . Solution: g (9) = 3(9) − 2 ✓ Substitute 9 for x in the function. g (9) = 27 − 2 ✓ Simplify the expression on the right side of the equation. g (9) = 25 g (9) = 5 Answer: Given g ( x) = 3x − 2 , Example 6: Given h( x) = g (9) = 5 4x + 8 , find the value of function if x = −5 2x − 4 Solution: 8 4(−5) + 8 2(−5) − 4 − 20 + 8 h(−5) = − 10 − 4 − 12 h(−5) = − 14 6 h(−5) = 7 h(−5) = Answer: Given h( x) = ✓ Substitute -5 for x in the function. ✓ Simplify the expression on the right side of the equation. (recall the concepts of integers and simplifying fractions) 4x + 8 6 , h(−5) = 2x − 4 7 Example 7: Evaluate f ( x) = 2 x if x = 3 . 2 Solution: 3 2 3 f =2 2 3 f = 23 2 3 f = 8 2 3 f = 4•2 2 3 f =2 2 2 3 for x in the function. 2 ✓ Substitute ✓ Simplify the expression on the right side of the equation. (get the cubed of 2 which is 8, then simplify) 3 2 Answer: Given f ( x) = 2 x , f = 2 2 Example 8: Evaluate the function h( x) = x + 2 where x is the greatest integer function given x = 2.4 . Solution: h(2.4) = 2.4 + 2 ✓ Substitute 2.4 for x in the function. ✓ Simplify the expression on the right side of the equation. (remember that in greatest integer function, value was rounded-off to the real number to the integer less than the number) h(2.4) = 2 + 2 h(2.4) = 4 Answer: Given h( x) = x + 2 , h(2.4) = 4 9 Example 9:Evaluate the function f ( x) = x − 8 where x − 8 means the absolute value of x − 8 if x = 3 . Solution: f (3) = 3 − 8 ✓ Substitute 3 for x in the function. ✓ Simplify the expression on the right side of the equation. (remember that any number in the absolute value sign is always positive) f (3) = − 5 f (3) = 5 Answer: Given f ( x) = x − 8 , f (3) = 5 Example 10: Evaluate the function f ( x) = x 2 − 2 x + 2 at f (2 x − 3) . Solution: f (2 x − 3) = (2 x − 3) 2 − 2(2 x − 3) + 2 f (2 x − 3) = (4 x 2 − 12 x + 9) − 4 x + 6 + 2 f (2 x − 3) = 4 x 2 − 12 x + 9 − 4 x + 6 + 2 f (2 x − 3) = 4 x 2 − 12 x − 4 x + 9 + 6 + 2 ✓ Substitute 2 x − 3 for x in the function. ✓ Simplify the expression on the right side of the equation. f (2 x − 3) = 4 x 2 − 16 x + 17 What’s More Your Turn! Independent Practice 1: Fill Me Evaluate the following functions by filling up the missing parts of the solution. 1. f ( x) = 3x − 5 , find f (2) Solution: f (2) = ___________________ f (2) = 6 − 5 f (2) = ___________________ 2. g ( x) = 3 2 x , find g(6) Solution: 10 g (6) = _________________ g (6) = 312 g (6) = _________________ 3. k (a) = a − 2 , find k (−9) Solution: k (−9) = ______________ k (−9) = 9 − 2 k (−9) = ______________ 4. p(a) = −4a − 2 , find p(2a) Solution: p(2a) = ______________ p(2a) = ______________ 5. g (t ) = t 2 − 2 , find g (−2) Solution: g (−2) = ________________ g (−2) = ________________ g (−2) = ________________ Independent Assessment 1: Evaluate! Evaluate the following functions. Write your answer and complete solution on separate paper. 1. Given w(n) = n − 1, find the value of the function if w = -1. 2. Given f ( x) = x − 3 , find f (9.3) . 3. Evaluate the function w( x) = − 2 x + 3 if x = -1. 4. Evaluate: 5. Given f ( x) = − x − 1 , find f (a 2 ) f ( x) = 4 x − 5 , find f (2 x + 3) Independent Practice 2: TRUE or SOLVE! 11 Analyze the following functions by evaluating its value. Write TRUE of the indicated answer and solution is correct, if not, rewrite the solution to arrive with the correct answer on the space provided. 1. Evaluate f (t ) = 2t − 3 ; f (t 2 ) Solution: Answer: f (t 2 ) = 2(t 2 ) − 3 f (t 2 ) = 2t 2 − 3 2. Given the function g ( x) = 5x − 13 , find g (9) . Solution: g (9) = 5(9) − 13 Answer: g (9) = 45 − 13 g (9) = 32 g (9) = 16 2 3. Given the function f ( x) = 5x − 7 , find the value of the function if x = −3 . 3x − 2 Solution: 5(−3) − 7 3(−3) − 2 − 15 − 7 f (−3) = −9−2 22 f (−3) = 11 f (−3) = −2 f (−3) = Answer: 4. Evaluate the function f ( x) = x 2 − 3x + 5 at f (3x − 1) . Solution: f (3x − 1) = (3x − 1) 2 − 3x + 5 f (3x − 1) = 9 x 2 − 6 x + 1 − 3x + 5 f (3x − 1) = 9 x 2 − 9 x + 6 5. Evaluate: g ( x) = 3 x if x = 4 3 Solution: 12 Answer: 4 4 g = 3 3 3 4 g = 3 34 3 4 g = 3 81 3 4 g = 3 27 • 3 3 4 g = 33 3 3 Answer: Independent Assessment 2: Find my Value! Evaluate the following functions. Write your solution on a separate paper. 1. g ( x) = 5x − 7 ; g ( x 2 + 1) Answer: _______________________ 2. h(t ) = x 2 + 2 x + 4 ; h(2) Answer: _______________________ 3. k ( x) = 3x 2 − 1 ; k (−3) 2x + 4 Answer: _______________________ 4. f ( x) = 2 x 2 + 5 x − 9 ; f (5x − 2) Answer: _______________________ 5. g ( p) = 4 x ; x = 3 2 Answer: _______________________ What I Have Learned A. Complete the following statements to show how you understood the different types of functions. Answer using your own words, 1. A polynomial function is _______________________________________________________ _________________________________________________________________________________. 2. An exponential function _______________________________________________________ _________________________________________________________________________________. 13 3. A rational function ____________________________________________________________ _________________________________________________________________________________. 4. An absolute value function ____________________________________________________ _________________________________________________________________________________. 5. A greatest integer function ____________________________________________________ _________________________________________________________________________________. B. Fill in the blanks to show how we evaluate functions. Evaluating function is the process of ___________________________ of the function at the _________________ assigned to a given variable. Just like in evaluating algebraic expressions, to evaluate function you just need to ________________________________ in the expression with the assigned value, then _________________________________ in the expression using the correct order of operations. Don’t forget to _______________________ your answer. What I Can Do In this part of the module, you will apply your knowledge on evaluating functions in solving real-life situations. Write your complete answer on the given space. 1. Mark charges ₱100.00 for an encoding work. In addition, he charges ₱5.00 per page of printed output. a. Find a function f(x) where x represents the number page of printed out. b. How much will Mark charge for 55-page encoding and printing work? 2. Under certain circumstances, a virus spreads according to the function: P(t ) = 1 1 + 15(2.1) −0.3t Where where P(t) is the proportion of the population that has the virus (t) days after the acquisition of virus started. Find p(4) and p(10), and interpret the results. 14 Assessment Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. Which of the following is not a polynomial function? a. f ( x) = 2 x − 10 b. g ( x) = 4 x 2 − 3x + 8 c. p( x) = x 3 − 7 d. s ( x) = 3 x − 4 − 9 2. What kind of function is being illustrated by f ( x) = a. b. c. d. 3x − 11 ? x+7 Rational Function Constant Function Greatest Integer Function Absolute Value Function 3. Find the function value given h( x) = 9 − 5x of x = 3m . a. 9 − 15m b. 9 − 15m c. 9 + 15m 2 d. 9 + 15m 4. Which of the following shows an exponential function? 2 a. f ( x) = 3 x + 8 b. f ( x) = 2 x 3 − 7 c. f ( x) = 3 x − 6 d. f ( x) = x − 8 5. Find the function value given a. b. c. d. h( x) = 3x − 8 , if x = 9a + 1 . 27 a + 5 27 a − 5 18a + 11 18a − 11 6. Which of the following is the value of the function f ( x) = 4 x − 8 + 2 given x = 2? 2 a. b. c. d. 8 9 10 11 15 7. Evaluate the function h( x) = x − 11 given x = 3.5. a. b. c. d. -8 8 -9 9 8. Give the value of the of the function c( x) = 3x 2 − 36 at a. b. c. d. c(5) . -21 14 111 39 9. Evaluate: h( x) = 5 x 3 − 3x + 9 given x = 3. a. b. c. d. 45 63 135 153 10. Find the value of the function a. f ( x) = 2 x 2 + 3 if x = 6. 75 b. 5 3 c. 15 d. 2 3 11. Evaluate the function f ( x) = 2 x 2 − 3x + 1 given x = 3x − 5 . a. f (3x − 5) = 18 x 2 − 69 x + 66 b. f (3x − 5) = 18 x 2 − 63 x + 51 c. f (3x − 5) = 18 x 2 + 69 x − 66 d. f (3x − 5) = 18 x 2 + 63 x − 51 12. Given g(x) = x2 − 3 , determine g(5). 2 a. 11 b. 7 2 c. -11 d. − 7 2 13. Evaluate the function g ( x) = 3 x if x = a. b. 3 5 . 3 243 243 c. 9 3 d. 33 9 16 14. Given g ( x) = a. x 2 − 2x + 5 , determine g (4) . x+3 5 7 5 7 13 c. 7 13 d. − 7 b. − 15. For what values of x can we not evaluate the function f ( x) = a. b. c. d. x+4 ? x2 − 9 ±4 ±3 ±2 ±1 Additional Activities Difference Quotient f ( x + h) − f ( x) this quantity is called difference quotient. Specifically, the difference h quotient is used in the discussion of the rate of change, a fundamental concept in calculus. Example: Find the difference quotient for each of the following function. A. f(x) = 4x - 2 B. f(x) = x2 Solution: A. f(x) = 4x - 2 f ( x + h) = 4( x + h) − 2 = 4 x + 4h − 2 f ( x + h) − f ( x) 4 x + 4h − 2 − (4 x − 2) = h h 4 x + 4 h − 2 − 4 x + 2) = h 4h = h =4 17 B. f(x) = x2 f ( x + h) = ( x + h) 2 = x 2 + 2hx + h 2 f ( x + h) − f ( x) x 2 + 2hx + h 2 − ( x) 2 = h h 2 2 2 x + 2hx + h − ( x) = h 2 2hx + h = h = 2x + h YOUR TURN! Find the value of f ( x + h) − f ( x) , h ≠ 0 for each of the following function. h 1. f ( x) = 3x + 4 2. g ( x) = x 2 + 3 18 What I Know 1. A 2. D 3. C 4. B 5. A 6. B 7. D 8. A 9. C 10.B 11.A 12.C 13.D 14.A 15.C 19 What's More Assessment Independent Practice 1 1. D 2. A 3. A 4. C 5. B 6. C 7. A 8. D 9. C 10.B 11.A 12.A 13.D 14.C 15.B 1. 2. 3. 4. 5. Independent Assessment 1 1. -2 2. 6 3. 5 Independent Assessment 2 Independent Practice 2 1. TRUE 4. 5. 2. 3. 2 4. 5. TRUE 1. 2. 3. -13 4. 5. 8 Answer Key References Books: CHED. General Mathematics Learner's Materials. Pasig City: Department of Education - Bureau of Learning Resources, 2016. Orines, Fernando B. Next Cantury Mathematics 11. Quezon City: Phoenix Publishing House, 2016. Oronce, Orlando A. General Mathematics, 1st Ed. Quezon City: Rex Book Store Inc., 2016. Online Sources: http://www.math.com/school/subject2/lessons/S2U2L3DP.html) https://www.toppr.com/guides/maths/relations-and-functions/types-offunctions/ 20 For inquiries or feedback, please write or call: Department of Education - Bureau of Learning Resources (DepEd-BLR) Ground Floor, Bonifacio Bldg., DepEd Complex Meralco Avenue, Pasig City, Philippines 1600 Telefax: (632) 8634-1072; 8634-1054; 8631-4985 Email Address: blr.lrqad@deped.gov.ph * blr.lrpd@deped.gov.ph