General Mathematics Quarter 1 Module 2: Evaluating Functions • DepEd Tambayan

Q: What grade and quarter is this General Mathematics module for?

It is a Senior High School General Mathematics module for Quarter 1, Module 2. The module does not print a specific grade level.

Q: What does evaluating a function mean?

The module defines it as finding the value of a function at the number assigned to a given variable. You replace the variable with the given value, then simplify using the correct order of operations.

Q: What types of functions does the module cover?

It covers constant, identity, polynomial (linear, quadratic, cubic), power, rational, exponential, logarithmic, absolute value, and greatest integer functions.

Q: Does the module have an answer key?

Yes. It includes answers for the pretest, the practice and assessment activities, and the post-test.

Q: How do you evaluate a greatest integer function?

Round the number down to an integer, then finish the computation. For example, the module shows that for h(x) = ‖x‖ + 2, h(2.4) = 2 + 2 = 4.

This is the General Mathematics Quarter 1 Module 2 self-learning module (SLM) for Senior High School, titled “Evaluating Functions.” It is an Alternative Delivery Mode module from DepEd Region 4-A CALABARZON (First Edition, 2021), written for learners studying at home or in class. After working through it, learners can name the common types of functions and find the value of a function for any given input, including expressions such as h + 1 or 2x − 3.

On this page 9 sections

What learners will learn

The module states that after going through it, learners are expected to:

recall the process of substitution;
identify the various types of functions; and
evaluate functions.

genmath_q1_mod2_evaluatingfunctions_v2

What the module covers

The module has one lesson, Lesson 1: Evaluating Functions. It moves through these parts:

What I Know: a 15-item multiple-choice pretest.
What’s In: a review of evaluating algebraic expressions when x = 3.
What’s New: a table of the common types of functions.
What is It: the meaning of evaluating a function and ten worked examples.
What’s More: four practice sets (Fill Me, Evaluate!, TRUE or SOLVE!, and Find my Value!).
What I Have Learned: sentence-completion items to sum up the lesson.
What I Can Do: two real-life problems.
Assessment: a 15-item multiple-choice post-test.
Additional Activities: the difference quotient.
Answer Key.

Review: evaluating an expression

The lesson starts with something learners already know. To find the value of an algebraic expression, you do two things:

Replace each letter with its assigned value. Put the number in parentheses to keep your work clear. The value stays the same everywhere the letter appears in that problem.
Perform the operations in the correct order. Do exponents first, then multiplication and division, and finally addition and subtraction.

For example, with x = 3, the expression x² + 4x − 10 becomes (3)² + 4(3) − 10 = 9 + 12 − 10 = 11.

Types of functions

The module introduces these types of functions, each with an example.

Type	Form or description	Example from the module
Constant	Same output for every input; f(x) = b	y = 7
Identity	Returns the input value; f(x) = x	f(2) = 2
Linear	Polynomial of degree one; y = mx + b	y = 2x + 5
Quadratic	Polynomial of degree two; y = ax² + bx + c, a ≠ 0	y = 3x² + 2x + 5
Cubic	Polynomial of degree three; f(x) = ax³ + bx² + cx + d, a ≠ 0	y = 5x³ + 3x² + 2x + 5
Power	y = axᵇ, where b is any real constant	f(x) = 8x⁵
Rational	p(x)/q(x), both polynomials, q(x) ≠ 0	f(x) = (x² − 3x + 2)/(x² − 4)
Exponential	The variable is in the exponent; growth when the base is greater than 1	y = 2ˣ
Logarithmic	Inverse of the exponential function; y = log_b x, x > 0, b > 0, b ≠ 1	y = log₇ 49
Absolute value	f(x) = \|x\|; equals x when x ≥ 0 and −x when x < 0	y = \|x − 4\| + 2
Greatest integer	Rounds the number down to an integer	f(x) = ‖x‖ + 1

How to evaluate a function

The module defines evaluating a function as finding the value of the function at the number assigned to a given variable. You use the same two steps as for expressions: substitute, then simplify using the order of operations. The worked examples include:

Linear: f(x) = 2x − 4, so f(3) = 2(3) − 4 = 2.
Negative input: g(x) = 3x² + 7, so g(−3) = 3(9) + 7 = 34.
Expression as input: f(x) = 5x + 1, so f(h + 1) = 5(h + 1) + 1 = 5h + 6. Use the distributive property, then combine like terms.
Square root: g(x) = √(3x − 2), so g(9) = √25 = 5.
Rational: h(x) = (4x + 8)/(2x − 4), so h(−5) = −12/−14 = 6/7.
Fractional exponent: f(x) = 2ˣ, so f(3/2) = √(2³) = √8 = 2√2.
Greatest integer: h(x) = ‖x‖ + 2, so h(2.4) = 2 + 2 = 4.
Absolute value: f(x) = |x − 8|, so f(3) = |−5| = 5.
Binomial input: f(x) = x² − 2x + 2, so f(2x − 3) = 4x² − 16x + 17.

Real-life problems and the difference quotient

In What I Can Do, learners write a function for an encoding job: Mark charges ₱100.00 plus ₱5.00 per printed page. They then find his charge for 55 pages. A second problem uses a function for the spread of a virus, where learners find P(4) and P(10) and interpret the results.

The Additional Activities introduce the difference quotient, [f(x + h) − f(x)]/h, which the module says is used in studying rate of change in calculus. It shows that for f(x) = 4x − 2 the difference quotient is 4, and for f(x) = x² it is 2x + h. Learners then try f(x) = 3x + 4 and f(x) = x² + 3.

How to use this module

Learners: Answer the What I Know pretest first to check what you already know. Then go through the parts in order and study the worked examples before trying the practice sets. Write all answers on a separate sheet of paper, not on the module. Finish with the Assessment. The module provides answer keys for each activity and test, and asks you to use them honestly.
Parents: Help your child keep to the order of the parts. Encourage them to check answers only after finishing each activity. If they have trouble, contact their teacher.
Teachers: Use the pretest and post-test, which have the same format, to measure progress. The What I Can Do problems can serve as a performance task.

Frequently asked questions

What grade and quarter is this General Mathematics module for?

It is a Senior High School General Mathematics module for Quarter 1, Module 2. The module does not print a specific grade level.

What does evaluating a function mean?

The module defines it as finding the value of a function at the number assigned to a given variable. You replace the variable with the given value, then simplify using the correct order of operations.

What types of functions does the module cover?

It covers constant, identity, polynomial (linear, quadratic, cubic), power, rational, exponential, logarithmic, absolute value, and greatest integer functions.

Does the module have an answer key?

Yes. It includes answers for the pretest, the practice and assessment activities, and the post-test.

How do you evaluate a greatest integer function?

Round the number down to an integer, then finish the computation. For example, the module shows that for h(x) = ‖x‖ + 2, h(2.4) = 2 + 2 = 4.

Full text of the module

The text below is a transcription of the PDF above, as printed.

General Mathematics Quarter 1 – Module 2: Evaluating Functions

Senior High School

[illustration: students reading and studying outdoors and indoors, with a teacher, a globe, and a student working by lamp light]

CO_Q1_General Mathematics SHS Module 2

General Mathematics Alternative Delivery Mode Quarter 1 – Module 2: Evaluating Functions First Edition, 2021

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, Azalea Gallano and Dianne Arlyn C. Pastrana

Illustrator: Dianne C. Jupiter

Layout Artist: Sayre M. Dialola and Noel Rey T. Estuita

Management Team: Francis Cesar B. Bringas, Job S. Zape, Jr., Ramonito Elumbaring, Reicon C. Condes, Elaine T. Balaogan, Fe M. Ong-ongowan, Hermogenes M. Panganiban, Phillip B. Gallendez, Josephine T. Natividad, Anicia J. Villaruel, Dexter M. Valle

Printed in the Philippines by ________________________

Department of Education – Region 4-A CALABARZON

Office Address: Gate 2 Karangalan Village, Brgy. San Isidro, Cainta, Rizal Telefax: 02-8682-5773/8684-4914/8647-7487 E-mail Address: lrmd.calabarzon@deped.gov.ph

General Mathematics Quarter 1 – Module 2: Evaluating Functions

Senior High School

Introductory Message

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

Thank you.

What I Need to Know

[illustration: hand pointing with a finger]

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:

recall the process of substitution;
identify the various types of functions; and
evaluate functions.

What I Know

[illustration: hand holding a pencil]

Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper.

Which of the following is a polynomial function?
a. f(x) = 2x² − 10x + 7
b. g(x) = 4|x² − 3x + 8|
c. p(x) = ‖x³ − 7‖
d. s(x) = 2^m − 1
What kind of function is being illustrated by f(x) = |2x³ − 3x| + 5?
a. Rational Function
b. Constant Function
c. Greatest Integer Function
d. Absolute Value Function
Find the function value given h(x) = 17 + 8x of x = 4d.
a. 17 − 32d
b. 17 − 32d²
c. 17 + 32d
d. 17 + 32d²
Which of the following shows a logarithmic function?
a. f(x) = 8x³ + 8
b. f(x) = log₉ 81
c. f(x) = |3^x − 6|
d. f(x) = ‖x − 1‖ − 8
Find the function value given h(x) = 7x − 11, if x = 8m + 3.
a. 56m + 10
b. 56m − 10
c. 56m² + 10
d. 56m² − 10
Which of the following is the value of the function f(x) = |3x² − 15x + 5| + 3 given x = 3?
a. 25
b. 16
c. 19
d. 10
Evaluate the function h(x) = ‖x‖ + 31 given x = 2.5.
a. 34
b. -34
c. -33
d. 33
Give the value of the of the function c(x) = 5x³ − 18 at c(3).
a. 117
b. 27
c. 153
d. 63
Evaluate: h(x) = 5x² − 8x + 12 given x = 5.
a. 22
b. 145
c. 97
d. -3
Find the value of the function h(x) = √(5x² − 4) if x = 4.
a. √80
b. 2√19
c. √16
d. 4
Evaluate the function f(x) = 3x² − 5x + 2 given x = 2x + 5.
a. 12x² + 50x + 52
b. 12x² + 65x + 77
c. 12x² − 50x + 52
d. 12x² − 65x + 77
Given h(x) = (2x² − 5)/3, determine h(5).
a. -15
b. −5/3
c. 15
d. 5/3
Evaluate the function k(x) = 5^x if x = 2/3.
a. ∛5
b. √25
c. √5
d. ∛25
Given g(x) = (2x² − 3x + 7)/(3x − 4), determine g(2).
a. 9/2
b. −9/2
c. 8/7
d. −8/7
For what values of x can we not evaluate the function f(x) = (3x + 7)/(x² − 4)?
a. ±4
b. ±3
c. ±2
d. ±1

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.

[illustration: right triangle with legs labeled 6 and 8 and the hypotenuse labeled x]

Here it is!

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!

Lesson 1 Evaluating Functions

What’s In

[illustration: hand holding a puzzle piece]

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.

x − 9
3x + 7
x² + 4x − 10
2x² − 6x + 26
3x³ − 6

If in the first activity, you do the same process in order to arrive with these answers, then this module would be very easy for you.

Solutions:

Given the following expressions, find its value if x = 3.

x − 9
= x − 9
= (3) − 9
= −6

Since x = 3, replace x by 3 in the expression, then subtract 9.
3x + 7
= 3x + 7
= 3(3) + 7
= 9 + 7
= 16

Following the steps, substitute x by 3. Multiply it by the numerical coefficient 3 and add 7.

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.

Replace each letter in the expression with the assigned value.
First, replace each letter in the expression with the value that is assign to it. To make your calculations clear and accurate, enclose the numbers you’re substituting in parentheses. The value that is 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 may differ from every problem.
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!
x² + 4x − 10

= x² + 4x − 10
= (3)² + 4(3) − 10
= 9 + 12 − 10
= 11

After replacing x by 3, get the square of 3 which is 9. Add it to the product of 4 and 3. Lastly, subtract 10 from its sum.

2x² − 6x + 26

= 2x² − 6x + 26
= 2(3)² − 6(3) + 26
= 18 − 18 + 26
= 26

Simply each term inside the parenthesis which results to 18. Subtract 18 from it and 26.

3x³ − 6

= 3x³ − 6
= 3(3)³ − 6
= 3(27) − 6
= 81 − 6
= 75

Get the cube of 3 which is 27. Then, multiply it to 3 to get 81. Finally, subtract 6.

What’s New

Types of Functions

Read and analyze some of the common types of functions that you might encounter as you go on with this module.

Types of Function	Description	Example
Constant Function	A constant function 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).	y = 7
Identity Function	TThe identity function returns the same value and uses as its argument. It can be expressed as f (x) = x , for all values of x.	f (2) = 2
Polynomial Function	A polynomial function is defined by y = a⁰ + a¹x + a²x² + … + aⁿxⁿ, where n is a non-negative integer and a⁰, a¹, a² ,…, n ∈ R.
● Linear Function	The polynomial function with degree one. It is in the form y = mx + b	y = 2x + 5
● Quadratic Function	If the degree of the polynomial function is two, then it is a quadratic function. It is expressed as y = ax² + bx + c, where a ≠ 0 and a, b, c are constant and x is a variable.	y = 3x² + 2x + 5
● Cubic Function	A cubic polynomial function is a polynomial of degree three and can be denoted by f(x) = ax³ + bx² + cx + d, where a ≠ 0 and a, b, c, and d are constant & x is a variable.	y = 5x³ + 3x² + 2x + 5
Power Function	A power function is in the form y = axᵇ where b is any real constant number. Many of our parent functions such as linear and quadratic functions are functions.	f (x) = 8x⁵

[transcriber’s note: table continued from the previous page]

Types of Function	Description	Example
Rational Function	A rational function can be represented by a rational fraction say, p(x) / q(x) in which numerator and denominator are polynomial functions of x, where q(x) ≠ 0.	f(x) = (x² − 3x + 2) / (x² − 4)
Exponential function	This function is in the form y = aᵇˣ, where x is an exponent and a and b are constants. (Note: only b is raised to the power x; not a.) If the base b is greater than 1, then the result is exponential growth.	y = 2ˣ
Logarithmic Function	Logarithmic functions are the inverses of exponential functions and vice versa. 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, x > 0, where b > 0 and b ≠ 1	y = log₇ 49
Absolute Value Function	he absolute value of any number, c, is represented in the form of \|c\|. If any 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.	y = \|x − 4\| + 2
Greatest Integer Function	If a function f: R→ R is defined by f(x) = [x], x ∈ X, round-off it to the integer less than the number. Suppose that the given interval is in the form of (x, x+1), the value of greatest integer function is x which is an integer.	f(x) = ‖x‖ + 1 where ‖x‖ is the greatest integer function

What is It

Learning the different types of function is essential in this lesson. Because you will encounter those types as we proceed with evaluating functions.

When we say 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, you just need to: a.) replace each letter in the expression with the assigned value and b.) perform the correct order of operations in the expression.

Look at these examples!

Example 1: Given f (x) = 2x − 4 , find the value of the function if x = 3.

Solution:

f (3) = 2(3) − 4 f (3) = 6 − 4 f (3) = 2

● Substitute x by 3 in the function. ● Simplify the expression on the right side of the equation.

Answer: Given f (x) = 2x − 4 , f (3) = 2

Example 2: Given g(x) = 3x² + 7 , find g(−3) .

Solution:

g(−3) = 3(−3)² + 7 g(−3) = 3(9) + 7 g(−3) = 27 + 7 g(−3) = 34

● Substitute x by -3 in the function. ● Simplify the expression on the right side of the equation.

Answer: Given g(x) = 3x² + 7 , g(−3) = 34

Example 3: Given p(x) = 3x² + 5x − 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

p(−1) = 3(−1)² + 5(−1) − 2 p(−1) = 3(1) − 5 − 2 p(−1) = 3 − 5 − 2 p(−1) = −4

● Treat each of these like two separate problems. In each case, substitute x by its given value and simplify. Start with x = 0, then x = −1.

Answer: Given p(x) = 3x² + 5x − 2 , p(0) = −2 , p(−1) = −4

Example 4: Given 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

● This time, substitute x by (h + 1). ● Use the distributive property on the right side, then combine like terms to simplify.

Answer: Given 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) g(9) = √(27 − 2) g(9) = √25 g(9) = 5

✓ Substitute 9 for x in the function. ✓ Simplify the expression on the right side of the equation.

Answer: Given g(x) = √(3x − 2) , g(9) = 5

Example 6: Given h(x) = (4x + 8) / (2x − 4) , find the value of function if x = −5

Solution:

h(−5) = (4(−5) + 8) / (2(−5) − 4) h(−5) = (−20 + 8) / (−10 − 4) h(−5) = −12 / −14 h(−5) = 6 / 7

● Substitute -5 for x in the function. ● Simplify the expression on the right side of the equation. (recall the concepts of integers and the process on how to simplify fractions)

Answer: Given h(x) = (4x + 8) / (2x − 4) , h(−5) = 6 / 7

Example 7: Evaluate f(x) = 2ˣ if x = 3/2 .

Solution:

f(3/2) = 2^(3/2) f(3/2) = √(2³) f(3/2) = √8 f(3/2) = √(4 • 2) f(3/2) = 2√2

● Substitute 3/2 for x in the function. ● Simplify the expression on the right side of the equation. (get the cube of 2 which is 8, then simplify)

Answer: Given f(x) = 2ˣ , f(3/2) = 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 h(2.4) = 2 + 2 h(2.4) = 4

● Substitute x by 2.4 in the function. ● Simplify the expression on the right side of the equation. (Remember that in greatest integer function, value IS rounded-off to the integer less than the number)

Answer: Given h(x) = ‖x‖ + 2 , h(2.4) = 4

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| f(3) = |−5| f(3) = 5

✓ 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)

Answer: Given f(x) = |x − 8| , f(3) = 5

Example 10: Evaluate the function f(x) = x² − 2x + 2 at f(2x − 3).

Solution:

f(2x − 3) = (2x − 3)² − 2(2x − 3) + 2 f(2x − 3) = (4x² − 12x + 9) − 4x + 6 + 2 f(2x − 3) = 4x² − 12x + 9 − 4x + 6 + 2 f(2x − 3) = 4x² − 12x − 4x + 9 + 6 + 2 f(2x − 3) = 4x² − 16x + 17

✓ Substitute 2x − 3 for x in the function. ✓ Simplify the expression on the right side of the equation.

What’s More

Your Turn!

Independent Practice 1: Fill Me

Evaluate the following functions by filling up the missing parts of the solution.

f (x) = 3x − 5 , find f (2)

Solution:

f (2) = _____________________
f (2) = 6 − 5
f (2) = _____________________
g(x) = 3|2x| , find g(6)

Solution:

g(6) = _____________________
g(6) = 3|12|
g(6) = _____________________
k(a) = |a| − 2 , find k(−9)

Solution:

k(−9) = _____________________
k(−9) = 9 − 2
k(−9) = _____________________
p(a) = −4a − 2 , find p(2a)

Solution:

p(2a) = _____________________
p(2a) = _____________________
g(t) = t² − 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.

Given w(n) = n − 1, find the value of the function if = -1.
[transcriber’s note: the variable is missing from the printed text, shown here as “if = -1”]
Given f (x) = ‖x‖ − 3, find f (9.3) .
Evaluate the function w(x) = |− 2x + 3| if x = -1.
Evaluate: f (x) = −x − 1, find f (a²)
Given f (x) = 4x − 5 , find f (2x + 3)

Independent Practice 2: TRUE or SOLVE!

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.

Evaluate f (t) = 2t − 3 ; f (t²)

Solution:

f (t²) = 2(t²) − 3
f (t²) = 2t² − 3

Answer:

Solution: g(9) = √(5(9) − 13) g(9) = √(45 − 13) g(9) = √32 g(9) = 16√2

Answer:

Given the function f(x) = (5x − 7)/(3x − 2), find the value of the function if x = −3.

Solution: f(−3) = (5(−3) − 7)/(3(−3) − 2) f(−3) = (−15 − 7)/(−9 − 2) f(−3) = 22/11 f(−3) = −2

Answer:

Evaluate the function f(x) = x² − 3x + 5 at f(3x − 1).

Solution: f(3x − 1) = (3x − 1)² − 3x + 5 f(3x − 1) = 9x² − 6x + 1 − 3x + 5 f(3x − 1) = 9x² − 9x + 6

Answer:

Evaluate: g(x) = 3^x if x = 4/3

Solution: g(4/3) = 3^(4/3) g(4/3) = ∛(3⁴) g(4/3) = ∛81 g(4/3) = ∛(27 • 3) g(4/3) = 3∛3

Answer: