Operations on Functions

Cris Cruz
27 Aug 202008:13

Summary

TLDRThis video script introduces the fundamental operations of functions, including addition, subtraction, multiplication, and division, with the stipulation that division by zero is undefined. It demonstrates these operations step by step using specific functions, such as 'v(x) - p(x)', 'f(x) * p(x)', 'v(x) / g(x)', and 'f(x) + g(x)'. The script also includes examples of factoring and simplifying expressions, concluding with evaluating functions at a specific point, x = 3, to yield numerical results.

Takeaways

  • ✏️ The objective is to perform fundamental operations with functions.
  • ➕ The sum of two functions, f(x) and g(x), is given by f(x) + g(x).
  • ➖ The difference between two functions is given by f(x) - g(x).
  • ✖️ The product of two functions is f(x) multiplied by g(x).
  • ➗ The quotient of two functions is f(x) divided by g(x), with the condition that g(x) is not equal to zero.
  • 🔍 For subtraction, substitute the given functions and combine like terms.
  • 🧮 For multiplication, use the distributive property to expand and simplify the expression.
  • 🔄 For division, factor the numerator and denominator, then simplify by canceling common factors.
  • 🔢 When adding functions, combine like terms after substituting the given values.
  • ✍️ For specific values, substitute the given value of x into the functions and simplify.

Q & A

  • What are the fundamental operations that can be performed on functions f and g?

    -The fundamental operations on functions f and g include the sum (f + g), difference (f - g), product (f * g), and quotient (f / g), provided that g(x) is not equal to zero to avoid division by zero.

  • What is the definition of the sum of two functions, f and g?

    -The sum of functions f and g, denoted as f + g, is defined as f(x) + g(x) for all x in the domain of both f and g.

  • How is the difference between two functions, f and g, represented?

    -The difference between functions f and g is represented as f - g, which is equal to f(x) - g(x) for all x in the domain of both f and g.

  • What is the definition of the product of two functions, f and g?

    -The product of functions f and g is defined as f * g, which equals f(x) * g(x) for all x in the domain of both functions.

  • What is the definition of the quotient of two functions, f and g, and what condition must be met?

    -The quotient of functions f and g is defined as f / g, which equals f(x) / g(x) for all x in the domain of both functions, provided that g(x) is not zero to avoid an undefined function.

  • In the given transcript, what is the expression for v(x) - p(x)?

    -The expression for v(x) - p(x) is x^2 - 4x - 5 - (2x - 7), which simplifies to x^2 - 6x + 2 after combining like terms.

  • How is the product of f(x) and p(x) calculated in the transcript?

    -The product f(x) * p(x) is calculated by multiplying (x + 3) with (2x - 7), resulting in 2x^2 - 7x + 6x - 21, which simplifies to 2x^2 - x - 21.

  • What is the quotient of v(x) divided by g(x) in the transcript, and how is it simplified?

    -The quotient v(x) / g(x) simplifies to (x^2 - 4x - 5) / (x^2 - 25), which factors and simplifies to (x - 5)(x + 1) / (x - 5)(x + 5), and further simplifies to (x + 1) / (x + 5) after canceling out the common factor (x - 5).

  • What is the expression for f(x) + g(x) in the transcript, and how is it simplified?

    -The expression for f(x) + g(x) is (x + 3) + (x^2 - 25), which simplifies to x^2 + x - 22 after combining like terms.

  • How is the value of f(x) + g(x) evaluated when x is substituted with 3 in the transcript?

    -When x is substituted with 3, the value of f(x) + g(x) becomes 3^2 + 3 - 22, which simplifies to 9 + 3 - 22, resulting in -10.

  • In the transcript, what is the process for evaluating f(3) + g(3), and what is the result?

    -The process involves substituting 3 for x in both f(x) and g(x), resulting in (3 + 3) + (3^2 - 25), which simplifies to 6 + 9 - 25, and the final result is -10.

Outlines

00:00

📚 Fundamental Operations on Functions

This paragraph introduces the basic arithmetic operations that can be performed on functions. It defines the sum, difference, product, and quotient of two functions, f and g, within their common domain. The paragraph also highlights the importance of ensuring the denominator is not zero in the quotient operation to avoid undefined results. It then proceeds with examples of subtraction and multiplication of functions, demonstrating the process of replacing function values and simplifying expressions to find the resulting function.

05:02

🔍 Detailed Example of Function Operations

The second paragraph provides a step-by-step walkthrough of performing arithmetic operations on functions, specifically focusing on subtraction, multiplication, and division. It begins with an example of subtracting function p from function v, then moves on to multiplying function f by function p. The paragraph also includes an example of dividing function v by function g, emphasizing the importance of factoring to simplify the division. Each step is clearly explained, from the initial definition of the operation to the final simplification of the expression, resulting in the new function representations.

Mindmap

Keywords

💡Functions

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs. In the video, functions such as f(x) and g(x) are discussed as mathematical expressions that depend on the variable x. The theme of the video revolves around performing basic operations like addition, subtraction, multiplication, and division on these functions.

💡Sum of Functions

The sum of two functions, f(x) + g(x), is a mathematical operation where the values of two functions are added together for each input x. The video illustrates this by showing how to combine f(x) and g(x) to produce a new function that represents their sum.

💡Difference of Functions

The difference of functions, represented as f(x) - g(x), involves subtracting the values of one function from another for each input x. The video explains this operation by showing how to subtract one function's expression from another, resulting in a new function.

💡Product of Functions

The product of functions, denoted as f(x) * g(x), is the multiplication of two functions' values for each input x. The video demonstrates how to multiply two functions by distributing and combining like terms to find the product.

💡Quotient of Functions

The quotient of functions, written as f(x) / g(x), is the division of one function by another for each input x. The video covers the important condition that the denominator, g(x), must not equal zero, as this would make the function undefined.

💡Factoring

Factoring involves breaking down a polynomial into simpler components (factors) that, when multiplied, give back the original polynomial. In the video, factoring is used to simplify the expression when dividing functions, allowing for the cancellation of common factors.

💡Combining Like Terms

Combining like terms is a process in algebra where terms with the same variables are added or subtracted. The video emphasizes this process during operations such as the subtraction and multiplication of functions, to simplify the resulting expressions.

💡Binomials

A binomial is a polynomial with two terms. The video uses binomials when multiplying functions, showing how to apply the distributive property to multiply and then combine like terms.

💡Substitution

Substitution is the process of replacing a variable with a specific value. The video demonstrates substitution by evaluating functions like f(x) and g(x) at specific values of x, such as substituting x = 3 to find the corresponding output.

💡Undefined Function

A function is undefined at points where it does not produce a valid output, often due to division by zero. The video mentions this concept in the context of dividing functions, where the quotient f(x)/g(x) is undefined if g(x) equals zero.

Highlights

Introduction to fundamental operations in functions, including addition, subtraction, multiplication, and division.

Definition of the sum of two functions, f and g, as f(x) + g(x).

Explanation of the difference between two functions, f(x) - g(x).

Description of the product of two functions, f(x) * g(x).

Clarification on the quotient of two functions, f(x) / g(x), with the condition that g(x) ≠ 0.

Example calculation of the difference between functions v(x) - p(x).

Step-by-step solution for v(x) - p(x), including substitution and rearrangement of terms.

Example of multiplying functions f(x) and p(x), with detailed steps for binomial multiplication.

Combining like terms after multiplying functions f(x) and p(x).

Division of functions v(x) by g(x), with factoring and simplification.

Cancellation of common factors in the division of functions v(x) / g(x).

Addition of functions f(x) and g(x), with a focus on rearranging and combining terms.

Substitution of a specific value (3) into the function f(x) + g(x) to find f(3) + g(3).

Verification of the result for f(3) + g(3) through direct substitution into f(x) and g(x).

Final confirmation of the result for f(3) + g(3) being -10 through two different methods.

Transcripts

play00:00

our objective is to perform the

play00:02

fundamental operations

play00:04

in functions let f and g be functions

play00:08

each function is defined for all x in

play00:10

the domains of both

play00:12

f and g letter a the sum of

play00:15

f and g is f plus g of x equals f of

play00:19

x plus g of x letter b

play00:22

difference of f and g is f minus g

play00:26

of x is equal to f of x minus

play00:29

g of x letter c product of

play00:32

f and g is f times g of x

play00:36

is equal to f of x times g of x

play00:39

and the quotient of f and g

play00:43

is f over g of x equals

play00:46

f of x divided by g of x provided that g

play00:50

of x is not equal to zero because when g

play00:53

of x

play00:53

our denominator is equal to zero the

play00:56

function will become

play00:58

undefined let us consider the following

play01:01

functions

play01:02

and let us solve number one v minus p

play01:06

of x our first step is definition of

play01:10

subtraction

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functions so this will become v of x

play01:15

minus p of x

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copy the left hand side of the equation

play01:21

our next step is

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replace v of x and p of x by the given

play01:26

values

play01:27

so our v of x is x squared

play01:30

minus four x minus five minus

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our p of x is two x

play01:37

minus seven make sure you enclose that

play01:40

in

play01:40

parenthesis copy this our third step

play01:44

will be

play01:45

rearranging the terms so this will

play01:47

become x

play01:48

squared minus 4x

play01:51

minus 2x minus 5

play01:54

this will become plus 7 combine like

play01:58

terms

play01:59

x squared negative 4x minus 2x

play02:03

negative 6x negative 5 plus 7

play02:07

positive 2. and this is our answer for

play02:10

v minus p of x

play02:14

next one let us try f times

play02:17

p of x so first step again

play02:20

definition of this time multiplication

play02:24

of functions so this will become

play02:27

f of x times

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p of x copy this

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next step we'll replace f of

play02:36

x and p of x by the given values

play02:40

so our f of x is

play02:43

x plus 3 times our p

play02:47

of x is 2x minus

play02:50

7 copy this third step

play02:54

multiply the binomials so we have

play02:58

x times 2x is 2x squared

play03:02

x times negative seven is negative seven

play03:06

x three times two x

play03:09

is positive six x and three

play03:12

times negative seven is negative twenty

play03:15

one

play03:16

let us combine like terms so this will

play03:19

still become

play03:19

2x squared negative 7x plus 6x

play03:24

negative x copy minus

play03:27

21. this will be our final answer for

play03:31

f times p of x

play03:34

okay another example this time v

play03:38

divided by g of x so our first step

play03:41

definition of this time division

play03:45

so this will become v of x

play03:48

divided by g of x

play03:52

copy this we replace

play03:56

v of x and g of x by the given values

play04:00

so our v of x x squared minus four

play04:03

x minus five divided by

play04:07

all over our g of x

play04:10

x squared minus twenty

play04:13

copy this then we factor the numerator

play04:17

and the denominator

play04:19

so the knowledge in factoring trinomials

play04:22

as well as binomials

play04:24

is an advantage in this lesson

play04:27

the factors of x squared minus 4x minus

play04:30

5

play04:31

are x minus 5 and x

play04:34

plus 1 all over the factors of

play04:37

x squared minus 25 are x minus five

play04:41

and x plus five cancel out the common

play04:45

factors

play04:46

so x minus five and x minus five

play04:50

so we will be left with x plus one

play04:53

all over x plus 5. so this is our answer

play04:57

for

play04:57

v divided by g of x

play05:01

okay next let us try f plus g

play05:05

of x so our first step definition of

play05:08

this time

play05:09

addition functions so we have f

play05:12

of x plus g of x

play05:16

copy this then we replace again

play05:19

our f of x will be replaced with x plus

play05:22

3

play05:23

plus our g of x is x

play05:26

squared minus 25 copy this

play05:30

and then rearrange the terms this will

play05:33

be the first one

play05:34

x squared plus x plus 3

play05:38

minus 25 and then we combine like terms

play05:42

x squared plus x 3 minus 25

play05:46

is negative 22. so this is our answer

play05:50

for

play05:51

f plus g of x so take note of this

play05:54

answer

play05:55

x squared plus x minus 22 let's have the

play05:58

next one

play06:00

so i'm still evaluating here f plus g

play06:03

but this time

play06:03

instead of having x here i have three

play06:07

so our previous answer x squared plus x

play06:10

minus 22

play06:12

is actually the same as here so let's

play06:14

proceed

play06:15

copy this our next step will be

play06:18

to substitute 3 for x so since we have

play06:21

here

play06:22

3 we're going to substitute 3 in all our

play06:25

x's here

play06:26

so this will become 3 squared

play06:30

plus 3 copy minus 22

play06:34

and then we simplify 3 squared is 9

play06:38

9 plus 3 is 12 and 12 minus 22

play06:42

is negative 10. so that will be our

play06:45

answer for

play06:46

f plus g of 3. now there is another

play06:50

solution here

play06:51

so we should also arrive with negative

play06:54

10.

play06:54

so we'll see first step

play06:58

so this will be f of 3

play07:02

plus g of three

play07:05

so copy this our next step will be to

play07:08

substitute three

play07:10

for x in both f of x

play07:13

and g of x so we're going to substitute

play07:16

3

play07:17

in our f of x so substituting 3

play07:21

here this will become 3 plus 3

play07:24

then plus now for g

play07:28

substituting 3 here this will become

play07:32

3 squared minus 25

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copy f plus g of 3 and then we simplify

play07:40

3 plus 3 is 6 3 squared

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is 9 and minus

play07:48

25 copy this let us simplify further

play07:52

6 plus 9 15 copy

play07:56

minus 25 15 minus 25

play08:00

is negative 10. so we have the same

play08:04

answer as the previous one this is our

play08:07

answer for

play08:08

f plus g of 3.

play08:12

gets

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Algebraic FunctionsFunction OperationsMathematics TutorialSubtraction RulesMultiplication MethodDivision ProcessCombining Like TermsFactoring TrinomialsBinomial FactorsAlgebraic SimplificationEducational Content