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Get in touch with us  # Basic Operation to Practise with Relations and Functions

## Relations and Functions:

### Relation:

A relation is a set of ordered pairs.

#### Domain:

The set of the first components of each ordered pair is called the domain.

#### Range:

The set of the second components of each ordered pair is called the range.

### Function:

A function is a relation in which each possible input value leads to exactly one output value.

Comparison of relation and function:

Question: Given a function f(x)=x2+2x, evaluate f(4).

Solution:

Given a function

f(x)=x2+2x

f(4)=(4)2+2×4

=16+8

=24

So, the value of f(4) is 24.

Question: Given the function

h(x)=√x−4 , solve

g(x)=2

Solution:

Given the function

g(x)=√x−4

To solve g(x)=2

√x−4=2

x−4=4

x=8

So, when

x=8

then the value of

g(x)=2

For any two functions

f(x) and g(x), we define a new function (f+g)(x) by the rule

(f+g)(x)=f(x)+g(x)

Example:

What is the sum of

f(x)=2×2−3x+10

and

g(x)=5x−7gx=5x−7?

Solution:

To define the sum of two functions, add them

(f+g)(x)f+gx

=f(x)+g(x)=fx+g(x)

=(2×2−3x+10)+(5x−7)=2×2−3x+10+(5x−7)

(Substitute the rule for each function)

=2×2−3x+10+5x−7=2×2−3x+10+5x−7

==

2×2−3x+5x+10−72×2−3x+5x+10−7 (Group-like terms)

=2×2+2x+3=2×2+2x+3

(Combine like terms)

The domain of

ff

is

{x∣x is a real number}{x∣x is a real number}.

The domain of

gg

is

{x∣x is a real number}{x∣x is a real number}.

So, the domain of

f+gf+g

is

{x∣x is a real number}{x∣x is a real number}.

The sum of the two functions

f(x)=2×2−3x+10fx=2×2−3x+10

and

g(x)=5x−7gx=5x−7 is

2×2+2x+32×2+2x+3.

### Subtraction on Functions:

For any two functions

f(x)f(x)

and

g(x)g(x), we define a new function

(f−g)(x)(f−g)(x) by the rule

(f−g)(x)=f(x)−g(x)f−gx=fx−g(x)

Example:

What is the difference of

f(x)=x2−5x−6fx=x2−5x−6

and

g(x)=3x+7gx=3x+7?

Solution:

To define the difference between two functions, subtract them

(f−g)(x)f−gx

=f(x)−g(x)=fx−g(x)

=(x2−5x−6)−(3x+7)=x2−5x−6−(3x+7)

(Substitute the rule for each function)

=x2−5x−6−3x−7=x2−5x−6−3x−7

(Use distributive property)

==

x2−5x−3x−6−7×2−5x−3x−6−7 (Group-like terms)

=x2−8x−13=x2−8x−13

(Combine like terms)

The domain of

ff

is

{x∣x is a real number}{x∣x is a real number}.

The domain of

gg

is

{x∣x is a real number}{x∣x is a real number}.

So, the domain of

f−g

is

{x∣x is a real number}.

The difference between the two functions

f(x)=x2−5x−6

and

g(x)=3x+7 is

x2−8x−13.

### Multiply Functions:

##### Multiplication of Functions:

For any two functions

f(x)

and

g(x), we define a new function

(f.g)(x) by the rule

(f.g)(x)=f(x).g(x)

Example:

Suppose demand, d, for a company’s product at cost, x, is predicted by the function

d(x)=−0.25×2+1000

, and the price, p, that the company charge for the product is given by

p(x)=x+16. Find the company’s revenue function.

Solution:

The company’s revenue is equal to the price of the product multiplied by the demand for the product.

Revenue =price×demandRevenue =price×demand

The demand is the function

d(x)d(x)

. The price is the function

p(x)p(x).

### Domains

Price:

p(x)=x+16px=x+16

p(x):0≤xpx:0≤x (Price cannot be negative)

Demand:

d(x)=−0.25×2+1000dx=−0.25×2+1000

d(x):x≤2010−−√dx:x≤2010 (Demand cannot be negative)

Revenue:

R(x)=(p.d)(x)=p(x).d(x)Rx=p.dx=px.d(x)

R(x):0≤x≤2010−−√Rx:0≤x≤2010

(Domain is the intersection of the domains of

pp

&

dd)

=(x+16)(−0.25×2+1000)=(x+16)(−0.25×2+1000)

=−0.25×2+1000x−4×2+16000 =−0.25×2+1000x−4×2+16000

=−4.25×2+1000x+16000=−4.25×2+1000x+16000

The company’s revenue function is

−4.25×2+1000x+16000−4.25×2+1000x+16000

### Divide Functions:

##### Division of Functions:

For any two functions

f(x)f(x)

and

g(x)g(x), we define a new function

(fg)(x)fg(x) by the rule

(fg)(x)=f(x)g(x)fg(x)=f(x)g(x)

Example:

Define the quotient

fgfg

of the functions

f(x)=x−3fx=x−3 and

g(x)=x2−x−6gx=x2−x−6?

Solution:

To define the quotient of two functions, divide them

(fg)(x)fg(x)

=f(x)g(x)=f(x)g(x)

=x−3×2−x−6=x−3×2−x−6 Substitute the rule for each function.

=x−3×2−3x+2x−6=x−3×2−3x+2x−6

==

x−3x(x−3)+2(x−3) x−3xx−3+2(x−3)

=x−3(x−3)(x+2)=x−3(x−3)(x+2)

Factor the denominator.

=1x+2=1x+2

Simplify.

The quotient of

fgfg

is

1x+21x+2.

The domain of

fgfg

is the set of all values for which

f, g,f, g, and

fgfg are defined, so

gg cannot be 0.

This is the set of all real numbers

xx

such that

x≠3x≠3 and

x≠−2x≠−2.

### Compose Functions:

Example:

Let

f(x)=3x−1fx=3x−1

and

g(x)=2xgx=2x. Identify the rule for the following functions.

(a)

f(g(5))f(g(5))

(b)

f(g(x))f(g(x))

Solution:

(a) Find the value of

f(g(5))f(g(5))

g(5)=2×5=10g5=2×5=10

Apply the rule for

gg.

f(g(5))=f(10)fg5=f(10)

Apply the rule for

ff to the result.

=3×10−1=3×10−1

Use the rule for

ff.

=29=29

Simplify.

f(g(5))=29fg5=29

(b) Find the value of

f(g(x))f(g(x))

g(x)=2xgx=2x

Apply the rule for

gg.

f(g(x))=f(2x)fgx=f(2x)

Apply the rule for

ff to the result.

=3×2x−1=3×2x−1

Use the rule for

ff.

=6x−1=6x−1

Simplify.

f(g(x))=6x−1fgx=6x−1

##### Conclusion:

When we apply a rule for one function to the rule for another function, we get an entirely new function

### Composite Function:

It is the result of applying the rule for one function to the rule of another function.

(f∘g)(x)=f(g(x))𝒇∘𝒈𝒙=𝒇(𝒈(𝒙))

(g∘f)(x)=g(f(x))𝒈∘𝒇𝒙=𝒈(𝒇(𝒙))

The operation

∘∘

that forms a composite function is called a composition of functions.

The domain of

f∘gf∘g

is the set of all real numbers

xx in the domain of

gg such that

g(x)g(x) is in the domain of

ff.

So, the domain of the composition is the intersection of the domains of

gg

and

f∘gf∘g, but not

ff.

Here the product of functions

fgfg

is not the same as the function composition

f(g(x))f(g(x)), because, in general,

f(x)g(x)≠f(g(x))𝒇𝒙𝒈(𝒙)≠𝒇(𝒈(𝒙)).

Example:

### What is the rule for the composition

(f∘g)(x)(f∘g)(x)

where

f(x)=2x+1fx=2x+1 and

g(x)=3−xgx=3−x?

Given,

f(x)=2x+1fx=2x+1

and

g(x)=3−xgx=3−x

(f∘g)(x)=f(g(x))f∘gx=f(g(x))

=f(3−x)=f(3−x)

=2(3−x)+1=23−x+1

=6−2x+1=6−2x+1

=−2x+7=−2x+7

The rule for the composition

(f∘g)(x)(f∘g)(x)

is

−2x+7−2x+7, and the domain is all real numbers.

Example:

What is the rule for the composition

(g∘f)(x)(g∘f)(x)

where

f(x)=x3fx=x3 and

g(x)=x+1gx=x+1?

Given,

f(x)=x3fx=x3

and

g(x)=x+1gx=x+1

(g∘f)(x)=g(f(x))g∘fx=g(f(x))

=g(x3)=g(x3)

=x3+1=x3+1

The rule for the composition

(g∘f)(x)(g∘f)(x)

is

x3+1×3+1, and the domain is all real numbers.

### Use a Composite Function Model:

Example:

‘Clothes U Wear’ posts discounts on social media. The store allows customers to use multiple discounts. They simply need to tell the cashier in which order they would like the discounts to be applied.

On your next trip to ‘Clothes U Wear’, in which order should the customer ask for the discounts to get a better yield?

Solution:

Write the functions to model the discounts.

Let

xx

represents the price of the purchase.

f(x)=x−5fx=x−5

g(x)gx

=x−10% of x=x−10% of x

=x−0.1x=x−0.1x

=0.9x=0.9x

Case1:

10%10%

off, then

\$5\$5 off

⇒⇒ Identify the rule for

f∘gf∘g.

(f∘g)(x)=f(g(x))f∘gx=f(g(x))

=f(0.9x)=f(0.9x)

=0.9x−5=0.9x−5

Case 2:

\$5 off, then 10% off

⇒⇒

Identify the rule for

g∘fg∘f.

(g∘f)(x)=g(f(x))g∘fx=g(f(x))

=g(x−5)=g(x−5)

=0.9(x−5)=0.9(x−5)

=0.9x−4.5=0.9x−4.5

When comparing case 1 and case 2, the discount applied in case 1 yields a better deal than the discount applied in case 2.

Suppose the customer buys a trouser that costs \$50.00 before the discounts.

Applying the discounts as

(f∘g)(x)(f∘g)(x)

The new cost

=(0.9)(50)−5=0.950−5

=45−5=45−5

=40=40

, or

\$40.00 \$40.00

Applying the discounts as

(g∘f)(x)(g∘f)(x)

The new cost

=(0.9)(50)−4.5=0.950−4.5

=45−4.5=45−4.5

=40.5=40.5

, or

\$40.50 \$40.50

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