Question

F(x) = 4x + 4 g(x) = 3x +7

1) find (f+g)(x)
2) find (f-g) (x)
3) find (f/g) (x)
4) what the domain for the answer in question 3. Use set builder notation or interval notation
5) find (f o g) (x)

Answer 1

Part 1)
`(f+g)(x)=7x+11`

Part 2)
`(f-g)(x)=x-3`

Part 3)
`(f/g)(x)=((4x+4))/((3x+7))`

Part 4) In interval notation the domain is (-∞,-7/3) ∪ (-7/3,∞)

Part 5)
`(f o g) (x)=12x+32`

Explanation:

we have

`f(x)=4x+4`

`g(x)=3x+7`

Part 1) Find (f+g)(x)

we know that

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

substitute the given functions

`(f+g)(x)=(4x+4)+(3x+7)`

Combine like terms

`(f+g)(x)=7x+11`

Part 2) Find (f-g)(x)

we know that

`(f-g)(x)=f(x)-g(x)`

substitute the given functions

`(f-g)(x)=(4x+4)-(3x+7)`

`(f-g)(x)=4x+4-3x-7`

Combine like terms

`(f-g)(x)=x-3`

Part 3) Find (f/g)(x)

we know that

`(f/g)(x)=(f(x))/(g(x))`

substitute the given functions

`(f/g)(x)=((4x+4))/((3x+7))`

Part 4) What the domain for the answer in question 3

we have

`(f/g)(x)=((4x+4))/((3x+7))`

we know that

The denominator of the quotient cannot be equal to zero

so

`3x+7=0`

`3x=-7`

`x=-(7)/(3)`

The domain is all real numbers except the number x=-7/3

In interval notation the domain is (-∞,-7/3) ∪ (-7/3,∞)

Part 5) Find (f o g) (x)

we know that

`(f o g) (x)=f(g(x))`

substitute

`f(g(x))=4(3x+7)+4`

`f(g(x))=12x+28+4`

`f(g(x))=12x+32`

therefore

`(f o g) (x)=12x+32`