Answer:
Part 1) [tex](f+g)(x)=7x+11[/tex]
Part 2) [tex](f-g)(x)=x-3[/tex]
Part 3) [tex](f/g)(x)=\frac{(4x+4)}{(3x+7)}[/tex]
Part 4) In interval notation the domain is (-∞,-7/3) ∪ (-7/3,∞)
Part 5) [tex](f o g) (x)=12x+32[/tex]
Step-by-step explanation:
we have
[tex]f(x)=4x+4[/tex]
[tex]g(x)=3x+7[/tex]
Part 1) Find (f+g)(x)
we know that
[tex](f+g)(x)=f(x)+g(x)[/tex]
substitute the given functions
[tex](f+g)(x)=(4x+4)+(3x+7)[/tex]
Combine like terms
[tex](f+g)(x)=7x+11[/tex]
Part 2) Find (f-g)(x)
we know that
[tex](f-g)(x)=f(x)-g(x)[/tex]
substitute the given functions
[tex](f-g)(x)=(4x+4)-(3x+7)[/tex]
[tex](f-g)(x)=4x+4-3x-7[/tex]
Combine like terms
[tex](f-g)(x)=x-3[/tex]
Part 3) Find (f/g)(x)
we know that
[tex](f/g)(x)=\frac{f(x)}{g(x)}[/tex]
substitute the given functions
[tex](f/g)(x)=\frac{(4x+4)}{(3x+7)}[/tex]
Part 4) What the domain for the answer in question 3
we have
[tex](f/g)(x)=\frac{(4x+4)}{(3x+7)}[/tex]
we know that
The denominator of the quotient cannot be equal to zero
so
[tex]3x+7=0[/tex]
[tex]3x=-7[/tex]
[tex]x=-\frac{7}{3}[/tex]
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
[tex](f o g) (x)=f(g(x))[/tex]
substitute
[tex]f(g(x))=4(3x+7)+4[/tex]
[tex]f(g(x))=12x+28+4[/tex]
[tex]f(g(x))=12x+32[/tex]
therefore
[tex](f o g) (x)=12x+32[/tex]
