Part 1) ![(f+g)(x)=7x+11](/tpl/images/0256/2252/27246.png)
Part 2) ![(f-g)(x)=x-3](/tpl/images/0256/2252/6202e.png)
Part 3) ![(f/g)(x)=\frac{(4x+4)}{(3x+7)}](/tpl/images/0256/2252/bf91d.png)
Part 4) In interval notation the domain is (-∞,-7/3) ∪ (-7/3,∞)
Part 5) ![(f o g) (x)=12x+32](/tpl/images/0256/2252/3b4e4.png)
Step-by-step explanation:
we have
![f(x)=4x+4](/tpl/images/0256/2252/ef2de.png)
![g(x)=3x+7](/tpl/images/0256/2252/22b42.png)
Part 1) Find (f+g)(x)
we know that
![(f+g)(x)=f(x)+g(x)](/tpl/images/0256/2252/21507.png)
substitute the given functions
![(f+g)(x)=(4x+4)+(3x+7)](/tpl/images/0256/2252/541ea.png)
Combine like terms
![(f+g)(x)=7x+11](/tpl/images/0256/2252/27246.png)
Part 2) Find (f-g)(x)
we know that
![(f-g)(x)=f(x)-g(x)](/tpl/images/0256/2252/92267.png)
substitute the given functions
![(f-g)(x)=(4x+4)-(3x+7)](/tpl/images/0256/2252/12901.png)
![(f-g)(x)=4x+4-3x-7](/tpl/images/0256/2252/e3690.png)
Combine like terms
![(f-g)(x)=x-3](/tpl/images/0256/2252/6202e.png)
Part 3) Find (f/g)(x)
we know that
![(f/g)(x)=\frac{f(x)}{g(x)}](/tpl/images/0256/2252/d001a.png)
substitute the given functions
![(f/g)(x)=\frac{(4x+4)}{(3x+7)}](/tpl/images/0256/2252/bf91d.png)
Part 4) What the domain for the answer in question 3
we have
![(f/g)(x)=\frac{(4x+4)}{(3x+7)}](/tpl/images/0256/2252/bf91d.png)
we know that
The denominator of the quotient cannot be equal to zero
so
![3x+7=0](/tpl/images/0256/2252/71630.png)
![3x=-7](/tpl/images/0256/2252/8296b.png)
![x=-\frac{7}{3}](/tpl/images/0256/2252/4c4a5.png)
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))](/tpl/images/0256/2252/6fb91.png)
substitute
![f(g(x))=4(3x+7)+4](/tpl/images/0256/2252/92f48.png)
![f(g(x))=12x+28+4](/tpl/images/0256/2252/c5e25.png)
![f(g(x))=12x+32](/tpl/images/0256/2252/2a3f2.png)
therefore
![(f o g) (x)=12x+32](/tpl/images/0256/2252/3b4e4.png)