Hello! I need some assistance with this homework question, pleaseQ4

Given the functions f(x) and g(x) defined as:
[tex]\begin{gathered} f(x)=3x+5 \\ g(x)=5x-9 \end{gathered}[/tex](a)
We use the definition of the addition of two functions:
[tex](f+g)(x)=f(x)+g(x)[/tex]Then:
[tex]\begin{gathered} (f+g)(x)=3x+5+5x-9 \\ \Rightarrow(f+g)(x)=8x-4\ldots(1) \end{gathered}[/tex]This is a polynomial, so the domain is {x | x is any real number}
(b)
Using the definition of subtraction of two functions:
[tex](f-g)(x)=f(x)-g(x)[/tex]Then:
[tex]\begin{gathered} (f-g)(x)=3x+5-5x+9 \\ \Rightarrow(f-g)(x)=-2x+14\ldots(2) \end{gathered}[/tex]This is a polynomial, so the domain is {x | x is any real number}
(c)
We use the definition of the product between two functions:
[tex](f\cdot g)(x)=f(x)\cdot g(x)[/tex]Then:
[tex]\begin{gathered} (f\cdot g)(x)=(3x+5)\cdot(5x-9)=15x^2-27x+25x-45 \\ \Rightarrow(f\cdot g)(x)=15x^2-2x-45\ldots(3) \end{gathered}[/tex]This is a polynomial, so the domain is {x | x is any real number}
(d)
We use the definition of the quotient between two functions:
[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}[/tex]Then:
[tex](\frac{f}{g})(x)=\frac{3x+5}{5x-9}\ldots(4)[/tex]This is a rational expression, so the domain is the set of all the numbers such that the denominator is not 0. Finding those values:
[tex]\begin{gathered} 5x-9=0 \\ \Rightarrow x=\frac{9}{5} \end{gathered}[/tex]The domain is {x | x ≠ 9/5}
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