Explanation
In this problem, we have the following functions:
[tex]\begin{gathered} f(x)=3x+2, \\ g(x)=-2x+1. \end{gathered}[/tex]
(a) The function (f + g)(x) is given by:
[tex](f+g)(x)=f(x)+g(x)=(3x+2)+(-2x+1)=x+3.[/tex]
(b) Using the previous result, and replacing x = 6, we get:
[tex](f+g)(6)=6+3=9.[/tex]
(c) The function (f · g)(x) is given by:
[tex]\begin{gathered} (f\cdot g)(x)=f(x)\cdot g(x) \\ =(3x+2)\cdot(-2x+1) \\ =3x\cdot(-2x)+3x\cdot1+2\cdot(-2x)+2\cdot1, \\ =-6x^2+3x-4x+2, \\ =-6x^2-x+2. \end{gathered}[/tex]
(d) The function (f - g)(x) is given by:
[tex](f-g)(x)=f(x)-g(x)=(3x+2)-(-2x+1)=5x-1.[/tex]
(e) The function (f/g)(x) is given by:
[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}=\frac{3x+2}{-2x+1}.[/tex]
The domain of this function is all the real numbers except the numbers that make zero the denominator because we can't divide by zero. The denominator is 0 when:
[tex]\begin{gathered} -2x+1=0, \\ 2x=1, \\ x=\frac{1}{2}. \end{gathered}[/tex]
So the domain of the function is:
[tex]Domain={}{}\text{ }\lbrace x|x\ne\frac{1}{2}\rbrace\text{ or }(-\infty,\frac{1}{2})\cup(\frac{1}{2},\infty).[/tex]Answer
(a) (f + g)(x) = x + 3
(b) (f + g)(6) = 9
(c) (f · g)(x) = -6x² - x +2
(d) (f - g)(x) = 5x - 1
(e) Domain = {x | x ≠ 1/2} = (-∞, 1/2) U (1/2, ∞)
