Explanation
The domain of a function is the set of all possible inputs for the function.
so
Let
[tex]\begin{gathered} f(x)=x^2+2x \\ g(x)=3x^2-1 \end{gathered}[/tex]Step 1
a) (f+g)(x)
to add the function , add like terms
[tex]\begin{gathered} f(x)+g(x)=(f+g)(x)=x^2+2x+3x^2-1 \\ (f+g)(x)=x^2+2x+3x^2-1 \\ (f+g)(x)=4x^2+2x-1 \end{gathered}[/tex]so
[tex]a)(f+g)(x)=4x^2+2x-1[/tex](b) domain of the function:
the domain of a quadratic function f(x) is the set of x -values for which the function is defined,so
[tex]\text{Domain:(-}\infty,\infty)[/tex]Step 2
c) (f-g)(x)
subtract the functions
[tex]\begin{gathered} f(x)-g(x)=(f-g)(x)=x^2+2x-(+3x^2-1) \\ (f-g)(x)=x^2+2x-3x^2+1 \\ (f-g)(x)=-2x^2+2x+1 \end{gathered}[/tex]d)Domain:For any linear function the domain and range are always all real numbers
[tex]\mleft(-\infty,\infty\mright)[/tex]Step 3
e) (fg)(x)
[tex]\begin{gathered} f(x)=x^2+2x \\ g(x)=3x^2-1 \\ \text{then} \\ (fg)(x)=(x^2+2x)(3x^2-1) \\ \text{apply distributive property} \\ (fg)(x)=(x^2\cdot3x^2)-(1\cdot x^2)+(2x\cdot3x^2)-(2x\cdot1) \\ (fg)(x)=3x^4-x^2+6x^3-2x \\ \text{reorder} \\ (fg)(x)=3x^4+6x^3-x^2-2x \end{gathered}[/tex]f) again we have a polynomial so, The domain of all polynomial functions is all real numbers:
[tex]\text{ Domain}\mleft(-\infty,\infty\mright)[/tex]Step 4
g) (f/g)(x)
[tex]\begin{gathered} f(x)=x^2+2x \\ g(x)=3x^2-1 \\ \text{then} \\ (\frac{f}{g})(x)=\frac{(x^2+2x)}{\mleft(3x^2-1\mright)} \\ \end{gathered}[/tex]
h) domain, we see this function is not definded when the denominator equals zero, we need to find that number
[tex]\begin{gathered} 3x^2-1=0 \\ \text{add 1 in both sides} \\ 3x^2-1+1=0+1 \\ 3x^2=1 \\ \text{divide both sides by 3} \\ \frac{3x^2}{3}=\frac{1}{3} \\ x^2=\frac{1}{3} \\ \sqrt{x^2}=\sqrt{\frac{1}{3}} \\ x=\frac{1}{\sqrt[]{3}} \\ so,\text{ the domain is} \\ (-\infty,\frac{1}{\sqrt[]{3}})\cup(\frac{1}{\sqrt[]{3}},\infty) \end{gathered}[/tex]I hope this helps you
