[tex]\begin{gathered} (f\text{ + g)(x) = x + 4 + }\sqrt[]{2x\text{ -6}} \\ \text{domain: \lbrack{}3, }\infty) \end{gathered}[/tex]Explanation:[tex]\begin{gathered} 31_{})\text{ f(x) = }\sqrt[]{2x\text{ - 6}} \\ g(x)\text{ = x + 4} \\ \\ (f+g)(x)\text{ =f(x) + g(x)} \end{gathered}[/tex][tex]\begin{gathered} f(x)+\text{ g(x) = }\sqrt[]{2x\text{ -6}}\text{ + (x + 4)} \\ f(x)+\text{ g(x) = x + 4 + }\sqrt[]{2x\text{ -6}} \\ (f\text{ + g)(x) = x + 4 + }\sqrt[]{2x\text{ -6}} \end{gathered}[/tex][tex]\begin{gathered} To\det er\min e\text{ the domain, we n}ed\text{ to find the values of x that makes the function solvable:} \\ we\text{ consider: }\sqrt[]{2x\text{ - 6}} \\ \text{when x = 3} \\ \sqrt[]{2(3)\text{ - 6}}\text{ = }\sqrt[]{0}\text{ = 0} \\ \text{when x = 4} \\ \sqrt[]{2(4)\text{ - 6}}\text{ = }\sqrt[]{2}\text{ } \\ \text{when x = 2} \\ \sqrt[]{2(2)\text{ - 6}}\text{ = }\sqrt[]{-2}\text{ } \\ we\text{ can't find the square root of a negative number except we introduce complex number } \end{gathered}[/tex][tex]\begin{gathered} So\text{ from x = 2 towards the negative side, we won't be able to find the squareroot of the value} \\ \text{Hence, x have to be equal or greater than 3} \\ \\ \text{Domain: x }\ge\text{ 3} \\ In\text{ interval notation, } \\ \text{domain: \lbrack{}3, }\infty) \end{gathered}[/tex]
