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[tex]\begin{gathered} \text{Given:} \\ f(x)=\sqrt[]{x+2} \\ g(x)=3x-2 \end{gathered}[/tex][tex]\begin{gathered} (f+g)(2)=? \\ \text{Solve first for }(f+g)(x) \\ (f+g)(x)=f(x)+g(x) \\ (f+g)(x)=\sqrt[]{x+2}+3x-2 \\ \text{Now solve for }(f+g)(2) \\ (f+g)(x)=\sqrt[]{x+2}+3x-2 \\ (f+g)(2)=\sqrt[]{2+2}+3(2)-2 \\ (f+g)(2)=\sqrt[]{4}+6-2 \\ (f+g)(2)=2+6-2 \\ (f+g)(2)=6 \end{gathered}[/tex][tex]\begin{gathered} (\frac{f}{g})(0)=? \\ \text{Solve first for }(\frac{f}{g})(x) \\ (\frac{f}{g})(x)=\frac{f(x)}{g(x)} \\ (\frac{f}{g})(x)=\frac{\sqrt[]{x+2}}{3x-2} \\ \\ \text{Now solve for }(\frac{f}{g})(0) \\ (\frac{f}{g})(x)=\frac{\sqrt[]{x+2}}{3x-2} \\ (\frac{f}{g})(0)=\frac{\sqrt[]{0+2}}{3(0)+2} \\ (\frac{f}{g})(0)=\frac{\sqrt[]{2}}{2} \end{gathered}[/tex][tex]\begin{gathered} (f-g)(-1)=?_{} \\ \text{Solve for }(f-g)(x) \\ (f-g)(x)=f(x)-g(x) \\ (f-g)(x)=\sqrt[]{x+2}-(3x-2) \\ (f-g)(x)=\sqrt[]{x+2}-3x+2 \\ \\ \text{Now solve for }(f-g)(-1) \\ (f-g)(x)=\sqrt[]{x+2}-3x+2 \\ (f-g)(-1)=\sqrt[]{-1+2}-3(-1)+2 \\ (f-g)(-1)=\sqrt[]{1}+3+2 \\ (f-g)(-1)=1+3+2 \\ (f-g)(-1)=6 \end{gathered}[/tex]
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