Answer:
10:A 11:D
Step-by-step explanation:
10)
g([tex](\sqrt{x+1}^2+3)/(\sqrt{x+1} )[/tex]
rewrite [tex]\sqrt{x+1}^2[/tex] as x +1
g [tex](\sqrt{x+1}^2)[/tex] = [tex]\frac{x+1+3}{\sqrt{x+1} }[/tex]
add 1+3
g[tex](\sqrt{x+1}^2)[/tex]=[tex]\frac{x+4}{\sqrt{x+1} }[/tex]
combine and symplify
[tex]g{\sqrt{x+1} } *\frac{{x+4 (\sqrt{x+1}) } }{{x+1} }[/tex]
reorder factors of the second side
[tex]g{\sqrt{x+1} } *\frac{{\sqrt{x+1} (x+4) } }{{x+1} }[/tex]
take out the root x+1
and you are left with A
11.
f(g(x))
substitue g into f
f([tex]f{\sqrt{x+3} } =\((\sqrt{x+3})^2 +9}[/tex]
rewrite sqrroot x+3 as x+3
[tex]f{\sqrt{x+3} } =\(x+3 +9}[/tex]
Now add 3+9
[tex]f{\sqrt{x+3} } =\(x+12}[/tex]
we get x+12
Now for domain
[tex]f{\sqrt{x+3} } =\((\sqrt{x+3})^2 +9}[/tex]
set the radicand root sqr x+3 >= 0 to find the expression defined
[tex]x+3\geq 0[/tex]
subtract 3 from both sides
[tex]x\geq 3[/tex]
turn it into interval notation
[-3,+∞)
