Question

Please help for 10 and 11.

Answer 1

10:A 11:D

Explanation:

10)

g(
`(√(x+1)^2+3)/(√(x+1) )`

rewrite
`√(x+1)^2` as x +1

g
`(√(x+1)^2)` =
`(x+1+3)/(√(x+1) )`

add 1+3

g
`(√(x+1)^2)`=
`(x+4)/(√(x+1) )`

combine and symplify

`g{√(x+1) } *\frac{{x+4 (√(x+1)) } }{{x+1} }`

reorder factors of the second side

`g{√(x+1) } *\frac{{√(x+1) (x+4) } }{{x+1} }`

take out the root x+1

and you are left with A

11.

f(g(x))

substitue g into f

f(
`f{√(x+3) } =\((√(x+3))^2 +9}`

rewrite sqrroot x+3 as x+3

`f{√(x+3) } =\(x+3 +9}`

Now add 3+9

`f{√(x+3) } =\(x+12}`

we get x+12

Now for domain

`f{√(x+3) } =\((√(x+3))^2 +9}`

set the radicand root sqr x+3 >= 0 to find the expression defined

`x+3\geq 0`

subtract 3 from both sides

`x\geq 3`

turn it into interval notation

[-3,+∞)