Question

Let `s(x) = x ^2 + 2x + 3x` and #t(x) = sqrt(x+4), how do you find sot(6)?

Answer 1

`sot(6) = 10+5sqrt(10) = 5(2+sqrt(10))`

Explanation:

Since `s(color(blue)(x)) = color(blue)(x)^2+2color(blue)(x)+3color(blue)(x) = color(blue)(x)^2+5color(blue)(x)`

`sot(6)` (also written `s(color(blue)(t(6)))`)
`color(white)("XXX")=color(blue)(t(6))^2+5*color(blue)(t(6))`

And since `t(color(red)(x)) =sqrt(color(red)(x)+4)`
`color(white)("XXX")t(color(red)(6)) = sqrt(color(red)(6)+4) = sqrt(10)`

Therefore:
`color(white)("XXX")sot(6)=s(t(6)) = (sqrt(10))^2+5*sqrt(10) = 10+5sqrt(10)`

Answer 2

Assuming you meant: `s(x)=x^2+2x+3 " and "t(x)=sqrt(x+4)`
Find `s@t(6)`

`color(blue)(s@t(6)=13+2sqrt(10))`

Explanation:

Assumption: There is a typing error in the question and that it should be:

`s(x)=x^2+2x+3` and `t(x)=sqrt(x+4)`

Find `s@t(6)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)(Step 1)`

`t(x) -> t(6) = sqrt(6+4) =sqrt(10)`

`color(blue)(Step2)`

`s(x)->s@t(6) =(sqrt(10))^2+2(sqrt(10))+3`

`s@t(6)=13+2sqrt(10)`