Question

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

Answer 1

See below for explanation

Explanation:

We have two functions and we have to compose them in the order to find `s(t(6))`. (The composition of functions is not conmutative)

We note that `s(x)=x^2+5x`

We proceed as follows

`s(t(x))=s(sqrt(x+4))=(sqrt(x+4))^2+5sqrt(x+4)=x+4+5sqrt(x+4)`

Then `s(t(6))=6+4+5sqrt(6+4)=10+5sqrt10`

Answer 2

`s(t(6))= 10+5sqrt(10)`
This question asks for s(t(6)) not s(t)(6)

Explanation:

To solve for `s(t)(6)`
`s(x) = x^2 +2x +3*x`
`t(x) = sqrt(x+4)`
let's first find s(t).
Substitute t(x) into s(x):
`s(t(x))= (sqrt(x + 4))^2 +2sqrt(x+4)+3*sqrt(x+4)`
Simplifying this gives us:
`s(t(x))= ( x + 4)+2sqrt(x+4)+3*sqrt(x+4)`
Now, subbing 6 into the X values gives us:
`s(t(6))= ( 6+ 4)+2sqrt(6+4)+3*sqrt(6+4)`
`s(t(6))= 10+2sqrt(10)+3*sqrt(10)`
`s(t(6))= 10+5sqrt(10)`

Additional info:
Note that functions like s(t(x)) are called composite functions.