Question

Question #402e2

Answer 1

`f@g(2)=9`

Explanation:

`f@g` is read "f of g of x". It's a "composition function". All you do is put the second function in the first. So substitute `g(x)` for the `x` in `f(x)`, and you've got your new function. Here are two methods:
1. Find `g(2)` first, then plug that in to `f(x)`:
`g(2)=2(2)+1=5`
Plug that into `f(x): f(5)=5+4=9`.
or
2. Find `f@g(x)` then plug `2` in for `x`:
`f@g(x)=f(g(x))=f(2x+1)`
Plugging `(2x+1)` in for `x` gives us
`f(2x+1)=(2x+1)+4=2x+5`
and
If `f@g(x)=2x+ 5`, then `f@g(2)=2(2)+5=9`

The second way looks longer, but they're both handy to know. Hope this helps, and have fun!