#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!