When I work with function composition, I usually convert "( f o g)(x)" to the more intuitive " f (g(x))" form. This is not required, but I certainly find it helpful. In this case, I get:

(g o f )(1) = g( f(1))

This means that, working from right to left (or from the inside out), I am plugging x = 1 into f(x), evaluating f(x), and then plugging the result into g(x). I can do the calculations bit by bit, like this: Since f(1) = 2(1) + 3 = 2 + 3 = 5, and since g(5) = –(5)² + 5 = –25 + 5 = –20, then (g o f )(1) = g( f(1)) = g(5) = –20. Doing the calculations all together (which will be useful later on when we're doing things symbolically), it looks like this:

(g o f )(1) = g( f (1))
    = g(2(   ) + 3)   ... setting up to insert the original input
   = g(2(1) + 3)
   = g(2 + 3)
   = g(5)
   = –(   )² + 5    ... setting up to insert the new input
   = –(5)² + 5 
   = –25 + 5 
   = –20

Note how I wrote each function's rule clearly, leaving open parentheses for where the input (x or whatever) would go. This is a useful technique. Whichever method you use (bit-by-bit or all-in-one), the answer is:

I just computed (g o f )(1); the composition can also work in the other order:

First, I'll convert this to the more intuitive form, and then I'll simplify:

( f o g)(1) = f (g(1))

Working bit-by-bit, since g(1) = –(1)² + 5 = –1 + 5 = 4, and since f(4) = 2(4) + 3 = 8 + 3 = 11, then ( f o g)(1) = f (g(1)) = f(4) = 11. On the other hand, working all-in-one (right to left, or from the inside out), I get this:

( f o g)(1) = f (g(1))
    = f (–(    )² + 5)    ... setting up to insert the original input
    = f (–(1)² + 5)
    = f (–1 + 5)
    = f (4) 
    = 2(   ) + 3     ... setting up to insert the new input 
    = 2(4) + 3 
    = 8 + 3 
    = 11

Either way, the answer is:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

A verbal note:  "f o g" is not pronounced as "fogg" and "g o f " is not pronounced as "goff". They are pronounced as "f-compose-g" and "g-compose-f ", respectively. Don't make yourself sound ignorant by pronouncing these wrongly!

As you have seen above, you can plug one function into another. You can also plug a function into itself:

( f o f )(1) = f ( f (1))
    = f (2(   ) + 3)    ... setting up to insert the original input
    = f (2(1) + 3)
    = f (2 + 3)
    = f (5)
    = 2(   ) + 3     ... setting up to insert the new  input
    = 2(5) + 3
    = 10 + 3
    = 13

(g o g)(1) = g(g(1))
    = g(–(    )² + 5)    ... setting up to insert the original input
    = g(–(1)² + 5)
    = g(–1 + 5)
    = g(4)
    = –(   )² + 5    ... setting up to insert the new  input
    = –(4)² + 5
    = –16 + 5
    = –11

In each of these cases, I wrote out the steps carefully, using parentheses to indicate where my input was going with respect to the formula. If it helps you to do the steps separately, then calculate g(1) outside of the other g(x) as a separate step. That is, do the calculations bit-by-bit, first finding g(1) = 4, and then plugging 4 into g(x) to get g(4) = –11.