Answer:
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:
(g o f )(1) = g( f (1)) = –20
I just computed (g o f )(1); the composition can also work in the other order
Step-by-step explanation:
