Page 54 - Calculus Demystified

CHAPTER 1
Basics
We compute g of any argument by doubling it and adding 1. Thus equation 41
(∗∗) equals
2
t − 2
2 + 1
t + 1
2
2t − 4
= + 1
t + 1
2
2t + t − 3
= .
t + 1
One of the main points of this example is to see that f ◦ g is different from
g ◦ f . We compute f ◦ g:
(f ◦ g)(t) = f(g(t))
= f(2t + 1)
2
(2t + 1) − 2
=
(2t + 1) + 1
2
4t + 4t − 1
= .
2t + 2
So f ◦ g and g ◦ f are different functions.

√
You Try It: Let f(x) =|x| and g(x) = x/x. Calculate f ◦ g(x) and g ◦ f(x).
We say a few words about recognizing compositions of functions.

EXAMPLE 1.37
2
How can we write the function k(x) = (2x + 3) asthe composition of two
functions g and f ?

SOLUTION
Notice that the function k can be thought of as two operations applied in
sequence. First we double and add three, then we square. Thus define f(x) =
2
2x + 3 and g(x) = x . Then k(x) = (g ◦ f )(x).
We can also compose three (or more) functions: Define

(h ◦ g ◦ f )(x) = h(g(f (x))).

EXAMPLE 1.38
Write the function k from the last example as the composition of three
functions (instead of just two).