Page 53 - Calculus Demystified

CHAPTER 1
40
1.8.4 COMPOSITION OF FUNCTIONS Basics
Suppose that f and g are functions and that the domain of g contains the range of f .
This means that if x is in the domain of f then f(x) makes sense but also g may
be applied to f(x) (Fig. 1.46). The result of these two operations, one following
the other, is called g composed with f or the composition of g with f . We write

(g ◦ f )(x) = g(f (x)).

f (x) g( f (x))
x
Fig. 1.46

EXAMPLE 1.35
2
Let f(x) = x − 1 and g(x) = 3x + 4. Calculate g ◦ f .
SOLUTION
We have
2
(g ◦ f )(x) = g(f (x)) = g(x − 1). (∗)
Notice that we have started to work inside the parentheses: the first step was
2
to substitute the definition of f , namely x − 1, into our equation.
Now the definition of g says that we take g of any argument by multiplying
that argument by 3 and then adding 4. In the present case we are applying g to
2
x − 1. Therefore the right side of equation (∗) equals
2
3 · (x − 1) + 4.
2
This easily simplifies to 3x + 1. In conclusion,
2
g ◦ f(x) = 3x + 1.
EXAMPLE 1.36
2
Let f(t) = (t − 2 )/(t + 1 ) and g(t) = 2t + 1. Calculate g ◦ f and f ◦ g.
SOLUTION
We calculate that

2
t − 2
(g ◦ f )(t) = g(f (t)) = g . (∗∗)
t + 1