Calculus- Early Transcendentals, 2021a

52 Functions
Solution. The domain of f is D f = {x ∈ R}. The domain of g is D g = {x ∈ R : x ≥ 0}. The function
( f ◦ g)(x)= f (g(x)) is:
f (g(x)) = (√ x ) 2 = x.
Typically, h(x) =x would have a domain of {x ∈ R}, but since it came from a composed function, we
must consider g(x) when looking at the domain of f (g(x)). Thus, the domain of f ◦ g is {x ∈ R : x ≥ 0}.
♣
Example 2.9: Combining Two Functions
Let f (x)=x 2 + 3 and g(x)=x − 2. Findf + g, f − g, fg, f /g, f ◦ g and g ◦ f . Also, determine the
domains of these new functions.
Solution. For f + g we have:
( f + g)(x)= f (x)+g(x)=(x 2 + 3)+(x − 2)=x 2 + x + 1.
For f − g we have:
( f − g)(x)= f (x) − g(x)=(x 2 + 3) − (x − 2)=x 2 + 3 − x + 2 = x 2 − x + 5.
For fgwe have:
For f /g we have:
For f ◦ g we have:
( fg)(x)= f (x) · g(x)=(x 2 + 3)(x − 2)=x 3 − 2x 2 + 3x − 6.
( f
g
)
(x)= f (x)
g(x) = x2 + 3
x − 2 .
For g ◦ f we have:
( f ◦ g)(x)= f (g(x)) = f (x − 2)=(x − 2) 2 + 3 = x 2 − 4x + 7.
(g ◦ f )(x)=g( f (x)) = g(x 2 + 3)=(x 2 + 3) − 2 = x 2 + 1.
The domains of f + g, f − g, fg, f ◦ g and g ◦ f is {x ∈ R}, while the domain of f /g is {x ∈ R : x ≠ 2}.
♣
As in the above problem, f ◦ g and g ◦ f are generally different functions.
Exercises for 2.2
Exercise 2.2.1 Starting with the graph of y = √ x, the graph of y = 1/x, and the graph of y =
(the upper unit semicircle), sketch the graph of each of the following functions:
√
1 − x 2
(a) f (x)= √ x − 2
(b) f (x)=−1 − 1/(x + 2)
(c) f (x)=4 + √ x + 2
(d) y = f (x)=x/(1 − x)