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Sullivan AP˙Sullivan˙Chapter01 October 8, 2016 17:4
Section 1.3 • Continuity 109
NEED TO REVIEW? Composite
functions are discussed in Section P.3,
pp. 27--30.
The proofs of these properties are based on properties of limits. For example, the
proof of the continuity of f + g is based on the Limit of a Sum property. That is, if
lim f (x) and lim g(x) exist, then lim[ f (x) + g(x)] = lim f (x) + lim g(x).
x→c x→c x→c x→c x→c
EXAMPLE 7
Identifying Where Functions Are Continuous
Determine where each function is continuous:
(a) F(x) = x 2 + 5 −
x
x 2 + 4
(b) G(x) = x 3 + 2x + x 2
x 2 − 1
Solution First we determine the domain of each function.
(a) F is the difference of the two functions f (x) = x 2 + 5 and g(x) =
x
x 2 + 4 , each
of whose domain is the set of all real numbers. So, the domain of F is the set of all real
numbers. Since f and g are continuous on their domains, the difference function F is
continuous on its domain.
(b) G is the sum of the two functions f (x) = x 3 + 2x, whose domain is the set of all
real numbers, and g(x) = x 2
, whose domain is {x|x = −1, x = 1} . Since f and g
x 2 − 1
are continuous on their domains, G is continuous on its domain, {x|x = −1, x = 1} . ■
NOW WORK Problem 45.
The continuity of a composite function depends on the continuity of its components.
THEOREM Continuity of a Composite Function
If a function g is continuous at c and a function f is continuous at g(c), then the
composite function ( f ◦ g)(x) = f (g(x)) is continuous at c. That is,
lim( f ◦ g)(x) = lim
x→c x→c
f (g(x)) = f [lim g(x)] = f (g(c))
x→c
EXAMPLE 8 Identifying Where Functions Are Continuous
Determine where each function is continuous:
(a) F(x) = √ x 2 + 4 (b) G(x) = √ x 2 − 1 (c) H(x) = x 2 − 1
x 2 − 4 + √ x − 1
Solution (a) F = f ◦ g is the composite of f (x) = √ x and g(x) = x 2 + 4. f is
continuous for x ≥ 0 and g is continuous for all real numbers. The domain of F is all
real numbers and F = ( f ◦ g)(x) = √ x 2 + 4 is continuous for all real numbers. That
is, F is continuous on its domain.
(b) G is the composite of f (x) = √ x and g(x) = x 2 − 1. f is continuous for x ≥ 0
and g is continuous for all real numbers. The domain of G is {x|x ≥ 1}∪{x|x ≤−1}
and G = ( f ◦ g)(x) = √ x 2 − 1 is continuous on its domain.
(c) H is the sum of f (x) = x 2 − 1
x 2 − 4 and the function g(x) = √ x − 1. The domain
of f is {x|x = −2, x = 2}; f is continuous on its domain. The domain of g is
x ≥ 1. The domain of H is {x|1 ≤ x < 2} ∪{x|x > 2}; H is continuous on its
domain. ■
NOW WORK Problem 47.
Teaching Tip
Students should have a strong knowledge
of what all of the basic functions (as shown
in Section P.2) look like. The following is
a suggested list of functions the students
should be familiar with. Students should
also be comfortable with transformations of
these functions.
y = x y = sinx y = e
2
y = x y = cos x y = e
−x
3
y = x y = tanx y = lnx
1
y = | x|
y = x y =
x
Alternate Example
Identifying Where Functions Are
Continuous
Which of the following functions are
continuous at x = 0?
1
(a) fx ( ) =
x
(b) g(x) = tan x
Solution
(a) Since x = 0 is not in the domain of f,
1
fx ( ) = is discontinuous at x = 0.
x
(b) The function gx ( ) = tan x is defined
and continuous on − π < x < π , so
2 2
gx ( ) = tanx
is continuous at x = 0.
x
NEED TO REVIEW? Inverse functions
are discussed in Section P.4, pp. 37--40.
Recall that for any function f that is one-to-one over its domain, its inverse f −1
is also a function, and the graphs of f and f −1 are symmetric with respect to the line
y = x. It is intuitive that if f is continuous, then so is f −1 . See Figure 31 on page 110.
The following theorem, whose proof is given in Appendix B, confirms this.
Section 1.3 • Continuity 109
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