Thomas Calculus 13th [Solutions]

Section 2.5 Continuity 89
49. The function can be extended: f (0) 2.3. 50. The function cannot be extended to be continuous at
x 0. If f (0) 2.3, it will be continuous from the
right. Or if f (0) 2.3, it will be continuous from
the left.
51. The function cannot be extended to be continuous
at x 0. If f (0) 1, it will be continuous from the
right. Or if f (0) 1, it will be continuous from
the left.
52. The function can be extended: f (0) 7.39.
53. f ( x ) is continuous on [0, 1] and f (0) 0, f (1) 0
by the Intermediate Value Theorem f ( x ) takes on
every value between f (0) and f (1) the equation
f ( x ) 0 has at least one solution between x 0
and x 1.
54. cos x x (cos x) x 0. If x 2 , cos
0. If
2 2
x 2 , cos 0. Thus cos x x 0 for
2 2
some x between
2 and according to the Intermediate Value Theorem, since the function cos x x is
2
continuous.
3
55. Let f ( x) x 15x 1, which is continuous on [ 4, 4]. Then f ( 4) 3, f ( 1) 15, f (1) 13, and f (4) 5.
By the Intermediate Value Theorem, f ( x ) 0 for some x in each of the intervals 4 x 1, 1 x 1, and
1 x 4. That is, x 3 15x 1 0 has three solutions in [ 4, 4]. Since a polynomial of degree 3 can have at
most 3 solutions, these are the only solutions.
2 2
56. Without loss of generality, assume that a b . Then F( x) ( x a) ( x b)
x is continuous for all values of x,
so it is continuous on the interval [ a, b ]. Moreover F( a)
a and F( b) b . By the Intermediate Value Theorem,
since a a b b , there is a number c between a and b such that F( x)
a b .
2
2
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