Page 68 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 3] FUNCTIONS, LIMITS, AND CONTINUITY 59

Suppose that f ðaÞ < 0 and f ðbÞ > 0. Since f ðxÞ is continuous there must be an interval ða; a þ hÞ, h > 0,
for which f ðxÞ < 0. The set of points ða; a þ hÞ has an upper bound and so has a least upper bound which
we call c. Then f ðcÞ @ 0. Now we cannot have f ðcÞ < 0, because if f ðcÞ were negative we would be able to
find an interval about c (including values greater than c)for which f ðxÞ < 0; but since c is the least upper
bound, this is impossible, and so we must have f ðcÞ¼ 0as required.
If f ðaÞ > 0 and f ðbÞ < 0, a similar argument can be used.

3 2
3.33. (a) Given f ðxÞ¼ 2x 3x þ 7x 10, evaluate f ð1Þ and f ð2Þ.(b) Prove that f ðxÞ¼ 0 for some
real number x such that 1 < x < 2. (c) Show how to calculate the value of x in (b).
3
3
2
2
(a) f ð1Þ¼ 2ð1Þ 3ð1Þ þ 7ð1Þ 10 ¼ 4, f ð2Þ¼ 2ð2Þ 3ð2Þ þ 7ð2Þ 10 ¼ 8.
(b)If f ðxÞ is continuous in a @ x @ b and if f ðaÞ and f ðbÞ have opposite signs, then there is a value of x
between a and b such that f ðxÞ¼ 0(Problem 3.32).
To apply this theorem we need only realize that the given polynomial is continuous in 1 @ x @ 2,
since we have already shown in (a)that f ð1Þ < 0 and f ð2Þ > 0. Thus there exists a number c between 1
and 2 such that f ðcÞ¼ 0.
3
2
(c) f ð1:5Þ¼ 2ð1:5Þ 3ð1:5Þ þ 7ð1:5Þ 10 ¼ 0:5. Then applying the theorem of (b) again, we see that the
required root lies between 1 and 1.5 and is ‘‘most likely’’ closer to 1.5 than to 1, since f ð1:5Þ¼ 0:5 has a
value closer to 0 than f ð1Þ¼ 4 (this is not always a valid conclusion but is worth pursuing in practice).
2
3
Thus we consider x ¼ 1:4. Since f ð1:4Þ¼ 2ð1:4Þ 3ð1:4Þ þ 7ð1:4Þ 10 ¼ 0:592, we conclude
that there is a root between 1.4 and 1.5 which is most likely closer to 1.5 than to 1.4.
Continuing in this manner, we find that the root is 1.46 to 2 decimal places.
3.34. Prove Theorem 10, Page 48.
Given any > 0, we can find x such that M f ðxÞ < by definition of the l.u.b. M.
1 1 1
Then > ,sothat is not bounded and hence cannot be continuous in view of

M f ðxÞ M f ðxÞ
Theorem 4, Page 47. However, if we suppose that f ðxÞ 6¼ M,then since M f ðxÞ is continuous, by
1
hypothesis, we must have also continuous. In view of this contradiction, we must have
M f ðxÞ
f ðxÞ¼ M for at least one value of x in the interval.
Similarly, we can show that there exists an x in the interval such that f ðxÞ¼ m (Problem 3.93).

Supplementary Problems

FUNCTIONS
3.35. Give the largest domain of definition for which each of the following rules of correspondence support the
construction of a function.
p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 p ffiffiffiffiffiffiffiffiffiffiffiffi 3 2
(a) ð3 xÞð2x þ 4Þ, (b) ðx 2Þ=ðx 4Þ, (c) sin 3x, (d)log ðx 3x 4x þ 12Þ.
10
Ans. (a) 2 @ x @ 3, (b)all x 6¼ 2, (c)2m =3 @ x @ ð2m þ 1Þ =3, m ¼ 0; 1; 2; ... ;
(d) x > 3, 2 < x < 2.
3x þ 1 1 2
3.36. If f ðxÞ¼ , x 6¼ 2, find: (a) 5f ð 1Þ 2f ð0Þþ 3f ð5Þ ; (b) f f ð Þg ; (c) f ð2x 3Þ;
x 2 6 2
(d) f ðxÞþ f ð4=xÞ, x 6¼ 0; (e) f ðhÞ f ð0Þ , h 6¼ 0; ( f ) f ðf f ðxÞg.
h
6x 8 7
5
5
Ans. (a) 61 (b) 1 (c) , x 6¼ 0, ,2 (d) , x 6¼ 0; 2(e) , h 6¼ 0; 2
18 25 2x 5 2 2 2h 4
10x þ 1
( f ) , x 6¼ 5; 2
x þ 5