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Since by hypothesis f x and g y are continuous at x0 and y0 , respectively, we lim f x f lim x f x0
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which proves that h x gf f x g is continuous at x x0 .
3.32. Prove Theorem 8, Page 48.
CHAP. 3]
FUNCTIONS, LIMITS, AND CONTINUITY
Suppose that f a < 0 and f b > 0. Since f x is continuous there must be an interval a; a h , h > 0, for which f x < 0. The set of points a; a h has an upper bound and so has a least upper bound which we call c. Then f c @ 0. Now we cannot have f c < 0, because if f c were negative we would be able to nd an interval about c (including values greater than c) for which f x < 0; but since c is the least upper bound, this is impossible, and so we must have f c 0 as required. If f a > 0 and f b < 0, a similar argument can be used.
3.33. (a) Given f x 2x3 3x2 7x 10, evaluate f 1 and f 2 . (b) Prove that f x 0 for some real number x such that 1 < x < 2. (c) Show how to calculate the value of x in (b).
(a) f 1 2 1 3 3 1 2 7 1 10 4, f 2 2 2 3 3 2 2 7 2 10 8. (b) If f x is continuous in a @ x @ b and if f a and f b have opposite signs, then there is a value of x between a and b such that f x 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 1 < 0 and f 2 > 0. Thus there exists a number c between 1 and 2 such that f c 0. (c) f 1:5 2 1:5 3 3 1:5 2 7 1:5 10 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 1:5 0:5 has a value closer to 0 than f 1 4 (this is not always a valid conclusion but is worth pursuing in practice). Thus we consider x 1:4. Since f 1:4 2 1:4 3 3 1:4 2 7 1:4 10 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 nd that the root is 1.46 to 2 decimal places.
3.34. Prove Theorem 10, Page 48.
Given any  > 0, we can nd x such that M f x <  by de nition of the l.u.b. M. 1 1 1 > , so that is not bounded and hence cannot be continuous in view of Then M f x  M f x Theorem 4, Page 47. However, if we suppose that f x 6 M, then since M f x is continuous, by 1 hypothesis, we must have also continuous. In view of this contradiction, we must have M f x f x 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 x m (Problem 3.93).
Supplementary Problems
FUNCTIONS 3.35. Give the largest domain of de nition for which each of the following rules of correspondence support the construction of a function. p p (a) 3 x 2x 4 , (b) x 2 = x2 4 , (c) sin 3x, (d) log10 x3 3x2 4x 12 . Ans. (a) 2 @ x @ 3, (d) x > 3, 2 < x < 2. 3.36. If f x (b) all x 6 2, (c) 2m=3 @ x @ 2m 1 =3, m 0; 1; 2; . . . ;
5f 1 2f 0 3f 5 ; (b) f f 1 g2 ; (c) f 2x 3 ; 2 6 f h f 0 , h 6 0; ( f ) f f f x g. (d) f x f 4=x , x 6 0; (e) h 6x 8 7 1 Ans. (a) 61 (b) 25 (c) , x 6 0, 5, 2 (d) 5, x 6 0; 2 (e) , h 6 0; 2 18 2 2 2x 5 2h 4 10x 1 , x 6 5; 2 (f) x 5 (a)
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