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THE MEAN-VALUE THEOREM AND THE SIGN OF THE DERIVATIVE
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It is dif cult to compute a value of x in (1, 3) for which f (x) = 12x 5 40x 4 + 24x 3 3x 2 + 12x 11 = 0 However, f (x) is itself a continuous function such that f (1) = 12 40 + 24 3 + 12 11 = 6 < 0 f (3) = 2916 3240 + 648 27 + 36 11 = 322 > 0 Hence, the intermediate-value theorem assures us that there must be some number c between 1 and 3 for which f (c) = 0.
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17.5 Show that f (x) = 2x 3 + x 4 = 0 has exactly one real solution.
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Since f (0) = 4 and f (2) = 16 + 2 4 = 14, the intermediate-value theorem guarantees that f has a zero between 0 and 2; call it x0 . Because f (x) = 6x 2 + 1 > 0, f (x) is increasing everywhere (Theorem 17.3). Therefore, when x > x0 , f (x) > 0; and when x < x0 , f (x) < 0. In other words, there is no zero other than x0 .
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17.6 Prove Rolle s theorem (Theorem 17.1).
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Case 1: Case 2: f (x) = 0 for all x in [a, b]. Then f (x) = 0 for all x in (a, b), since the derivative of a constant function is 0. f (x) > 0 for some x in (a, b). Then, by the extreme-value theorem (Theorem 14.2), an absolute maximum of f on [a, b] exists, and must be positive [since f (x) > 0 for some x in (a, b)]. Because f (a) = f (b) = 0, the maximum is achieved at some point c in the open interval (a, b). Thus, the absolute maximum is also a relative maximum and, by Theorem 14.1, f (c) = 0. f (x) < 0 for some x in (a, b). Let g(x) = f (x). Then, by Case 2, g (c) = 0 for some c in (a, b). Consequently, f (c) = g (c) = 0.
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17.7 Prove the mean-value theorem (Theorem 17.2).
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Let g(x) = f (x) f (b) f (a) (x a) f (a) b a
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THE MEAN-VALUE THEOREM AND THE SIGN OF THE DERIVATIVE
[CHAP. 17
Then g is continuous over [a, b] and differentiable on (a, b). Moreover, g(a) = f (a) f (b) f (a) (a a) f (a) = f (a) 0 f (a) = 0 b a f (b) f (a) g(b) = f (b) (b a) f (a) = f (b) (f (b) f (a)) f (a) b a = f (b) f (b) + f (a) f (a) = 0
By Rolle s theorem, applied to g, there exists c in (a, b) for which g (c) = 0. But, g (x) = f (x) whence, 0 = g (c) = f (c) f (b) f (a) b a f (b) f (a) b a or f (c) = f (b) f (a) b a
17.8 Prove Theorem 17.3.
Assume that f (x) > 0 for all x in (a, b) and that a < u < v < b. We must show that f (u) < f (v). By the mean-value theorem, applied to f on the closed interval [u, v], there is some number c in (u, v) such that f (c) = f (v) f (u) v u or f (v) f (u) = f (c)(v u)
But f (c) > 0 and v u > 0; hence, f (v) f (u) > 0, f (u) < f (v). The case f (x) < 0 is handled similarly.
17.9 Prove Corollary 17.5.
By the extreme-value theorem, f has an absolute maximum value f (d) at some argument d in [a, b], and an absolute minimum value f (c) at some argument c in [a, b]. If f (c) = f (d) = k, then f is constant on [a, b], and its range is the single point k. If f (c) = f (d), then the intermediate-value theorem, applied to the closed subinterval bounded by d and c, ensures that f assumes every value between f (c) and f (d). The range of f is then the closed interval [f (c), f (d)] (which includes the values assumed on that part of [a, b] that lies outside the subinterval).
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