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17.1 Verify Rolle s theorem for f (x) = x 3 3x 2 x + 3 on the interval [1, 3].
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f is differentiable everywhere and, therefore, also continuous. Furthermore, f (1) = (1)3 3(1)2 1 + 3 = 1 3 1 + 3 = 0 f (3) = (3)3 3(3)2 3 + 3 = 27 27 3 + 3 = 0 so that all the hypotheses of Rolle s theorem are valid. There must then be some c in (1, 3) for which f (c) = 0. Now, by the quadratic formula, the roots of f (x) = 3x 2 6x 1 = 0 are 62 4(3)( 1) 6 = x= 2(3) 2 6 4 3 =1 3 = 6 3 6 36 + 12 6 48 6 16 3 = = 6 6 6
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THE MEAN-VALUE THEOREM AND THE SIGN OF THE DERIVATIVE Consider the root c = 1 + 2 3. Since 3 < 3, 3 1<1+ Thus, c is in (1, 3) and f (c) = 0. 2 2 3 < 1 + (3) = 1 + 2 = 3 3 3
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17.2 Verify the mean-value theorem for f (x) = x 3 6x 2 4x + 30 on the interval [4, 6].
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f is differentiable and, therefore, continuous for all x. f (6) = (6)3 6(6)2 4(6) + 30 = 216 216 24 + 30 = 6 f (4) = (4)3 6(4)2 4(4) + 30 = 64 96 16 + 30 = 18 whence, f (6) f (4) 6 ( 18) 24 = = = 12 6 4 6 4 2
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We must therefore nd some c in (4, 6) such that f (c) = 12. Now, f (x) = 3x 2 12x 4, so that c will be a solution of 3x 2 12x 4 = 12 By the quadratic formula, 144 4(3)( 16) 12 144 + 192 12 336 = = 2(3) 6 6 12 4 21 12 16 21 = = 2 2 21 = 3 6 6 Choose c = 2 + 2 21. Since 4 < 21 < 5, 3 x= 12 4<2+ Thus, c is in (4, 6) and f (c) = 12. 8 2 2 2 = 2 + (4) < 2 + 21 < 2 + (5) < 2 + 4 = 6 3 3 3 3 or 3x 2 12x 16 = 0
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17.3 Determine when the function f (x) = x 3 6x 2 + 9x + 2 is increasing and when it is decreasing, and sketch its graph.
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We have f (x) = 3x 2 12x + 9 = 3(x 2 4x + 3) = 3(x 1)(x 3). The crucial points are 1 and 3 [see Fig. 17-5(a)]. (i) When x < 1, both (x 1) and (x 3) are negative and so f (x) > 0 in ( , 1). (ii) When x moves from ( , 1) into (1, 3), the factor (x 1) changes from negative to positive, but (x 3) remains negative. Hence, f (x) < 0 in (1, 3). (iii) When x moves from (1, 3) into (3, ), (x 3) changes from negative to positive, but (x 1) remains positive. Hence, f (x) > 0 in (3, ). Thus, by Theorem 17.3, f is increasing for x < 1, decreasing for 1 < x < 3, and increasing for x > 3. Note that f (1) = 6, f (3) = 2, lim f (x) = + , and lim f (x) = . A rough sketch of the graph is shown in Fig. 17-5(b).
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x + x
17.4 Verify Rolle s theorem for f (x) = 2x 6 8x 5 + 6x 4 x 3 + 6x 2 11x + 6 on [1, 3].
f is differentiable everywhere, and f (1) = 2 8 + 6 1 + 6 11 + 6 = 0 f (3) = 1458 1944 + 486 27 + 54 33 + 6 = 0
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