1 2 sec2 u 1du 1 1 tan u u c tan 3 3 c 3 3 1

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Unformatted text preview: ) = 0; f (0) = 0 f (0) = f (4) = 0; f (3) = 0; [0, 4], c = 3 3. f (2) = f (4) = 0, f (x) = 2x − 6, 2c − 6 = 0, c = 3 4. f (0) = f (2) = 0, f (x) = 3x − 6x + 2, 3c − 6c + 2 = 0; c = 2 2 6± √ √ 36 − 24 = 1 ± 3/3 6 5. f (π/2) = f (3π/2) = 0, f (x) = − sin x, − sin c = 0, c = π x2 − 4x + 1 c2 − 4c + 1 , = 0, c2 − 4c + 1 = 0 (x − 2)2 (c − 2)2 √ √ √ 4 ± 16 − 4 = 2 ± 3, of which only c = 2 − 3 is in (−1, 1) c= 2 6. f (−1) = f (1) = 0, f (x) = 7. f (0) = f (4) = 0, f (x) = 8. 1 1 1 1 − √ , − √ = 0, c = 1 2 2x2 2c f (1) = f (3) = 0, f (x) = − 2 4 2 4 + 2 , − 3 + 2 = 0, −6 + 4c = 0, c = 3/2 3 x 3x c 3c Exercise Set 6.5 9. 202 6 3 f (8) − f (0) = = = f (1.54); c = 1.54 8−0 8 4 10. 42 − 12 = 3, c = 1 6 − (−4) 11. f (−4) = 12, f (6) = 42, f (x) = 2x + 1, 2c + 1 = 12. f (−1) = −6, f (2) = 6, f (x) = 3x2 + 1, 3c2 + 1 = in (−1, 2) 13. f (4) − f (0) = 1.19 = f (0.77) 4−0 6 − (−6) = 4, c2 = 1, c = ±1 of which only c = 1 is 2 − (−1) 1 2−1 1√ 1 ,√ = = , c + 1 = 3/2, c + 1 = 9/4, c = 5/4 f (0) = 1, f (3) = 2, f (x) = √ 3−0 3 2 x+1 2 c+1 14. f (3) = 10/3, f (4) = 17/4, f (x) = 1 − 1/x2 , 1 − 1/c2 = √ which only c = 2 3 is in (3, 4) 15. f (−5) = 0, f (3) = 4, f (x) = − √ √ 17/4 − 10/3 = 11/12, c2 = 12, c = ±2 3 of 4−3 √ 1 x c 4−0 = , −2c = 25 − c2 , , −√ = 2 2 3 − (−5) 2 25 − x 25 − c √ 4c2 = 25 − c2 , c2 = 5, c = − 5 √ √ (we reject c = 5 because it does not satisfy the equation −2c = 25 − c2 ) 16. f (2) = 1, f (5) = 1/4, f (x) = −1/(x − 1)2 , c = −1 (reject), or c = 3 17. (a) 1 1 1/4 − 1 = − , (c − 1)2 = 4, c − 1 = ±2, = 2 (c − 1) 5−2 4 f (−2) = f (1) = 0 (b) c = −1.29 6 -2 1 -2 (c) 18. x0 = −1, x1 = −1.5, x2 = −1.32, x3 = −1.290, x4 = −1.2885843 f (−2) − f (1) −16 − 5 = = 7 so y − 5 = 7(x − 1), −2 − 1 −3 y = 7x − 2 (a) m = (b) f (x) = 3x2 + 4 = 7 has solutions x = ±1; discard x = 1, so c = −1 (d) 5 -2 1 (c) y − f (−1) = 7(x − (−1)) or y = 7x + 2 -20 19. (a) f (x) = sec2 x, sec2 c = 0 has no solution 20. (a) f (−1) = 1, f (8) = 4, f (x) = (b) tan x is not continuous on [0, π ] 2 −1/3 x 3 4−1 1 2 −1/3 = c = , c1/3 = 2, c = 8 which is not in (−1, 8). 3 8 − (−1) 3 (b) x2/3 is not differentiable at x = 0, which is in (−1, 8). 203 21. Chapter 6 Two x-intercepts of f determine two solutions a and b of f (x) = 0; by Rolle’s Theorem there exists a point c between a and b such that f (c) = 0, i.e. c is an x-intercept for f . (b) 22. (a) f (x) = sin x = 0 at x = nπ , and f (x) = cos x = 0 at x = nπ + π/2, which lies between nπ and (n + 1)π , (n = 0, ±1, ±2, . . .) f (x1 ) − f (x0 ) is the average rate of change of y with respect to x on the interval [x0 , x1 ]. By the x1 − x0 Mean-Value Theorem there is a value c in (x0 , x1 ) such that the instantaneous rate of change f (c) = f (x1 ) − f (x0 ) . x1 − x0 23. Let s(t) be the position function of the automobile for 0 ≤ t ≤ 5, then by the Mean-Value Theorem there is at least one point c in (0, 5) where s (c) = v (c) = [s(5) − s(0)]/(5 − 0) = 4/5 = 0.8 mi/min = 48 mi/h. 24. Let T (t) denote the temperature at time with t = 0 denoting 11 AM, then T (0) = 76 and T (12) = 52. (a) By the Mean-Value Theorem there is a value c between 0 and 12 such that T (c) = [T (12) − T (0)]/(12 − 0) = (52 − 76)/(12) = −2◦ F/h. (b) Assume that T (t1 ) = 88◦ F where 0 < t1 < 12, then there is at least one point c in (t1 , 12) where T (c) = [T (12) − T (t1 )]/(12 − t1 ) = (52 − 88)/(12 − t1 ) = −36/(12 − t1 ). But 12 − t1 < 12 so T (c) < −36/12 = −3◦ F/h. 25. Let f (t) and g (t) denote the distances from the first and second runners to the starting point, and let h(t) = f (t) − g (t). Since they start (at t = 0) and finish (at t = t1 ) at the same time, h(0) = h(t1 ) = 0, so by Rolle’s Theorem there is a time t2 for which h (t2 ) = 0, i.e. f (t2 ) = g (t2 ); so they have the same velocity at time t2 . 26. f (x) = x6 − 2x2 + x satisfies f (0) = f (1) = 0, so by Rolle’s Theorem f (c) = 0 for some c in (0, 1). 27. (a) By the Constant Difference Theorem f (x) − g (x) = k for some k ; since f (x0 ) = g (x0 ), k = 0, so f (x) = g (x) for all x. (b) Set f (x) = sin2 x + cos2 x, g (x) = 1; then f (x) = 2 sin x cos x − 2 cos x sin x = 0 = g (x). Since f (0) = 1 = g (0), f (x) = g (x) for all x. (a) By the Constant Difference Theorem f (x) − g (x) = k for some k ; since f (x0 ) − g (x0 ) = c, k = c, so f (x) − g (x) = c for all x. (b) Set f (x) = (x − 1)3 , g (x) = (x2 + 3)(x − 3). Then f (x) = 3(x − 1)2 , g (x) = (x2 + 3) + 2x(x − 3) = 3x2 − 6x + 3 = 3(x2 − 2x + 1) = 3(x − 1)2 , so f (x) = g (x) and hence f (x) − g (x) = k . Expand f (x) and g (x) to get h(x) = f (x) − g (x) = (x3 − 3x2 + 3x − 1) − (x3 − 3x2 + 3x − 9) = 8. (c) h(x) = x3 − 3x2 + 3x − 1 − (x3 − 3x2 + 3x − 9) = 8 28. 29. 30. f (y ) − f (x) = f (c), y−x so |f (x) − f (y )| = |f (c)||x − y | ≤ M |x − y |; if x > y exchange x and y ;...
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