C the particle is speeding up because the slope

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Unformatted text preview: √ 10, √ 10, no limit, +∞, (b) 5, 10, 0, 0, 10, −∞, +∞ (b) −1, +1, −1, −1, no limit, −1, +1 22. 1 25. 0 23. 26. k 2 does not exist 27. 3−k The numerator satisfies: |2x + x sin 3x| ≤ |2x| + |x| = 3|x|. Since the denominator grows like x2 , the limit is 0. √ √ x2 1 x2 + 4 − 2 x2 + 4 + 2 √ √ 30. (a) = =√ , so 2+4+2 2 ( x2 + 4 + 2) 2+4+2 x2 x x x √ 1 x2 + 4 − 2 1 = lim √ = lim x→0 x→0 x2 4 x2 + 4 + 2 28. Supplementary Exercises 2 x f (x) (b) 1 0.236 64 0.1 0.2498 0.01 0.2500 0.001 0.2500 0.0001 0.25000 0.00001 0.00000 The division may entail division by zero (e.g. on an HP 42S), or the numerator may be inaccurate (catastrophic subtraction, e.g.). (c) in the 3d picture, catastrophic subtraction 31. x f (x) 0.1 2.59 0.01 2.70 0.001 2.717 0.0001 2.718 0.00001 2.7183 0.000001 2.71828 32. x f (x) 3.1 5.74 3.01 5.56 3.001 5.547 3.0001 5.545 3.00001 5.5452 3.000001 5.54518 33. x f (x) 1.1 0.49 1.01 0.54 1.001 0.540 1.0001 0.5403 1.00001 0.54030 1.000001 0.54030 34. x f (x) 0.1 99.0 0.01 9048.8 35. x f (x) 100 0.48809 0.001 368063.3 0.0001 4562.7 104 0.49876 1000 0.49611 0.00001 3.9 × 10−34 105 0.49961 106 0.49988 0.000001 0 107 0.49996 36. For large values of x (not much more than 100) the computer can’t handle 5x or 3x , yet the limit is 5. 37. δ = 0.07747 39. (a) 38. $2,001.60, $2,009.66, $2,013.62, $2013.75 x3 − x − 1 = 0, x3 = x + 1, x = √ 3 x + 1. (b) y 2 -1 1 x -1 y (c) (d) (b) x1 40. 0, −1, −2, −9, −730 x x2 x3 y (a) 1, 1.26, 1.31, 1.322, 1.324, 1.3246, 1.3247 20 10 x -1 41. x= √ 5 1 x1 2 x2 3 x + 2; 1.267168 42. x = cos x; 0.739085 (after 33 iterations!). CHAPTER 3 The Derivative EXERCISE SET 3.1 1. (a) (4)2 /2 − (3)2 /2 7 f (4) − f (3) = = 4−3 1 2 f (x1 ) − f (3) x2 /2 − 9/2 = lim = lim 1 x1 →3 x1 →3 x1 − 3 x1 − 3 msec = (b) mtan x2 − 9 (x1 + 3)(x1 − 3) x1 + 3 1 = lim = lim =3 x1 →3 2(x1 − 3) x1 →3 x1 →3 2(x1 − 3) 2 = lim (c) mtan = lim x1 →x0 f (x1 ) − f (x0 ) x1 − x0 (d) y 10 x2 /2 − x2 /2 0 = lim 1 x1 →x0 x1 − x0 Tangent x2 − x2 1 0 x1 →x0 2(x1 − x0 ) x1 + x0 = x0 = lim x1 →x0 2 = lim 2. Secant 5 x 23 − 13 f (2) − f (1) = =7 2−1 1 f (x1 ) − f (1) x3 − 1 (x1 − 1)(x2 + x1 + 1) 1 = lim 1 = lim = lim x1 →1 x1 →1 x1 − 1 x1 →1 x1 − 1 x1 − 1 (a) msec = (b) mtan = lim (x2 + x1 + 1) = 3 1 x1 →1 (c) mtan = lim x1 →x0 f (x1 ) − f (x0 ) x1 − x0 y (d) 9 x3 − x3 0 = lim 1 x1 →x0 x1 − x0 Tangent = lim (x2 + x1 x0 + x2 ) 1 0 x1 →x0 = 3x2 0 3. (a) Secant 5 x 1/3 − 1/2 1 f (3) − f (2) = =− 3−2 1 6 f (x1 ) − f (2) 1/x1 − 1/2 = lim = lim x1 →2 x1 →2 x1 − 2 x1 − 2 msec = (b) mtan 2 − x1 −1 1 = lim =− x1 →2 2x1 (x1 − 2) x1 →2 2x1 4 = lim (c) mtan = lim x1 →x0 f (x1 ) − f (x0 ) x1 − x0 (d) 1/x1 − 1/x0 x1 →x0 x1 − x0 x0 − x1 = lim x1 →x0 x0 x1 (x1 − x0 ) y 4 = lim = lim x1 →x0 Secant 1 −1 =− 2 x0 x1 x0 1 65 x Tangent Exercise Set 3.1 4. (a) 66 f (2) − f (1) 1/4 − 1 3 = =− 2−1 1 4 f (x1 ) − f (1) 1/x2 − 1 1 = lim = lim x1 →1 x1 →1 x1 − 1 x1 − 1 msec = (b) mtan = lim 1 − x2 −(x1 + 1) 1 = −2 = lim − 1) x1 →1 x2 1 x1 →1 x2 (x1 1 (c) mtan = lim x1 →x0 = lim x1 →x0 f (x1 ) − f (x0 ) x1 − x0 (d) y 1/x2 − 1/x2 1 0 x1 − x0 x2 − x2 0 1 2 x2 (x − x ) x1 →x0 x 0 01 1 1 = lim 2 2 −(x1 + x0 ) =− 3 = lim 2 x2 x1 →x0 x0 1 x0 5. (a) mtan = lim x1 →x0 Tangent x Secant f (x1 ) − f (x0 ) (x2 + 1) − (x2 + 1) 1 0 = lim x1 →x0 x1 − x0 x1 − x0 x2 − x2 1 0 = lim (x1 + x0 ) = 2x0 x1 →x0 x1 − x0 x1 →x0 = 2(2) = 4 = lim (b) 6. mtan (a) mtan = lim x1 →x0 f (x1 ) − f (x0 ) (x2 + 3x1 + 2) − (x2 + 3x0 + 2) 1 0 = lim x1 →x0 x1 − x0 x1 − x0 (x2 − x2 ) + 3(x1 − x0 ) 1 0 = lim (x1 + x0 + 3) = 2x0 + 3 x1 →x0 x1 →x0 x1 − x0 = 2(2) + 3 = 7 = lim (b) 7. mtan (a) mtan = lim x1 →x0 f (x1 ) − f (x0 ) = lim x1 →x0 x1 − x0 = lim √ x1 →x0 1 1 mtan = √ = 2 21 (a) mtan = lim (b) 9. √ x1 − x0 x1 − x0 1 1 √ =√ x1 + x0 2 x0 (b) 8. √ mtan = − (a) mtan = (50 − 10)/(15 − 5) = 40/10 = 4 m/s √ √ 1/ x1 − 1/ x0 f (x1 ) − f (x0 ) = lim x1 →x0 x1 →x0 x1 − x0 x1 − x0 √ √ x0 − x1 1 −1 = lim √ √ = − 3/2 = lim √ √ √ √ x1 →x0 x0 x1 (x1 − x0 ) x1 →x0 x0 x1 ( x1 + x0 ) 2x0 1 1 =− 16 2(4)3/2 (b) Velocity (m/s) 4 10 20 t (s) 67 Chapter 3 (10 − 10)/(3 − 0) = 0 cm/s t = 0, t = 2, and t = 4.2 (horizontal tangent line) (c) maximum: (d) 11. (a) (b) 10. (3 − 18)/(4 − 2) = −7.5 cm/s (slope of estimated tangent line to curve at t = 3) t = 1 (slope > 0) minimum: t = 3 (slope < 0) From the figure: s t0 t1 (a) t t2 The particle is moving faster at time t0 because the slope of the tangent to the curve at t0 is greater than that at t2 . (b) The initial velocity is 0 because the slope of a horizontal line is 0. (c) The particle is speeding up because the slope increases as t increases from t0 to t1 . (d) The particle is slo...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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