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Download Calculus: Early Transcendentals (6th) -- Student Solutions by James Stewart PDF

By James Stewart

Offers thoroughly worked-out suggestions to all odd-numbered workouts in the textual content, giving scholars how to money their solutions and confirm that they took the proper steps to reach at a solution.

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X→0+ x→0+ (ii) Since sgn x = −1 for x < 0, lim sgn x = lim −1 = −1. x→0− x→0− (iii) Since lim sgn x 6= lim sgn x, lim sgn x does not exist. x→0− x→0 x→0+ (iv) Since |sgn x| = 1 for x 6= 0, lim |sgn x| = lim 1 = 1. x→0 47. (a) (i) lim F (x) = lim x→1+ x→1+ (ii) lim F (x) = lim x→1− x→1− x2 − 1 x2 − 1 = lim = lim (x + 1) = 2 |x − 1| x→1+ x − 1 x→1+ x→0 (c) x2 − 1 x2 − 1 = lim = lim − (x + 1) = −2 |x − 1| x→1− − (x − 1) x→1− (b) No, lim F (x) does not exist since lim F (x) 6= lim F (x). x→1 x→1− x→1+ 49.

X→0 ⎧ cos x ⎪ ⎨ 19. f (x) = 0 ⎪ ⎩ 1 − x2 if x < 0 if x = 0 if x > 0 lim f (x) = 1, but f (0) = 0 6= 1, so f is discontinuous at 0. x→0 21. F (x) = x is a rational function. So by Theorem 5 (or Theorem 7), F is continuous at every number in its domain, x2 + 5x + 6 x | x2 + 5x + 6 6= 0 = {x | (x + 3)(x + 2) 6= 0} = {x | x 6= −3, − 2} or (−∞, −3) ∪ (−3, −2) ∪ (−2, ∞). 23. By Theorem 5, the polynomials x2 and 2x − 1 are continuous on (−∞, ∞). By Theorem 7, the root function √ continuous on [0, ∞). By Theorem 9, the composite function 2x − 1 is continuous on its domain, [ 12 , ∞).

Ln x2 − 2x − 2 ≤ 0  ln 3 ln 2  ln 4 ln 3  ln 5 ln 4  ···  ln 32 ln 31  = ln 25 5 ln 2 ln 32 = = =5 ln 2 ln 2 ln 2 ⇒ x2 − 2x − 2 ≤ e0 = 1 ⇒ x2 − 2x − 3 ≤ 0 ⇒ (x − 3)(x + 1) ≤ 0 ⇒ x ∈ [−1, 3]. Since the argument must be positive, x2 − 2x − 2 > 0 ⇒ √  √     x− 1− 3 x− 1+ 3 > 0 ⇒ √ √   √  √      x ∈ −∞, 1 − 3 ∪ 1 + 3, ∞ . The intersection of these intervals is −1, 1 − 3 ∪ 1 + 3, 3 . 15. Let d be the distance traveled on each half of the trip. Let t1 and t2 be the times taken for the first and second halves of the trip.

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