By Howard Anton
Read or Download Student Solutions Manual Set for Calculus Early Transcendentals Single Variable 8th Edition PDF
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Additional resources for Student Solutions Manual Set for Calculus Early Transcendentals Single Variable 8th Edition
Example text
30. (a) x = 2 + t, y = −1 + 2t (a) (5/2, 0) (b) (9/4, −1/2) (b) Set t = 0 to get (x0 , y0 ); t = 1 for (x1 , y1 ). (d) x = 2 − t, y = 4 − 6t (b) (c) x = at, y = b(1 − t) t = 3/4 (c) (11/4, 1/2) The two branches corresponding to −1 ≤ t ≤ 0 and 0 ≤ t ≤ 1 coincide. y − y0 y1 − y0 t − t0 to obtain = . t1 − t0 x − x0 x1 − x0 (c) x = 3 − 2(t − 1), y = −1 + 5(t − 1) (b) from (x0 , y0 ) to (x1 , y1 ) (a) Eliminate 5 0 5 -2 31. 7 32. 36 (a) If a = 0 the line segment is vertical; if c = 0 it is horizontal.
01 if |x − 3| < 9. 10. 05 x+1 x+1 1 1 ,δ= 9000 9000 11. 001 if |x − 4| < 12. 004 if δ < 1 then | x − 3| √ = √ <√ 4 x+3 | x + 3| 8+3 13. if δ ≤ 1 then 14. 05 16. |(4x − 5) − 7| = |4x − 12| = 4|x − 3| < 17. 05 if |x − 5| < 1, δ = 1 x 5 5|x| 20 15. |3x − 15| = 3|x − 5| < if |x − 3| < if |x − 2| < 1 2 ,δ= 1 2 1 4 ,δ= 1 4 if |x − 5| < 1 3 ,δ= 1 3 51 18. 19. Chapter 2 |2 − 3x − 5| = 3|x + 1| < x2 + x − 1 = |x| < x if |x + 1| < 1 3 ,δ= if |x| < , δ = 1 3 20. x2 − 9 − (−6) = |x + 3| < x+3 21. if δ < 1 then |2x2 − 2| = 2|x − 1||x + 1| < 6|x − 1| < 22.
Y 5 -5 -1 5 x 8. Suppose the radius of the uncoated ball is r and that of the coated ball is r + h. e. V = π(r + h)3 − πr3 = πh[3r2 + 3rh + h2 ] in3 . 3 3 3 9. (a) The base has sides (10 − 2x)/2 and 6 − 2x, and the height is x, so V = (6 − 2x)(5 − x)x ft3 . (b) From the picture we see that x < 5 and 2x < 6, so 0 < x < 3. 21 ft 10. {x = 0} and ∅ (the empty set) 11. impossible; we would have to solve 2(3x − 2) − 5 = 3(2x − 5) − 2, or −9 = −17 12. (a) 13. 1/(2 − x2 ) 15. (3 − x)/x (b) no; f (g(x)) can be defined at x = 1, whereas g, and therefore f ◦ g, requires x=1 14.