Veronica
2
The Derivative
4.
2.1 Concepts Review
1. tangent line 2. secant line 3.
f (c + h ) − f ( c ) h
4. average velocity
Problem Set 2.1
1. Slope =
5–3 2– 3 2 =4
Slope ≈ 1.5 5.
2. Slope = 3.
6–4 = –2 4–6
Slope ≈
5 2
Slope ≈ −2
6.
Slope ≈ –
3 2
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7. y = x 2 + 1 a., b.
d.
msec =
[(2.01)3 − 1.0] − 7 2.01 − 2 0.120601 = 0.01 = 12.0601
e.
mtan = lim
= lim
f (2 + h) – f (2) h h →0
[(2 + h)3– 1] – (23 − 1) h h →0
= lim
12h + 6h 2 + h3 h h→0
c. d.
m tan = 2 msec =
(1.01)2 + 1.0 − 2 1.01 − 1 0.0201 = .01 = 2.01
h(12 + 6h + h 2 ) h h→0 = 12 = lim
9. f (x) = x 2 – 1 f (c + h ) – f (c ) mtan = lim h h→0
= lim [(c + h)2 – 1] – (c 2 – 1) h h→0
e.
mtan = lim
= lim
f (1 + h) – f (1) h h →0
[(1 + h)2 + 1] – (12 + 1) h h →0
2 + 2h + h 2 − 2 h h →0 h(2 + h) =lim h h →0 = lim (2 + h) = 2 = lim
h →0
c 2 + 2ch + h 2 – 1 – c 2 + 1 h h→0 h(2c + h) = lim = 2c h h→0 At x = –2, m tan = –4 x = –1, m tan = –2 x = 1, m tan = 2 x = 2, m tan = 4 = lim
8. y = x – 1 a., b.
3
10. f (x) = x 3 – 3x f (c + h ) – f (c ) mtan = lim h h→0
= lim = lim [(c + h)3 – 3(c + h)] – (c3 – 3c) h h→0 c3 + 3c 2 h + 3ch 2 + h3 – 3c – 3h – c3 + 3c h h→0
c.
m tan = 12h(3c 2 + 3ch + h 2 − 3) = 3c 2 – 3 h h→0 At x = –2, m tan = 9 x = –1, m tan = 0 x = 0, m tan = –3 x = 1, m tan = 0 x = 2, m tan = 9 = lim
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Section 2.1
95
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11.
13. a. b. c.
16(12 ) –16(02 ) = 16 ft 16(22 ) –16(12 ) = 48 ft
Vave =
144 – 64 = 80 ft/sec 3–2
d.
f ( x) = mtan
1 x +1
16(3.01) 2 − 16(3)2 3.01 − 3 0.9616 = 0.01 = 96.16 ft/s Vave = f (t ) = 16t 2 ; v = 32c v = 32(3) = 96 ft/s Vave = (32 + 1) – (22 + 1) = 5 m/sec 3– 2
f (1 + h) – f (1) =lim h h→0
1 1
e.
− = lim 2+ h 2 h h →0 − 2(2h h) + = lim h h →0 1 = lim − h→0 2(2 + h) =– 1 4 1 1 y – = – ( x –1) 2 4 1 x –1 f (0 + h) − f (0) = lim h h →0 1 +1 = lim h −1 h h →0
= lim
h h −1 h →0
14. a.
b.
[(2.003)2 + 1] − (22 + 1) 2.003 − 2 0.012009 = 0.003 = 4.003 m/sec Vave =
Vave = [(2 + h) 2 + 1] – (22 + 1) 2+h–2
12. f (x) =
mtan
c.
4h + h 2 h = 4 +h = f (t ) =t2 + 1 f (2 + h) – f (2) v = lim h h →0 = lim = lim [(2 + h)2 + 1] – (22 + 1) h h →0
d.
h 1 = lim h →0 h − 1 = −1 y + 1 = –1(x – 0); y = –x – 1
4h + h 2 h h →0 = lim (4 + h)
h →0
=4
96
Section 2.1
Instructor’s Resource Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15. a.
v = lim = lim = lim
h →0
f (α + h) – f (α ) h
2(α + h) + 1 – 2α + 1 h h →0
h →0
2α + 2h + 1 – 2α + 1 h ( 2α + 2h + 1 – 2α + 1)( 2α + 2h + 1 + 2α + 1) h( 2α + 2h + 1 + 2α + 1)
= lim
h →0
= lim
=
2h 2α + 2h + 1 + 2α + 1)
2 = 1 2α +1
h →0 h(
2α + 1 + 2α + 1 1 2α + 1 = 1 2
ft/s
b.
2α + 1 = 2 3 2 The object reaches a velocity of 1 ft/s when t = 3 . 2 2
2 α + 1= 4; α =
16. f (t ) = – t2 + 4 t
v = lim
= lim [–(c + h)2 + 4(c + h)] – (– c 2 + 4c) h h →0
18. a. b. c.
1000(3)2 – 1000(2)2 = 5000 1000(2.5)2 – 1000(2)2 2250 = = 4500 2.5 – 2 0.5 f (t ) = 1000t 2 r = lim = lim 1000(2 + h)2 − 1000(2) 2 h...
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