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CHAPTER

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

94

Section 2.1

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper SaddleRiver, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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

Instructor’s Resource Manual

Section 2.1

95

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, inany form or by any means, without permission in writing from the publisher.

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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