Ensayos

2

CHAPTER

2.1 Concepts Review

The Derivative
4.

1. tangent line
2. secant line
3.

f (c + h ) − f ( c )
h

4. average velocity

Problem Set 2.1
1. Slope =

2. Slope =

5–3
2– 3
2

=4

6–4
= –2
4–6

Slope ≈ 1.5
5.

3.

Slope ≈

Slope ≈ −2

5
2

6.

Slope ≈ –

94

Section 2.1

3
2

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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

[(2.01)3 − 1.0] − 7
2.01 − 2
0.120601
=
0.01
= 12.0601

d.

msec =

e.

mtan = lim

a., b.

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

= lim

c.

m tan = 2

d.

msec =

e.

(1.01)2 + 1.0 − 2
1.01 − 1
0.0201
=
.01
= 2.01

f (1 + h) – f (1)
h
h →0

mtan = lim

[(1 + h)2 + 1] – (12 + 1)
h
h →0

= lim

2 + 2h + h 2 − 2
h
h →0
h(2 + h)
= lim
h
h →0
= lim (2 + h) = 2
= lim

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
[(c + h)2 – 1] – (c 2 – 1)
h
h→0

= lim

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

h →0

3

8. y = x – 1
a., b.

10. f ( x ) = x 3 – 3x
f (c + h ) – f (c )
mtan = lim
h
h→0
[(c + h)3 – 3(c +h)] – (c3 – 3c)
h
h→0

= lim

c3 + 3c 2 h + 3ch 2 + h3 – 3c – 3h – c3 + 3c
h
h→0

= lim

h(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

c.

m tan = 12

Instructor’s Resource Manual

Section 2.1

95

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

11.

13. a.

16(12 ) –16(02 ) = 16 ft

b.

16(22 ) –16(12 ) = 48 ft

c.

Vave =

d.

f ( x) =
mtan

1
x +1

f (1 + h) – f (1)
= lim
h
h→0

−
= lim 2+ h 2
h
h →0
−2(2h h)
+
= lim
h
h →0
1
= lim −
h→0 2(2 + h)
1

e.

b.

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

12. f ( x ) =
mtan

= lim

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

1

14. a.

h
h −1

h →0

h
1
= lim
h →0 h − 1= −1
y + 1 = –1(x – 0); y = –x – 1

Vave =

d.

(32 + 1) – (22 + 1)
= 5 m/sec
3– 2

[(2.003)2 + 1] − (22 + 1)
2.003 − 2
0.012009
=
0.003
= 4.003 m/sec
Vave =

Vave =

c.

144 – 64
= 80 ft/sec
3–2

[(2 + h) 2 + 1] – (22 + 1)
2+h–2

4h + h 2
h
= 4 +h
=

f (t ) = t2 + 1
f (2 + h) – f (2)
v = lim
h
h →0
[(2 + h)2 + 1] – (22 + 1)
h
h →0

= lim

4h + h 2
hh →0
= lim (4 + h)
= lim

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
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15. a.

f (α +h) – f (α )
h

v = lim

h →0

2(α + h) + 1 – 2α + 1
h
h →0

= lim

2α + 2h + 1 – 2α + 1
h

= lim

h →0

= lim

( 2α + 2h + 1 – 2α + 1)( 2α + 2h + 1 + 2α + 1)
h( 2α + 2h + 1 + 2α + 1)

h →0

2h

= lim

2α + 2h + 1 + 2α + 1)

h →0 h(

2

=

2α + 1 + 2α + 1
1

b.

2α + 1

=

=

1
2α + 1

ft/s

1
2

2α + 1 = 2
3
2
The object reaches a...