Calculus Single Variable Canadian 8th Edition by Adams Christopher Solution Manual

Page 1

SECTION 2.1 (PAGE 100)

CHAPTER 2.

ADAMS and ESSEX: CALCULUS 8

DIFFERENTIATION

7. Slope of y =

x + 1 at x = 3 is √

√ 4+h −2 4+h +2 m = lim ·√ h→0 h 4+h +2 4+h−4 = lim √ h→0 h h+h+2 1 1 = lim √ = . h→0 4 4+h +2

Section 2.1 Tangent Lines and Their Slopes (page 100) 1. Slope of y = 3x − 1 at (1, 2) is m = lim

h→0

3(1 + h) − 1 − (3 × 1 − 1) 3h = lim = 3. h→0 h h

The tangent line is y − 2 = 3(x − 1), or y = 3x − 1. (The tangent to a straight line at any point on it is the same straight line.)

Tangent line is y − 2 = 1 x

8. The slope of y = √ at x = 9 is

2. Since y = x/2 is a straight line, its tangent at any point

1 1 1 √ − h→0 h 3 9+h √ √ 3− 9+h 3+ 9+h = lim √ · √ h→0 3h 9 + h 3+ 9+h 9−9−h = lim √ √ h→0 3h 9 + h(3 + 9 + h) 1 1 =− =− . 3(3)(6) 54

m = lim

(a, a/2) on it is the same line y = x/2.

3. Slope of y = 2x 2 − 5 at (2, 3) is 2(2 + h)2 − 5 − (2(22 ) − 5) h→0 h 8 + 8h + 2h 2 − 8 = lim h→0 h = lim (8 + 2h) = 8

m = lim

The tangent line at (9, 13 ) is y = 1 y = 12 − 54 x.

h→0

Tangent line is y − 3 = 8(x − 2) or y = 8x − 13.

9. Slope of y =

4. The slope of y = 6 − x − x 2 at x = −2 is 6 − (−2 + h) − (−2 + h)2 − 4 h→0 h 3h − h 2 = lim = lim (3 − h) = 3. h→0 h→0 h

The tangent line at (−2, 4) is y = 3x + 10.

5. Slope of y = x 3 + 8 at x = −2 is m = lim

= lim 12 − 6h + h

2

h→0

m = lim

h→0

x2

10.

= 12

1 1 h h2 + 1

The tangent line at (0, 1) is y = 1.

h→0

−h = 0. h2 + 1

1 54 (x

− 9), or

2x at x = 2 is x +2

p 5 − (1 + h)2 − 2 h→0 h 5 − (1 + h)2 − 4 = lim p h→0 h 5 − (1 + h)2 + 2

m = lim

1 at (0, 1) is +1 − 1 = lim

1 Tangent line is y − 1 = (x − 2), 4 or x − 4y = −2. √ The slope of y = 5 − x 2 at x = 1 is

Tangent line is y − 0 = 12(x + 2) or y = 12x + 24.

6. The slope of y =

1 3

2(2 + h) −1 2 m = lim + h + 2 h→0 h 4 + 2h − 2 − h − 2 = lim h→0 h(2 + h + 2) h 1 = lim = . h→0 h(4 + h) 4

m = lim

(−2 + h)3 + 8 − (−8 + 8) h→0 h −8 + 12h − 6h 2 + h 3 + 8 − 0 = lim h→0 h

1 (x − 3), or x − 4y = −5. 4

−2 − h 1 = lim p =− h→0 5 − (1 + h)2 + 2 2

The tangent line at (1, 2) is y = 2 − 12 (x − 1), or y = 52 − 12 x.

40 Copyright © 2014 Pearson Canada Inc.


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