Solucionario Capitulo 2 James Steward 6 Edicion

2

LIMITS AND DERIVATIVES
(b) Using the values of t that correspond to the points closest to P (t = 10 and t = 20), we have −38.8 + (−27.8) = −33.3 2

2.1 The Tangent and Velocity Problems
1. (a) Using P (15, 250), we construct the following table:

t 5 10 20 25 30

Q (5, 694) (10, 444) (20, 111) (25, 28) (30, 0)

slope = mP Q
694−250 5−15 444−250 10−15 111−250 20−15 28−250 25−150−250 30−15

= − 444 = −44.4 10 = − 194 = −38.8 5 = − 139 = −27.8 5 = − 222 = −22.2 10

= − 250 = −16.6 15

(c) From the graph, we can estimate the slope of the tangent line at P to be
−300 9

= −33.3.

3. (a)

x (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) 0.5 0.9 0.99 0.999 1.5 1.1 1.01 1.001

Q (0.5, 0.333333) (0.9, 0.473684) (0.99, 0.497487) (0.999, 0.499750) (1.5, 0.6) (1.1, 0.523810)(1.01, 0.502488) (1.001, 0.500250)

mP Q 0.333333 0.263158 0.251256 0.250125 0.2 0.238095 0.248756 0.249875

(b) The slope appears to be 1 . 4 (c) y −
1 2

= 1 (x − 1) or y = 1 x + 1 . 4 4 4

5. (a) y = y(t) = 40t − 16t2 . At t = 2, y = 40(2) − 16(2)2 = 16. The average velocity between times 2 and 2 + h is

vave =

40(2 + h) − 16(2 + h)2 − 16 −24h − 16h2 y(2 + h) − y(2) = = = −24 −16h, if h 6= 0. (2 + h) − 2 h h (ii) [2, 2.1]: h = 0.1, vave = −25.6 ft/s (i) [2, 2.5]: h = 0.5, vave = −32 ft/s (iv) [2, 2.01]: h = 0.01, vave = −24.16 ft/s

(iii) [2, 2.05]: h = 0.05, vave = −24.8 ft/s

(b) The instantaneous velocity when t = 2 (h approaches 0) is −24 ft/s.
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CHAPTER 2

LIMITS AND DERIVATIVES

7. (a) (i) On the interval [1, 3], vave =

10.7 − 1.4 9.3s(3) − s(1) = = = 4.65 m/s. 3−1 2 2 s(3) − s(2) 10.7 − 5.1 = = 5.6 m/s. 3−2 1 25.8 − 10.7 15.1 s(5) − s(3) = = = 7.55 m/s. 5−3 2 2 17.7 − 10.7 s(4) − s(3) = = 7 m/s. 4−3 1 Using the points (2, 4) and (5, 23) from the approximate tangent line, the instantaneous velocity at t = 3 is about 23 − 4 ≈ 6.3 m/s. 5−2

(ii) On the interval [2, 3], vave = (iii) On the interval [3, 5], vave = (iv) On theinterval [3, 4], vave = (b)

9. (a) For the curve y = sin(10π/x) and the point P (1, 0):

x 2 1.5 1.4 1.3 1.2 1.1 (2, 0)

Q (1.5, 0.8660) (1.4, −0.4339) (1.3, −0.8230) (1.2, 0.8660) (1.1, −0.2817)

mP Q 0 1.7321 −1.0847 −2.7433 4.3301 −2.8173

x 0.5 0.6 0.7 0.8 0.9

Q (0.5, 0) (0.6, 0.8660) (0.7, 0.7818) (0.8, 1) (0.9, −0.3420)

mP Q 0 −2.1651 −2.6061 −5 3.4202

As x approaches 1, theslopes do not appear to be approaching any particular value. (b) We see that problems with estimation are caused by the frequent oscillations of the graph. The tangent is so steep at P that we need to take x-values much closer to 1 in order to get accurate estimates of its slope.

(c) If we choose x = 1.001, then the point Q is (1.001, −0.0314) and mP Q ≈ −31.3794. If x = 0.999, then Q is (0.999,0.0314) and mP Q = −31.4422. The average of these slopes is −31.4108. So we estimate that the slope of the tangent line at P is about −31.4.

SECTION 2.2 THE LIMIT OF A FUNCTION

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2.2 The Limit of a Function
1. As x approaches 2, f (x) approaches 5. [Or, the values of f (x) can be made as close to 5 as we like by taking x sufficiently

close to 2 (but x 6= 2).] Yes, the graphcould have a hole at (2, 5) and be defined such that f (2) = 3.
3. (a) lim f (x) = ∞ means that the values of f (x) can be made arbitrarily large (as large as we please) by taking x
x→−3

sufficiently close to −3 (but not equal to −3). (b) lim f(x) = −∞ means that the values of f (x) can be made arbitrarily large negative by taking x sufficiently close to 4
x→4+

through values larger than 4.
5.(a) f (x) approaches 2 as x approaches 1 from the left, so lim f (x) = 2.
x→1−

(b) f (x) approaches 3 as x approaches 1 from the right, so lim f (x) = 3.
x→1+

(c) lim f (x) does not exist because the limits in part (a) and part (b) are not equal.
x→1

(d) f (x) approaches 4 as x approaches 5 from the left and from the right, so lim f (x) = 4.
x→5

(e) f(5) is not defined, so it...