# Livro

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TX.10

Solução detalhada de todos os exercícios ímpares do livro: Cálculo, James Stewart - VOLUME I e VOLUME II

"O que sabemos é uma gota, o que ignoramos é um oceano." Isaac Newton
23/11/2010.TX

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1

FUNCTIONS AND MODELS

1.1 Four Ways to Represent a Function
1. (a) The point (−1, −2) is on the graph of f , so f (−1) = −2.

(b) When x = 2, y is about 2.8, so f (2) ≈ 2.8.
(c) f (x) = 2 is equivalent to y = 2. When y = 2, we have x = −3 and x = 1.
(d) Reasonable estimates for x when y = 0 are x = −2.5 and x = 0.3.
(e) The domain of f consists of all x-values on the graph of f . For this function, the domain is −3 ≤ x ≤ 3, or [−3, 3].
Therange of f consists of all y -values on the graph of f . For this function, the range is −2 ≤ y ≤ 3, or [−2, 3].

(f ) As x increases from −1 to 3, y increases from −2 to 3. Thus, f is increasing on the interval [−1, 3].
3. From Figure 1 in the text, the lowest point occurs at about (t, a) = (12, −85). The highest point occurs at about (17, 115).

Thus, the range of the vertical groundacceleration is −85 ≤ a ≤ 115. Written in interval notation, we get [−85, 115].

5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails

the Vertical Line Test.
7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−3, 2] and the range

is [−3, −2) ∪ [−1, 3].

9. Theperson’s weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The person’s weight

dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual
increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems.
11. The water will cool down almost tofreezing as the ice

13. Of course, this graph depends strongly on the

melts. Then, when the ice has melted, the water will

geographical location!

slowly warm up to room temperature.

15. As the price increases, the amount sold decreases.

17.

9

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¤

CHAPTER 1 FUNCTIONS AND MODELS

19. (a)

(b) From the graph, we estimate the number of
92 million in 1995 and 485 million in 1999.

21. f (x) = 3x2 − x + 2.

f (2) = 3(2)2 − 2 + 2 = 12 − 2 + 2 = 12.
f (−2) = 3(−2)2 − (−2) + 2 = 12 + 2 + 2 = 16.
f (a) = 3a2 − a + 2.
f (−a) = 3(−a)2 − (−a) + 2 = 3a2 + a + 2.
f (a + 1) = 3(a + 1)2 − (a + 1) + 2 = 3(a2 + 2a + 1) − a − 1 + 2 = 3a2 + 6a + 3 − a + 1 = 3a2 + 5a + 4.
2f (a) = 2 · f (a) = 2(3a2 − a + 2) = 6a2 −2a + 4.
f (2a) = 3(2a)2 − (2a) + 2 = 3(4a2 ) − 2a + 2 = 12a2 − 2a + 2.
f (a2 ) = 3(a2 )2 − (a2 ) + 2 = 3(a4 ) − a2 + 2 = 3a4 − a2 + 2.
[f (a)]2 = 3a2 − a + 2

2

= 3a2 − a + 2 3a2 − a + 2

= 9a4 − 3a3 + 6a2 − 3a3 + a2 − 2a + 6a2 − 2a + 4 = 9a4 − 6a3 + 13a2 − 4a + 4.

f (a + h) = 3(a + h)2 − (a + h) + 2 = 3(a2 + 2ah + h2 ) − a − h + 2 = 3a2 + 6ah + 3h2 − a − h + 2.
23. f (x) = 4 + 3x −x2 , so f (3 + h) = 4 + 3(3 + h) − (3 + h)2 = 4 + 9 + 3h − (9 + 6h + h2 ) = 4 − 3h − h2 ,

and

(4 − 3h − h2 ) − 4
h(−3 − h)
f (3 + h) − f (3)
=
=
= −3 − h.
h
h
h

1
a−x
1

f (x) − f (a)
x
a = xa = a − x = −1(x − a) = − 1
=
25.
x−a
x−a
x−a
xa(x − a)
xa(x − a)
ax
27. f (x) = x/(3x − 1) is deﬁned for all x except when 0 = 3x − 1

is

x ∈ R | x 6=

1
3

= −∞,1
3

1
3,∞

.

⇔ x = 1 , so the domain
3

29. f (t) = t + 3 t is deﬁned when t ≥ 0. These values of t give real number results for t, whereas any value of t gives a real

number result for 3 t. The domain is [0, ∞).
31. h(x) = 1

4
x2 − 5x is deﬁned when x2 − 5x > 0

x(x − 5) > 0. Note that x2 − 5x 6= 0 since that would result in

division by zero....

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