# Sol stewart 6th ed

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1

FUNCTIONS AND MODELS

1.1 Four Ways to Represent a Function
In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here. 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 estimatesfor 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]. The range 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. FromFigure 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 ground acceleration is −85 ≤ a ≤ 115. Written in interval notation, we get [−85, 115]. the Vertical Line Test.

5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails7. 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. The person’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 agradual 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 to freezing as the ice 13. Of course, this graph depends strongly on the

melts. Then, when the ice has melted, the water will slowly warm up to room temperature.

geographical location!

15. As the price increases, the amount solddecreases.

17.

9

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CHAPTER 1 FUNCTIONS AND MODELS

19. (a)

(b) From the graph, we estimate the number of cell-phone subscribers worldwide to be about 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

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

= 3a2 − a + 23a2 − a + 2

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

√ √ √ 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

x ∈ R | x 6=

1 3

= −∞,

1 3

1 3,∞

.

⇔ x = 1 , so the domain 3

division by zero. The expressionx(x − 5) is positive if x < 0 or x > 5. (See Appendix A for methods for solving inequalities.) Thus, the domain is (−∞, 0) ∪ (5, ∞).

√ 4 x2 − 5x is deﬁned when x2 − 5x > 0

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

SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION

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33. f (x) = 5 is deﬁned for all real numbers, so the domain is R, or (−∞, ∞).

The...

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