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1
CHAPTER 1
Functions
EXERCISE SET 1.
- (a) around 1943 (b) 1960; (c) no;you need the year’s population (d) war;marketing techniques (e) news of health risk;social pressure, antismoking campaigns, increased taxation
- (a) 1989;$35,600 (b) 1975, 1983;$32, (c) the first two years;the curve is steeper (downhill)
- (a) − 2. 9 , − 2. 0 , 2. 35 , 2. 9 (b) none (c) y = 0 (d) − 1. 75 ≤ x ≤ 2. 15 (e) ymax = 2.8 at x = − 2 .6; ymin = − 2 .2 at x = 1. 2
- (a) x = − 1 , 4 (b) none (c) y = − 1 (d) x = 0, 3 , 5 (e) ymax = 9 at x = 6; ymin = −2 at x = 0
- (a) x = 2, 4 (b) none (c) x ≤ 2;4 ≤ x (d) ymin = −1;no maximum value
- (a) x = 9 (b) none (c) x ≥ 25 (d) ymin = 1;no maximum value
- (a) Breaks could be caused by war, pestilence, flood, earthquakes, for example. (b) C decreases for eight hours, takes a jump upwards, and then repeats.
- (a) Yes, if the thermometer is not near a window or door or other source of sudden temperature change. (b) No;the number is always an integer, so the changes are in movements (jumps) of at least one unit.
- (a) The side adjacent to the building has length x, so L = x + 2y. Since A = xy = 1000, L = x + 2000/x. (b) x > 0 and x must be smaller than the width of the building, which was not given. (c) 120
80
20 80
(d) Lmin ≈ 89 .44 ft
- (a) V = lwh = (6 − 2 x)(6 − 2 x)x (b) From the figure it is clear that 0 < x < 3.
(c) 20
0
0 3
(d) Vmax ≈ 16 in^3
2 Chapter 1
- (a) V = 500 = πr^2 h so h =
πr^2
. Then
C = (0.02)(2)πr^2 + (0.01)2πrh = 0. 04 πr^2 + 0. 02 πr
πr^2 = 0. 04 πr^2 +
r
; Cmin ≈ 4 .39 at r ≈ 3. 4 , h ≈ 13. 8.
7
4
1.5 6
(b) C = (0.02)(2)(2r)^2 + (0.01)2πrh = 0. 16 r^2 +
r
. Since
- 04 π < 0 .16, the top and bottom now get more weight. Since they cost more, we diminish their sizes in the solution, and the cans become taller.
7
4
1.5 5.
(c) r ≈ 3 .1 cm, h ≈ 16 .0 cm, C ≈ 4 .76 cents
- (a) The length of a track with straightaways of length L and semicircles of radius r is P = (2)L + (2)(πr) ft. Let L = 360 and r = 80 to get P = 720 + 160π = 1222.65 ft. Since this is less than 1320 ft (a quarter-mile), a solution is possible. (b) P = 2L + 2πr = 1320 and 2r = 2x + 160, so L = 12 (1320 − 2 πr) = 12 (1320 − 2 π(80 + x)) = 660 − 80 π − πx.
450
0
0 100
(c) The shortest straightaway is L = 360, so x = 15.49 ft. (d) The longest straightaway occurs when x = 0, so L = 660 − 80 π = 408.67 ft.
EXERCISE SET 1.
- (a) f (0) = 3(0)^2 −2 = −2; f (2) = 3(2)^2 −2 = 10; f (−2) = 3(−2)^2 −2 = 10; f (3) = 3(3)^2 −2 = 25; f (
2)^2 − 2 = 4; f (3t) = 3(3t)^2 − 2 = 27t^2 − 2 (b) f (0) = 2(0) = 0; f (2) = 2(2) = 4; f (−2) = 2(−2) = −4; f (3) = 2(3) = 6; f (
f (3t) = 1/ 3 t for t > 1 and f (3t) = 6t for t ≤ 1.
- (a) g(3) =
= 2; g(−1) =
= 0; g(π) =
π + 1 π − 1
; g(− 1 .1) =
g(t^2 − 1) =
t^2 − 1 + 1 t^2 − 1 − 1
t^2 t^2 − 2 (b) g(3) =
3 + 1 = 2; g(−1) = 3; g(π) =
π + 1; g(− 1 .1) = 3; g(t^2 − 1) = 3 if t^2 < 2 and g(t^2 − 1) =
t^2 − 1 + 1 = |t| if t^2 ≥ 2.
- (a) x = 3 (b) x ≤ −
3 or x ≥
(c) x^2 − 2 x + 5 = 0 has no real solutions so x^2 − 2 x + 5 is always positive or always negative. If x = 0, then x^2 − 2 x + 5 = 5 > 0;domain: ( −∞, +∞). (d) x = 0 (e) sin x = 1, so x = (2n + 12 )π, n = 0, ± 1 , ± 2 ,...
4 Chapter 1
- (a) V = (8 − 2 x)(15 − 2 x)x (b) −∞ < x < +∞, −∞ < V < +∞ (c) 0 < x < 4
(d) minimum value at x = 0 or at x = 4;maximum value somewhere in between (can be approximated by zooming with graphing calculator)
- (a) x = 3000 tan θ (b) θ = nπ + π/2 for n an integer, −∞ < n < ∞
(c) 0 ≤ θ < π/2, 0 ≤ x < +∞ (d) 3000 ft 6000
0
0 6
- (i) x = 1, −2 causes division by zero (ii) g(x) = x + 1, all x
- (i) x = 0 causes division by zero (ii) g(x) =
x + 1 for x ≥ 0
- (a) 25 ◦F (b) 2 ◦F (c) − 15 ◦F
- If v = 48 then −60 = WCI = 1. 6 T − 55;thus T = (−60 + 55)/ 1. 6 ≈ − 3 ◦F.
- If v = 8 then −10 = WCI = 91.4 + (91. 4 − T )(0.0203(8) − 0. 304
8 − 0 .474);thus T = 91.4 + (10 + 91.4)/(0.0203(8) − 0. 304
8 − 0 .474) and T = 5◦F
- The WCI is given by three formulae, but the first and third don’t work with the data. Hence −15 = WCI = 91.4 + (91. 4 − 20)(0. 0203 v − 0. 304
v − 0 .474);set x =
v so that v = x^2 and obtain 0. 0203 x^2 − 0. 304 x − 0 .474 + (15 + 91.4)/(91. 4 − 20) = 0. Use the quadratic formula to find the two roots. Square them to get v and discard the spurious solution, leaving v ≈ 25.
- Let t denote time in minutes after 9:23 AM. Then D(t) = 1000 − 20 t ft.
EXERCISE SET 1.
- (e) seems best, though only (a) is bad.
-0.
y
-1 1
x
- (e) seems best, though only (a) is bad and (b) is not good.
-0.
y
-1 1
x
- (b) and (c) are good; (a) is very bad.
12
13
14
y
-1 1
x
Exercise Set 1.3 5
- (b) and (c) are good; (a) is very bad. - - - -
y
-2 -1 0 1 2
x
5. [− 3 , 3] × [0, 5]
2
y
x
6. [− 4 , 2] × [0, 3]
y
x
- (a) window too narrow, too short (b) window wide enough, but too short (c) good window, good spacing - -
y
-5 5 10 20
x
(d) window too narrow, too short
(e) window too narrow, too short
- (a) window too narrow (b) window too short (c) good window, good tick spacing - -
50
y
-16 -8 -4 4
x
(d) window too narrow, too short
(e) shows one local minimum only, window too narrow, too short
[− 5 , 14] × [− 60 , 40]
20
40
y
-5 5 10
x
10. [6, 12] × [− 100 , 100]
100
y
10 12
x
11. [− 0. 1 , 0 .1] × [− 3 , 3]
2
3
y
-0.
x
12. [− 1000 , 1000] × [− 13 , 13]
5
10
y
x 1000
13. [− 250 , 1050] × [− 1500000 , 600000]
y -1000 1000
x
Exercise Set 1.3 7
- (a)
-1 1
x
y
1
(b)
1
x
1
2
y
(c)
-1 (^1)
x
y (^) (d)
c 2 c
1
y
x
(e)
c
1
y
x C
(f )
1
x
1
y
1
x
y
-1 1
- The portions of the graph of y = f (x) which lie below the x-axis are reflected over the x-axis to give the graph of y = |f (x)|.
- Erase the portion of the graph of y = f (x) which lies in the left-half plane and replace it with the reflection over the y-axis of the portion in the right-half plane (symmetry over the y-axis) and you obtain the graph of y = f (|x|).
- (a) for example, let a = 1. 1
a
y
x
(b)
1
2
3
y
1 2 3
x
- They are identical.
x
y
5
10
15
y
-1 1 2 3
x
8 Chapter 1
- This graph is very complex. We show three views, small (near the origin), medium and large:
(a) y
x
-1 2
(b)
y (^) x 10
(c)
1000
y
40
x
- (a)
1
y
-3 -1 1 3
x
(b)
-0.
1
-3 -1 1 3
x
y
(c)
y
-1 1 2 3
x
(d)
1
-2 -1 1
x
y
- y
1
2
x
- (a) stretches or shrinks the graph in the y-direction; reflects it over the x-axis if c changes sign
2
4 c = 2
c = 1
c = –1.
-1 1
x
y
(b) As c increases, the parabola moves down and to the left. If c increases, up and right.
2
8
-1 2
x
c = 2 c = 1
c = –1.
y
(c) The graph rises or falls in the y-direction with changes in c.
4 2
6
8
-2 1 2
x
c = 2 c = 0. c = -
y
10 Chapter 1
- (a) y x -3 1
(b)
x 1
y
1
(c)
x 1
y
1 (d)
x 1
y 1
- (a)
1
y
x -2 -1 1 2
(b)
1
y
1
x
(c)
1 y
-1 1 2 3
x
(d)
1
y
-1 1 2 3
x
- 1 y
2
x
- Translate right 2 units, and up one unit.
10
-2 2 6
x
y
- Translate left 1 unit, reflect over x-axis, and translate up 2 units. y
x
1
- Translate left 1 unit, stretch vertically by a factor of 2, reflect over x-axis, translate down 3 units. - -
-6 -2 2 6
y x
Exercise Set 1.4 11
- Translate right 3 units, compress vertically by a factor of 12 , and translate up 2 units. y
x
2
4
- y = (x + 3)^2 − 9; translate left 3 units and down 9 units. -
10
15
-8 -4 -2 2
x
y
- y = (x + 3)^2 − 19; translate left 3 units and down 19 units. y x
- y = −(x − 1)^2 + 2; translate right 1 unit, reflect over x-axis, translate up 2 units.
2
-2 -1 1 3
x
y
- y = 12 [(x − 1)^2 + 2]; translate left 1 unit and up 2 units, compress vertically by a factor of 12. y
x
1
2
1
- Translate left 1 unit, reflect over x-axis, translate up 3 units.
1
2
2 8
x
y
- Translate right 4 units and up 1 unit. y
x
1
4
4 10
- Compress vertically by a factor of 12 , translate up 1 unit.
2
y
1 2 3
x
- Stretch vertically by a factor of
3 and reflect over x-axis. y (^) x
2
Translate right 3 units.
10
y
4 6
x
- Translate right 1 unit and reflect over x-axis. y
x
2 -2 2
- Translate left 1 unit, reflect over x-axis, translate up 2 units.
2
6
y
-3 -1 1 2
x
Exercise Set 1.4 13
- y
x
2
- (f + g)(x) = x^2 + 2x + 1, all x; (f − g)(x) = 2x − x^2 − 1, all x; (f g)(x) = 2x^3 + 2x, all x; (f /g)(x) = 2x/(x^2 + 1), all x
- (f + g)(x) = 3x − 2 + |x|, all x; (f − g)(x) = 3x − 2 − |x|, all x; (f g)(x) = 3x|x| − 2 |x|, all x; (f /g)(x) = (3x − 2)/|x|, all x = 0
- (f + g)(x) = 3
x − 1, x ≥ 1;( f − g)(x) =
x − 1, x ≥ 1;( f g)(x) = 2x − 2, x ≥ 1; (f /g)(x) = 2, x > 1
- (f + g)(x) = (2x^2 + 1)/[x(x^2 + 1)], all x = 0; (f − g)(x) = − 1 /[x(x^2 + 1)], all x = 0; (f g)(x) = 1 /(x^2 + 1), all x = 0; (f /g)(x) = x^2 /(x^2 + 1), all x = 0
- (a) 3 (b) 9 (c) 2 (d) 2
- (a) π − 1 (b) 0 (c) −π^2 + 3π − 1 (d) 1
- (a) t^4 + 1 (b) t^2 + 4t + 5 (c) x^2 + 4x + 5 (d)
x^2
(e) x^2 + 2xh + h^2 + 1 (f ) x^2 + 1 (g) x + 1 (h) 9 x^2 + 1
- (a)
5 s + 2 (b)
x + 2 (c) 3
5 x (d) 1 /
x (e) 4
x (f ) 0 (g) 1 / 4
x (h) |x − 1 |
- (f ◦ g)(x) = 2x^2 − 2 x + 1, all x; (g ◦ f )(x) = 4x^2 + 2x, all x
- (f ◦ g)(x) = 2 − x^6 , all x; (g ◦ f )(x) = −x^6 + 6x^4 − 12 x^2 + 8, all x
- (f ◦ g)(x) = 1 − x, x ≤ 1;( g ◦ f )(x) =
1 − x^2 , |x| ≤ 1
- (f ◦ g)(x) =
x^2 + 3 − 3, |x| ≥
6;( g ◦ f )(x) =
x, x ≥ 3
- (f ◦ g)(x) =
1 − 2 x , x =
, 1;( g ◦ f )(x) = −
2 x
, x = 0, 1
- (f ◦ g)(x) =
x x^2 + 1
, x = 0; (g ◦ f )(x) =
x
- x−^6 + 1 46. x x + 1
- (a) g(x) =
x, h(x) = x + 2 (b) g(x) = |x|, h(x) = x^2 − 3 x + 5
- (a) g(x) = x + 1, h(x) = x^2 (b) g(x) = 1/x, h(x) = x − 3
- (a) g(x) = x^2 , h(x) = sin x (b) g(x) = 3/x, h(x) = 5 + cos x
14 Chapter 1
- (a) g(x) = 3 sin x, h(x) = x^2 (b) g(x) = 3x^2 + 4x, h(x) = sin x
- (a) f (x) = x^3 , g(x) = 1 + sin x, h(x) = x^2 (b) f (x) =
x, g(x) = 1 − x, h(x) = 3
x
- (a) f (x) = 1/x, g(x) = 1 − x, h(x) = x^2 (b) f (x) = |x|, g(x) = 5 + x, h(x) = 2x
- y
x
1
2
-3 -2 -1 1 2 3
- Note that f (g(−x)) = f (−g(x)) = f (g(x)), so f (g(x)) is even.
f ( g ( x ))
x
y
–3 –1 1
1
- Note that g(f (−x)) = g(f (x)), so g(f (x)) is even.
x
y
1 3
1
3
g ( f ( x ))
- f (g(x)) = 0 when g(x) = ±2, so x = ± 1 .4; g(f (x)) = 0 when f (x) = 0, so x = ±2.
- f (g(x)) = 0 at x = −1 and g(f (x)) = 0 at x = − 1
3(x + h)^2 − 5 − (3x^2 − 5) h
6 xh + 3h^2 h
= 6x + 3h;
3 w^2 − 5 − (3x^2 − 5) w − x
3(w − x)(w + x) w − x
= 3w + 3x
(x + h)^2 + 6(x + h) − (x^2 + 6x) h
2 xh + h^2 + 6h h = 2x + h + 6;
w^2 + 6w − (x^2 + 6x) w − x
= w + x + 6
1 /(x + h) − 1 /x h
x − (x + h) xh(x + h)
x(x + h)
1 /w − 1 /x w − x
x − w wx(w − x)
xw
1 /(x + h)^2 − 1 /x^2 h
x^2 − (x + h)^2 x^2 h(x + h)^2
2 x + h x^2 (x + h)^2
1 /w^2 − 1 /x^2 w − x
x^2 − w^2 x^2 w^2 (w − x)
x + w x^2 w^2
- (a) the origin (b) the x-axis (c) the y-axis (d) none
16 Chapter 1
-4 4
-3 3
- (a) Whether we replace x with −x, y with −y, or both, we obtain the same equation, so by Theorem 1.4.3 the graph is symmetric about the x-axis, the y-axis and the origin. (b) y = (1 − x^2 /^3 )^3 /^2 (c) For quadrant II, the same;for III and IV use y = −(1 − x^2 /^3 )^3 /^2. (For graphing it may be helpful to use the tricks that precede Exercise 29 in Section 1.3.)
2
5
y
1 2
x
- y
x
2
2
- (a) 1
y
C
x O c^ o
(b) 2
y
O
x C c^ o
- (a)
x 1
y
1
(b)
x 1
y
1
(c)
x 1
y
3
(d)
x
y 1
C^ c/
- Yes, e.g. f (x) = xk^ and g(x) = xn^ where k and n are integers.
- If x ≥ 0 then |x| = x and f (x) = g(x). If x < 0 then f (x) = |x|p/q^ if p is even and f (x) = −|x|p/q if p is odd;in both cases f (x) agrees with g(x).
Exercise Set 1.5 17
EXERCISE SET 1.
- (a)
(b) Yes;the first and third slopes above are negative reciprocals of each other.
- (a)
(b) Yes;there are two pairs of equal slopes, so two pairs of parallel lines.
- III < II < IV < I 4. III < IV < I < II
- (a)
= 2. Since the slopes connecting all pairs of points are equal, they lie on a line.
(b)
. Since the slopes connecting the pairs of points are not equal, the points do not lie on a line.
- The slope, m = −2, is obtained from
y − 5 x − 7
, and thus y − 5 = −2(x − 7).
(a) If x = 9 then y = 1. (b) If y = 12 then x = 7/2.
- The slope, m = 3, is equal to
y − 2 x − 1
, and thus y − 2 = 3(x − 1).
(a) If x = 5 then y = 14. (b) If y = −2 then x = − 1 /3.
- (a) Compute the slopes:
y − 0 x − 0
or y = x/2. Also
y − 5 x − 7
= 2 or y = 2x − 9. Solve simultaneously to obtain x = 6, y = 3.
- (a) The first slope is
1 − x and the second is
4 − x
. Since they are negatives of each other we
get 2(4 − x) = −5(1 − x) or 7x = 13, x = 13/7.
- (a) 27 ◦^ (b) 135 ◦^ (c) 63 ◦^ (d) 91 ◦
- (a) 153 ◦^ (b) 45 ◦^ (c) 117 ◦^ (d) 89 ◦
- (a) m = tan φ = −
3 /3, so φ = 150◦^ (b) m = tan φ = 4, so φ = 76◦
- (a) m = tan φ =
3, so φ = 60◦^ (b) m = tan φ = −2, so φ = 117◦
- y = 0 and x = 0 respectively
- y = − 2 x + 4
6
0
-1 1
- y = 5x − 3 2
–1 2
Exercise Set 1.5 19
- (a) The velocity is the slope, which is
= 9/10 ft/s.
(b) x = − 4 (c) The line has slope 9/10 and passes through (0, −4), so has equation x = 9t/ 10 − 4;at t = 2, x = − 2 .2. (d) t = 80/ 9
- (a) v =
= 2 m/s (b) x − 1 = 2(t − 2) or x = 2t − 3 (c) x = − 3
- (a) The acceleration is the slope of the velocity, so a =
ft/s^2.
(b) v − 3 = − 43 (t − 1), or v = − 43 t + 133 (c) v = 133 ft/s
- (a) The acceleration is the slope of the velocity, so a =
ft/s^2.
(b) v = 5 ft/s (c) v = 4 ft/s (d) t = 4 s
- (a) It moves (to the left) 6 units with velocity v = −3 cm/s, then remains motionless for 5 s, then moves 3 units to the left with velocity v = −1 cm/s.
(b) vave =
cm/s
(c) Since the motion is in one direction only, the speed is the negative of the velocity, so save = 109 cm/s.
- It moves right with constant velocity v = 5 km/h;then accelerates;then moves with constant, though increased, velocity again;then slows down.
- (a) If x 1 denotes the final position and x 0 the initial position, then v = (x 1 − x 0 )/(t 1 − t 0 ) = 0 mi/h, since x 1 = x 0. (b) If the distance traveled in one direction is d, then the outward journey took t = d/40 h. Thus save =
total dist total time
2 d t + (2/3)t
80 t t + (2/3)t
= 48 mi/h.
(c) t + (2/3)t = 5, so t = 3 and 2d = 80t = 240 mi round trip
- (a) down, since v < 0 (b) v = 0 at t = 2 (c) It’s constant at 32 ft/s^2.
- (a) v
t
20
60
20 80
(b) v =
10 t if 0 ≤ t ≤ 10 100 if 10 ≤ t ≤ 100 600 − 5 t if 100 ≤ t ≤ 120
- x
t
20 Chapter 1
- (a) y = 20 − 15 = 5 when x = 45, so 5 = 45k, k = 1/9, y = x/ 9
(b) y
x
(^2 )
(c) l = 15 + y = 15 + 100(1/9) = 26.11 in. (d) If ymax = 15 then solve 15 = kx = x/9 for x = 135 lb.
- (a) Since y = 0.2 = (1)k, k = 1/5 and y = x/ 5
(b) y
x
1
2 6
(c) y = 3k = 3/5 so 0.6 ft. (d) ymax = (1/2)3 = 1.5 so solve 1.5 = x/5 for x = 7.5 tons
- Each increment of 1 in the value of x yields the increment of 1.2 for y, so the relationship is linear. If y = mx + b then m = 1.2;from x = 0, y = 2, follows b = 2, so y = 1. 2 x + 2
- Each increment of 1 in the value of x yields the increment of − 2 .1 for y, so the relationship is linear. If y = mx + b then m = − 2 .1;from x = 0, y = 10.5 follows b = 10.5, so y = − 2. 1 x + 10. 5
- (a) With TF as independent variable, we have
TC − 100
TF − 212
, so TC =
(TF − 32).
(b) 5/ (c) Set TF = TC = 59 (TF − 32) and solve for TF : TF = TC = − 40 ◦^ (F or C). (d) 37 ◦^ C
- (a) One degree Celsius is one degree Kelvin, so the slope is the ratio 1/1 = 1. Thus TC = TK − 273 .15. (b) TC = 0 − 273 .15 = − 273. 15 ◦^ C
- (a)
p − 1 h − 0
, or p = 0. 098 h + 1 (b) when p = 2, or h = 1/ 0. 098 ≈ 10 .20 m
- (a)
R − 123. 4
T − 20
, so R = 0. 42 T + 115. (b) T = 32. 38 ◦C
- (a)
r − 0. 80 t − 0
, so r = − 0. 0125 t + 0. 8 (b) 64 days
- (a) Let the position at rest be y 0. Then y 0 + y = y 0 + kx;with x = 11 we get y 0 + kx = y 0 + 11k = 40, and with x = 24 we get y 0 + kx = y 0 + 24k = 60. Solve to get k = 20/13 and y 0 = 300/13. (b) 300 /13 + (20/13)W = 30, so W = (390 − 300)/20 = 9/2 g.
