/ =0. (c) Section P.3 Functions and Their Graphs. (c) g(-2) = 5-(-2) 2 = 5-4 = 1. (e) g(x) = 0 for x = -I, 1 and 2. 2.
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1 Section P,3 Functions and Their Graphs 3 Section P.3 Functions and Their Graphs. (a) Domain off." -4 < x < 4 ~ [-4, 4] Range off:-3 < y < 5 ~ [-3, 5] Domain of g: -3 < x < 3 ~ [-3, 3] Range of g: -4 < y < 4 ~ [-4, 4] (b) (-)=- g(3) = -4 (c) f(x) = g(x) for x = - (d) f(x) = for x :. (a) (e) g(x) = 0 for x = -I, and Domain of f. -5 _< x _< 5 ~ [-5, 5] Range off -4 _< y _< 4 ~ [-4, 4] Domain of g: -4 _< x _< 5 ~ [-4, 5] 4. (a) f(-4) = ~+5 = ~ = (b) f(ll) = ~+5 --,fi-g = 4 (c) f(-8)= xi~ + 5 = ~-~, undefined (d) f(x + Ax) = ",./x + kx + S 5. gto)" "= 5-o~ = 5 (b) g(x/~) 5 (x/~) = - =5-5=0 (c) g(-) = 5-(-) = 5-4 = (d) g(t-) = 5-(t-) = 5-(-t+) = 4 + t - t 6. (a) g(4) = 4(4) = 0 (b) g(-~) =\~(3(3\~- - 4) = ~.(_~.)9 5 = 458 (b) (c) (d) Rangeofg: -4 _< y _< ~ [-4, ] = - g(3) = f(x) = g(x)for x = - and x = 4 f(x) = forx =-4,4 (c) g(c) = c(c - 4) = C c (d) g(t+4) = (t+4)(t+4) = (t+4)t = t 3 +gfl +6t 7. (a) f(0) = cos((0)) = cos0 = (e) g(x) = 0for x = - 3. (a) S(0) = 7(0) - 4 = -4 (b) 7(-3)= 7(-3)- 4 = -5 (c) f(b) = 7(b)-4 = 7b-4 / =0 (c) z COS (a) f(z) = sin n- = 0 (d) f(x-) = 7(x-)-4 = 7x-ll (c) 9. f(x+ax)-f(x) = (x+ax) 3-x3 = x ~ +3xAx+3x(Ax) +(Ax)3_x, - 3x +3xAx+(Ax),Ax f(x) - f() 3x - - (3 - ) _ 3(x - ) x- = x- x--- = 3,x :~. f(x)- f() = (/~--i"- ) x- x- -~x- l+~x- -- -x (x- )x/tx - +-,./Tx- -, f(x) - f() = x 3 - x - 0 _ x(x + )(x - ) : (x + ), x x- x- x-
2 4 Chapter P Preparation for Calculus 3. f(x) = 4x Domain: (-~o, 00) Range: [0, 00) 4. g(x) = x~ - Domain: (-~o, 00) Range: [-5, 5. g(x) --,S~ Domain: 6x >_ 0 Range: [0, ~o) 6. h(x) = -x/-~ + 3 x > 0 ~ [0,~o) Domain: x + 3 > 0 ~ [-3, Range: (-~, O] 7. f(t) = sec-- 4 n t (n+l)cr ~ t 4n+ 4 Domain: all t 4n +, n an integer Range: (-00,- ] 8. h(t) = cot t Domain: all t = nn-, n aninteger Range: (-~o, 00) 3 9. f(x):- x Domain: all x Range: (-00, 0)w (0, 00) 0. g(x) = ~ Domain: (-0% ) t~ (, 00) Range: (-00, 0)w (0, 00). f(x): ~x x_>o and -x>_o x>_o and x_<l Domain: 0 < x _< ~ [0, ]. f(x) = x/x - 3x + x - 3x + > 0 (x- )(x- ) _~ 0 Domain: x >_ orx < Domain: (-00, w [, gt~ I = - cos x - cos x 0 cos x Domain: all x nn-, n an integer 4. h(x) = sin x - 0/) sinx-- 0 sinx - Domain: all x nor, nn-, n integer f(x) -ix + 3 x+3 0 x+3 0 Domain: all x -3 Domain: (-~,-3) ~ (-3, (x- )(x + ) 0 Domain: all x + Domain: (-c~,-) ~ (-, ) ~ (, 7. f(x) = {; ++, ;><00 (a) f(-) = (-) + = - (b) f(0) = (0)+ = (c) f() = () + = 6 (d) f(t ~ +0 = ( t +0+ = t ~ +4 (Note: t: + > 0 for all t) Domain: (-oo, oo) Rans~. (-~, 0 ~ [~, ~)
3 X s~. S(~) -- ~ + ~ ~ -< X +, x > (a) f(-) = (-) + = 6 (b) f(0) = 0 + = (c) () = + = (Note: s + > for all s) Domain: (-m, m) Range: [, o~) 9. f(x) = {~ _ + + l,, x x > < (a) f(-3) : -3 + = 4 (b) f() =-+ = 0 (c) f(3) =-3+ =- = s 4 + 8s a + 0 Section P.3 Functions and Their Graphs 5 3. g(x) 4 Domain: (-~, 0) c.) (0, oo) Range: (-0% 0) ~ (0, oo) h(x) = ~x - 6 y (d) f(b + )= -(b + )+ = _b Domain: (-~o, Range: (-0% 0] ~ [, oo) =~x+4, x_<5 30. s(~) {(x- ~)~, ~ ~ ~ (a) f(-3)= -~+ 4 = -,~ = (b) f(0)= ~,/-~+ 4 = (o) f(5) = ~+4 = 3 (d) f(lo) = 0o- 5): = 5 Domain: [-4, ~) Range: [0, ~) 3. f(x) = x Domain: (-0% oo) Range: (-o% oo) Domain: x- 6 > 0 x > 6 ~ [6, Range: [0, 34. f(x): ¼x Domain: (-~o, Range: (-~o, 35. f(x) = ~- x ~ Domain: [-3, 3] Range: [0, 3] y 4"
4 6 Chapter P Preparation for Calculus s(.) -- x + x Domain: [-, ] Range: I-, wi~ ~ [-,.83] y-intercept: (0, ) 40. d x-intercept: (-x/~, 0) (-,~, o) (0, g(t) : 3 sin ~-t Domain: Range: [-3, 3] h o~ = -s cos Domain: Range: [-5, 5] 5 4. x- y = 0 ~ y = +~X y = x _ ~ y = +~x y is not a function ofx. Some vertical lines intersect the graph twice. ~x~-y=o~ y=x/~x~-4 y is a function ofx. Vertical lines intersect the graph at most once. y is a function ofx. Vertical lines intersect the graph at most once. x + y = 4 y = +x/~- x y is not a function ofx. Some vertical lines intersect the graph twice. x +y = 6 ~ y = +x/6-x y is not a function ofx because there are two values ofy for some x. x +y = 6 ~ y = 6-x a y is a function ofx because there is one value ofy for each x. y is not a function ofx because there are two values ofy for some x. X xy- x + 4 y = 0 ~ y - xz +4 y is a function ofx because there is one value ofy for each x. 39. The student travels -0 - mi/min during the first 4 0 minutes. The student is stationary for the next minutes. 6- Finally, the student travels - mi/min during 0-6 the final 4 minutes y = f(x) - 5 is a vertical shift 5 units downward. Matches b. 5. y = f(x + 5) is a horizontal shift 5 units to the left. Matches d. y = -f(-x) - is a reflection in the y-axis, a reflection in the x-axis, and a vertical shift downward units. Matches c. 5. y = -f(x - 4) is a horizontal shift 4 units to the right, followed by a reflection in the x-axis. Matches a.
5 Section P.3 Functions and Their Graphs y = f(x + 6) + is a horizontal shift to the left 6 units, and a vertical shift upward units. Matches e. 54. y = f(x - ) + 3 is a horizontal shift to the right unit, and a vertical shift upward 3 units. Matches g. 55. (a) the graph is shifted 3 units to the left. (f) The graph is stretched vertically by a factor of -~ (b) The graph is shifted unit to the right. 56. (a) g(x) = f(x - 4) g(6) = f()= g(o) = f(-4)=-3 Shift f right 4 units 3- - " ~ (o, -3) t (c) The graph is shifted units upward i 4 (d) The graph is shifted 4 units downward. (b) g(x) = f(x + ) Shift f left units 4~ O, ) l (c) g(x) = f(x) + 4 Vertical shift upwards 4 units 6 (, 5) 4 (e) The graph is stretched vertically by a factor of 3. x _ 3 I I ~-x (d) x(x/= S(x/- Vertical shift down unit 3 (, 0) 3
6 Chapter P Preparation for Calculus (e) g(x) = f(x) g() = f()= g(-4) = f(-4)=-6 (, ) (f) g(x) = ½S(x) g() = -}f()= ½ g(-4) = -}f(-4)=--} " (a) y = ~x (a) (b) h(x) = sin(x + (st/)) + l is a horizontal shift a / units to the left, followed by a vertical shift unit upwards. h(x) = -sin(x - )is a horizontal shift unit to the right followed by a reflection about the x-axis. 59. (a) (go)) = ~(0) = 0 (b) g(f())= g()= 0 (~) g(y(0))= g(o)=-~ (d) f(g(-4))= f(5)= ~ (e) f(g(x)) = f(x -) = ~x - (f) g(f(x))= g(~x) : (.,,/-~) -l=x-, (x > O) 60. f(x) = sin x, g(x) = ax (a) f(g())= f(a ) = sin(rc)= 0 (b) flg(~))= f/~)= sin/~) = (c) g(f(o)) = g(o) = Vertical shift units upward (b) y =-~ -~ (e) y(g(x))= y(~x)= s~.(~x) (~ g(f(x)) = g(sin x) = x sin x 6, f(x) = x, g(x) = (f o g)(x) = f(g(x)) Domain: [0, ~o) = = =X,x>_O (go f)(x) = g(f(x)) = g(x ) = x/~-x = x Reflection about the x-axis (c) y =,/Tx- y Domain: No. Their domains are different. (fo g) = (go f) for x> Horizontal shift units to the right
7 ~z. s(x) -- x~ -~, g(.) = cos (fo g)(x) = f(g(x)) = f(cos x) = coszx- Domain: (-0% oo) Section P.3 Functions and Their Graphs 9 ( ) 64. (fog)(x) : S ~ - ~+ Domain: (-, oo) (go f)(x) : g(x --) = cos(x a -) Domain: (-0% No, fog, go f. 3, g(x) = x ~ - 63, f(x)= 7 (f o g)(x) = Domain: all x:~ + => (-co,-) u (-, )~ (, (go f)(x) = 3 f(g(x)) = f(x ~- ) - x~- g(s(.)) g = - = -~5-- - x Domain: all x + 0 ~ (-oo, O)~ (0, oo) No, fog ~ go f. ou can find the domain of g o f by determining the intervals where ( + x) and x are both positive, or both negative i _½ 0 i ~ 65. (a) (fo g)(3) = f(g(3)) = f(-)= 4 (b) g(f())= g()= - (c) g(f(5)) : g(-5), which is undefined (d) (fo g)(-3)= f(g(-3))= f(-)= 3 (e) (go f)(-) = g(f(-))= g(4)= (f) f(g(-)) = f(-4), which is undefined 66. (Ao r)(t): A(r(t))= A(0.6)= sr(0.6t) = 0.36n- (Ao r)(t)represents the area of the circle at time t. 67. F(x) = x/~x- Let h(x) = x, g(x) = x- and f(x) = x/--~x. Then, (fo go h)(x) = f(g(x)) = f((x)- ) = 4(x)- = x/~tx- = F(x). [Other answers possible] 68. F(x) = -4 sin(- x) Let f(x) = -4x, g(x) = sin x and h(x) = - x. Then, (fogo h)(x): f(g(- x)): f(sin(- x)):-4 sin(- x): F(x). [Other answers possible] f(--x) = (--X) (4 -- (--X) ) : X(-X ) : f(x) Even 70. f(-x) = ~ =-3~x =-f(x) Odd S(-x) = (-x)cos(-x) = -x cos x = -S(x) Odd S(-x) = sin(-x)= sin(-x)sin(-x)= (-sin x)(-sin x)= sin x Even ~
8 30 Chapter P Preparation for Calculus 73. (a) If f is even, then (-~, 4)is on the graph. (b) If f is odd, then (-~, -4)is on the graph. 74. (a) If f is even, then (-4, 9) is on the graph. (b) If f is odd, then (-4, -9) is on the graph. 75. f is even because the graph is symmetric about the y-axis, g is neither even nor odd. h is odd because the graph is symmetric about the origin. 76. (a) If f is even, then the graph is symmetric about the y-axis Slope- = ) y--- 7(x- 7 y - = --x For the line segment, you must restrict the domain. (5, 8) 4 6 3, ) x + y = 0 (b) If f is odd, then the graph is symmetric about the origin. y =--X y = _~-~ f(x) : _.~fs-~, x <_ 0 f 4 x (-6) Slope y - 4 = -~(x -(-~/) y-4 =-5x-0 y = -5x - 6 For the line segment, you must restrict the domain. 80. X + y = 36 y = 36 - x y =-x/36-x, -6_< x_< 6 y f(x) =-5x-6,- <_ x < 0 8. Answers will vary. Sample answer: Speed begins and ends at 0. The speed might be constant in the middle: ~~:~x Time (in hours)
9 Section P. 3 Functions and Their Graphs 3 8. Answers will vary. Sample answer: Height begins a few feet above 0, and ends at (a) For each time t, there corresponds a depth d. (b) Domain: 0 < t _< 5 Range: 0 < d < 30 (c) d Distance 83. Answers will vary. Sample answer: In general, as the price decreases, the store will sell more. (d) d(4) ~ 8. At time 4 seconds, the depth is approximately 8 cm. 89. (a) A Price (in dollars) ~r~ ~ Answers will vary. Sample answer: As time goes on, the value of the car will decrease ear (5 ~ 955) (b) A(0),~ 384 acres/farm y = ~C--X y = c - x x + y = c, acircle. For the domain to be [-5, 5], c = For the domain to be the set of all real numbers, you must require that x + 3cx + 6 ~: 0. So, the discriminant must be less than zero: (3c) - 4(6) < 0 9c < 4 8 C <-~ -~ff-~ < c < x/- ~ (b) ~ H = ~-f-x + k,l.6j ) = x x re(x) =l l+lx- Ifx < O, then f(x) :-x-(x-) =-x+. IfO < x <, then f(x) = x-(x-) =. If x_>, then f(x) = x+(x-) = x-. So, -x+, x_< 0 f(x) = ~, 0 < x <. x-, x > 87. (a) T(4) = 6,T(5) ~ 3 (b) If H(t) = T(t - ), then the changes in temperature will occur hour later. (c) If H(t) = T(t) -, then the overall temperature would be degree lower.
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