18. (a) S(t) = sin(0.5080t- 2.07) (b) ~oo. (c)

Transcription

1 Review Exercises fr Chapter P 5 1. (a) T =.985 x 10-p p + 5.8p (a) S(t) = sin(0.5080t-.07) ~ (c) Fr T = 00 F, p,~ 8.9 lb/in.. (d) The mdel is based n data up t 100 punds per square inch. (c) The mdel is a gd fit. 17. (a) es, is a functin f t. At each time t, there is ne and nl ne displacement. The amplitude is apprximatel (.5-1.5)/ = 0.5. The perid is apprximatel ( ) = 0.5. (c) One mdel is = 0.5 sin(x-t) +. (d) (0.15,.5) (0.75, 1.5) The mdel appears t fit the data. The mdel is a gd fit. (d) The average is the cnstant term in each mdel F fr Miami and 5.7 F fr Sracuse. (e) The perid fr Miami is ~ff0.91 = 1.8. The perid fr Sracuse is ~/ = 1.. In bth cases the perid is apprximatel 1, r ne ear. (f) Sracuse has greater variabilit because 5.7 > Answers will var. 0. Answers will var. 1. es, A 1 -< A. T see this, cnsider the tw triangles f areas A1 and A: Ct C Fr i = 1,, the angles satisf d + fli + i = ~. At least ne f c~ <_, fl~ <_ flz, ~ < 7" must hld. Assume a~ _< ~. Because 0~ -< ~r/ (acute triangle), and the sine functin increases n [0, a-/], u have A~ = ½blCl sin 1 < ½ bc sin q _< ½bzc sin a = A Review Exercises fr Chapter P 1. =5x-8 x = 0: = 5(0)- 8 = -8 ~ (0,-8)-intercept : 0:0 : 5x - 8 ~ x : } ~ (}, 0/x-intercept /. : (x)(x-) x : 0: = (0 - )(0 - ) : 1 ~ (0, 1)-intercept = 0:0 = (x - )(x - ) ~ x =, ~ (, 0), (, 0)x-intercepts

2 Chapter P Preparatin fr Calculus x- x=0:- - => 0, -intercept 0- = 0:0- x - -> x=~ (, 0)x-intercept. x-. x= x = 0 and = 0 are bth impssible. N intercepts. Smmetric with respect t -axis because (-x)- (-x) + = 0 x~- x ~ + = O. Smmetric with respect t -axis because = (-x) -(-x)~ + = x -x ~ +. 7 =-½x+ Slpe: -½ -intercept:-~ =-} x+5 Slpe: -} -intercept:-} 5 10, 0.0x = 0.5 x + 15 = 5 =- x+-~ Slpe: -intercept: (0,-}) - - -I x - = 1 Slpe: - =-x+1 = x- -intercept: (0,-) 11. = 9-8x-x: = -(x)(x+9) -intercept: (0, ~ 9) x-intercepts: (1, 0), (-9, O) 8 " - " I : x(-x) -intercept: (0, 0) x-intercepts: (0, 0), (, 0) " ", - 8

3 Review Exercises fr Chapter P 7 1. = x/~-- x Dmain: (-% ] 18. x-+l =O~ = x+l -x ~ = 7 ~ =x~ = x-- " (x+l)-x~ =7 =x+l 0 = x -x+ N real slutin. N pints f intersectin. The graphs f = x+land = x a +7dnt intersect. 19. Answers will var. Sample answer: u need factrs (x + )and (x - ). 15. = x 5 Xmin = -5 Xmax = 5 Xscl = 1 min = -0 max = 10 scl = 5 1. = 8..~x- Xmin = -0 Xmax = 0 Xscl = 10 min = -0 max = 0 scl = 10 Multipl b x t btain rigin smmetr. = x(x+)(x-) = x x 0. =kx (a) = k(1) ~ k = and = x 1= k() =:> k = -½ and = -½x (c) 0 = k(o) =:> and k will d! (d) = k() :=> k = 1 ~ = x x+ = ~ = ½(-5x-l) x-=-5~ =x Slpe = ~_ ~ Using a graphing utilit, the lines intersect at (, ). Analticall, ½(-5x-l) = x+5-5x = x+15 = 8x =x. Fr x =, = x+5 =+5 =. The line is hrizntal and has slpe 0., (-7, 8) (- 1, 8)

4 8 Chapter P Preparatin fr Calculus.. t (-8! -(-8) t t t - 5 = 5t = 1 1 -()_ t t 1 - = 9 + t -5 = t 5 t- +5 =¼x 1 +0 = 7x1 0 = 7x Because rn is undefined the line is vertical. 7. x =-8rx+8 = 0 (-8, 1) I I I Because m = 0, the line is hrizntal. - = 0(x - 5) =r-=0 8" - 9. (a) = 7x x+101 5x - = has slpe 5-5 = -5(x + ) - 15 = 5x+15 0 = 5x x + (5, ) = = --5(x + ) = -5x - 15 =0 (d) Slpe is undefined s the line is vertical. x+=0 x = - 0. (a) - = --}(x- ) =x+ x+ = 0 x + = 0 has slpe. Slpe f the perpendicular line is 1. -= l(x - ) =x+ 0=x-+ (c) m ) - = -~(x - =-x+ x+ = 0 (d) Because the line is hrizntal the slpe is 0. = -=0

5 Review Exercises fr Chapter P 9 1. The slpe is V = -850t + 1,500. V() = -850()+ 1,500 = $ (a) C = 9.5t t +,500 =.75t +,500 R = 0t (c) 0t =.75t +, t --,500 t ~ 50.8 hurs t break even, x - = = _+x/~ - Nt a functin because there are tw values f fr sme x. 1. x - = 0 Functin fx because there is ne value fr fr each x. - xi 5. - x is a functin fx because there is ne value f fr each x. ~ - - l- I I 1.,0 5. x = 9 - ~ Nt a functin fx since there are tw values f fr sme x _1 7. f(x) x (a) f(o) des nt exist. f(l+zlr)-f(1) _ l+ax 1 X q, x < 0 8. f(x)= x],x > Ax, Ax ~, 0 (a) f(-) = (-) + = 18 (because- < O) f(0)= 0[= (c) f(1) = 1 = 1 9. (a) Dmain: -x > 0 ~ - < x < r [-,] (c) Range: [0, ] Dmain: all x, 5r (-0% 5) t (5, ) Range: all ~ 0 r (-0% 0)c9 (0, ) Dmain: all x r (-0% ) Range: all r (-0% ) 0. f(x) = 1- x and g(x) = x + 1 (a) f(x)-g(x) = (1-x z)-(x+l) =-x x f(x)g(x)= (1- x)(x + 1)= x - x + x + 1 (e) g(f(x)) = g(1-x ) = (1-x )+1 = x

6 0 Chapter P Preparatin fr Calculus f(x) = x + c, c =,0, c 7 0. (a) Odd pwers: f(x)= x, g(x)= x, h(x)= x 5 g ~ ~,.-x f(x) = (x - c), c =, O, The graphs f f g, and h all rise t the right and fall t the left. As the degree increases, the graph rises and falls mre steepl. All three graphs pass thrugh the pints (0, 0), (1, 1), and (, ) and are smmetric with respect t the rigin. Even pwers: f(x) = x, g(x) = x, h(x) = x (c) f(x) = (x) +c,c =,0, c= (d) f(x) = cx, e =,0, ~ t 1c= The graphs f f g, and h all rise t the left and t the right. As the degree increases, the graph rises mre steepl. All three graphs pass thrugh the pints (0, 0), (1, 1), and (, 1)and are smmetric with respect t the -axis. All f the graphs, even and dd, pass thrugh the rigin. As the pwers increase, the graphs becme flatter in the interval < x < 1. = x 7 will lk like h(x) = x s, but rise and fall even mre steepl, = x 8 will lk like h(x) = x, but rise even mre steepl.. (a) f(x) = x(x - ) The leading cefficient is psitive and the degree is even s the graph will rise t the left and t the right. loo. f(x) = X - X g(x) = x(x - ) t [ (, -) (a) The graph f g is btained frm f b a vertical shift dwn 1 unit, fllwed b a reflectin in the x-axis: g(x) = -Ef(x)- 1] = -x + x + 1 The leading cefficient is psitive and the degree is dd s the graph will rise t the right and fall t the left. 800 The graph fg is btained frm f b a vertical shift upwards f 1 and a hrizntal shift f t the right. g(x) = f(x - ) (x - ) - (x- ) -b 1

7 Prblem Slving fr Chapter P 1 (c) h(x) = x(x - ) The leading cefficient is psitive and the degree is even s the graph will rise t the left and t the right (a) x + = -- 1-x A = x = x(lz-x) Dmain: 0 < x < 1 r (0, 1) O (c) Maximum area is A = in.. In general, the maximum area is attained when the rectangle is a square. In this case, x =.. 7. Fr cmpan (a) the prfit rse rapidl fr the first ear, and then leveled ff. Fr the secnd cmpan, the prfit drpped, and then rse again later. (a) (cubic), negative leading cefficient (quartic), psitive leading cefficient (c) (quadratic), negative leading cefficient (d) 5, psitive leading cefficient 8. (a) =.0x I a (c) The data pint (7, ) is prbabl an errr. Withut this pint, the new mdel is =.x (a) es, is a functin f t. At each time t, there is ne and nl ne displacement. The amplitude is apprximatel (0.5 - (-0.5))/ = 0.5. The perid is apprximatel cs/zrt~ 1 cs(5.7,) (c) One mdel is = ~- \1.1 ) ~ ~ (d) 0.5 [[ (0.5,-0.5) Prblem Slving fr Chapter P The mdel appears t fit the data. 1. (a) x - x + a - 8 = 0 (x -x+9)+(~ -8+1) =9+1 (x-) +(-) = 5 Center: (, ); Radius: 5 Slpe f line frm (0, 0) t (, ) is ~. Slpe f tangent line is. S, - ~(-0) 0 = -"x~ = --x Tangentline (c) Slpe f line frm (, 0) t (, -0 ) is - - Slpe f tangent line is S, 0 9 -~( x ) ~ -x Tangent line 9 (d) --x =-x-- 9 x= Intersectin: (, -~)