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Calculus_-_Early_Transcendentals_5e_-_J._Stewart_-__SOLUTION

Calculus_-_Early_Transcendentals_5e_-_J._Stewart_-__SOLUTION...

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Unformatted text preview: Stewart Calculus ET 5e 0534393217;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 . 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 . 2. (a) The point ( 4, 2 ) is on the graph of f , so f ( 4)= 2 . The point ( 3,4 ) is on the graph of g , so g(3)=4 . (b) We are looking for the values of x for which the y values are equal. The y values for f and g are equal at the points ( 2,1 ) and ( 2,2 ) , so the desired values of x are 2 and 2 . (c) f (x)= 1 is equivalent to y= 1 . When y= 1 , we have x= 3 and x=4 . (d) As x increases from 0 to 4 , y decreases from 3 to 1 . Thus, f is decreasing on the interval 0,4 . (e) The domain of f consists of all x values on the graph of f . For this function, the domain is 4 x 4 , or 4,4 . 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) The domain of g is 4,3 and the range is 0.5,4 . 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 ground acceleration is 85 a 115 . In Figure 11, the range of the north south acceleration is approximately 325 a 485 . In Figure 12, the range of the east west acceleration is approximately 210 a 200 . 4. Example 1: A car is driven at 60 mi / h for 2 hours. The distance d traveled by the car is a function of the time t . The domain of the function is {t |0 t 2} , where t is measured in hours. The range of the function is { d |0 d 120} , where d is measured in miles. Example 2: At a certain university, the number of students N on campus at any time on a particular day is a function of the time t after midnight. The domain of the function is {t |0 t 24} , where t is measured in hours. The range of the function is { N |0 N k} , where N is an integer and k is the largest number of students on campus at once. 1 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function Example 3: A certain employee is paid $8.00 per hour and works a maximum of 30 hours per week. The number of hours worked is rounded down to the nearest quarter of an hour. This employee’s gross weekly pay P is a function of the number of hours worked h . The domain of the function is 0,30 and the range of the function is { 0,2.00,4.00,... ,238.00,240.00} . 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. 6. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [ 2,2] and the range is [ 1,2] . 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] . 8. No, the curve is not the graph of a function since for x=0 , points on the curve. 1 , and 2 , there are infinitely many 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 a gradual increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems. 10. The salesman travels away from home from 8 to 9 A.M. and is then stationary until 10 : 00 . The salesman travels farther away from 10 until noon. There is no change in his distance from home until 1 : 00 , at which time the distance from home decreases until 3 : 00 . Then the distance starts increasing again, reaching the maximum distance away from home at 5 : 00 . There is no change from 5 until 6 , and then the distance decreases rapidly until 7 : 00 P.M., at which time the salesman reaches home. 11. The water will cool down almost to freezing as the ice melts. Then, when the ice has melted, the water will slowly warm up to room temperature. 2 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 12. The summer solstice (the longest day of the year) is around June 21, and the winter solstice (the shortest day) is around December 22. 13. Of course, this graph depends strongly on the geographical location! 14. The temperature of the pie would increase rapidly, level off to oven temperature, decrease rapidly, and then level off to room temperature. 15. 16. (a) (b) (c) 3 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function (d) 17. (a) (b) From the graph, we estimate the number of cell phone subscribers in Malaysia to be about 540 in 1994 and 1450 in 1996. 18. (a) (b) From the graph in part (a), we estimate the temperature at 11:00 A.M. to be about 84.5 C. 2 19. f (x)=3x x+2. 2 f (2)=3(2) 2+2=12 2+2=12. 2 f ( 2)=3( 2) ( 2)+2=12+2+2=16. 4 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 2 f (a)=3a a+2. 2 2 f ( a)=3( a) ( a)+2=3a +a+2. 2 2 2 2 f (a+1)=3(a+1) (a+1)+2=3(a +2a+1) a 1+2=3a +6a+3 a+1=3a +5a+4. 2 2 2 f (a)=2 f (a)=2(3a a+2)=6a 2a+4. 2 2 2 f (2a)=3(2a) (2a)+2=3(4a ) 2a+2=12a 2a+2. 2 22 2 4 2 4 2 f (a )=3(a ) (a )+2=3(a ) a +2=3a a +2. f (a) 2 2 2 ( 2 = 3a a+2 = 3a a+2 4 3 2 3 ) ( 3a2 a+2) 2 2 4 3 2 =9a 3a +6a 3a +a 2a+6a 2a+4=9a 6a +13a 4a+4. 2 2 2 2 2 f (a+h)=3(a+h) (a+h)+2=3(a +2ah+h ) a h+2=3a +6ah+3h a h+2. 4 3 4 3 2 r +3r +3r+1 . We ( r+1 ) = 3 3 wish to find the amount of air needed to inflate the balloon from a radius of r to r+1 . Hence, we need 4 4 3 4 3 2 2 to find the difference V ( r+1 ) V ( r ) = r +3r +3r+1 r = 3r +3r+1 . 3 3 3 ( 20. A spherical balloon with radius r+1 has volume V ( r+1 ) = ( 2 ) 2 ( 2 2 21. f (x)=x x , so f (2+h)=2+h (2+h) =2+h (4+4h+h )=2+h 4 4h h = 2 2 2 2 ) ) ( h2+3h+2) , 2 f (x+h)=x+h ( x+h ) =x+h (x +2xh+h )=x+h x 2xh h , and 2 2 2 2 f (x+h) f (x) x+h x 2xh h x+x h 2xh h h(1 2x h) = = = =1 2x h . h h h h x 2+h 2+h x+h , so f (2+h)= = , f (x+h)= , and x+1 2+h+1 3+h x+h+1 x+h x f (x+h) f (x) x+h+1 x+1 1 ( x+h ) ( x+1 ) x ( x+h+1 ) = = = . h h h ( x+h+1 ) ( x+1 ) ( x+h+1 ) ( x+1 ) 22. f (x)= 23. f (x)=x/(3x 1) is defined for all x except when 0=3x 1 { x R| x 1 3 } = /( , 2 1 3 1 , 3 ) x= 1 , so the domain is 3 . 2 24. f (x)=(5x+4) x +3x+2 is defined for all x except when 0=x +3x+2 1 , so the domain is { x R| x 2, 1} =( , 2) ( 2, 1) ( 1, ) . 0=(x+2)(x+1) x= 2 or 25. 5 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 3 f (t)= t + 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, 26. g(u)= u + 4 u is defined when u 0 and 4 u 0 / 4 ). u 4 . Thus, the domain is 0 u 4= 0,4 . 2 2 2 27. h(x)=1 x 5x is defined when x 5x>0 x(x 5)>0 . Note that x 5x 0 since that would result in division by zero. The expression x(x 5) is positive if x<0 or x>5 . (See Appendix A for ,0) (5, ) . methods for solving inequalities.) Thus, the domain is ( 2 2 28. h(x)= 4 x . Now y= 4 x 2 2 y =4 x 2 2 x +y =4 , so the graph is the top half of a circle of radius 2 with center at the origin. The domain is From the graph, the range is 0 y 2 , or 0,2 . { x|4 x2 0} = { x|4 29. f (x)=5 is defined for all real numbers, so the domain is R , or ( horizontal line with y intercept 5 . , 1 (x+3) is defined for all real numbers, so the domain is R , or ( 2 3 a line with x intercept 3 and y intercept . 2 2 x } =[ 2,2] . ) . The graph of f is a 30. F(x)= 31. f (t)=t 6t is defined for all real numbers, so the domain is R , or ( } ={ x|2 2 x , , ) . The graph of F is ) . The graph of f is a 2 parabola opening upward since the coefficient of t is positive. To find the t intercepts, let y=0 and 2 solve for t . 0=t 6t=t(t 6) t=0 and t=6 . The t coordinate of the vertex is halfway between the t 6 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 2 intercepts, that is, at t=3 . Since f (3)=3 6 3= 9 , the vertex is (3, 9) . 2 4 t (2+t)(2 t) 32. H(t)= , so for t 2 , H(t)=2+t . The domain is {t |t 2} . So the graph of H is = 2 t 2 t the same as the graph of the function f (t)=t+2 (a line) except for the hole at ( 2,4 ) . 33. g(x)= x 5 is defined when x 5 0 or x 5 , so the domain is 5, 2 y =x 5 ) . Since y= x 5 2 x=y +5 , we see that g is the top half of a parabola. 34. F(x)= 2x+1 = = { { 2x+1 (2x+1) if 2x+1 0 if 2x+1<1 1 2 1 if x< 2 if x 2x+1 2x 1 The domain is R , or ( , ). 35. 7 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function G(x)= 3x+ x x . Since x = G(x)= { { x if x 0 x if x<0 3x+x x 3x x x , we have if x>0 = if x<0 { 4x x 2x x Note that G is not defined for x=0 . The domain is ( 36. g(x)= x 2 x . Since x = { g(x)= x if x 0 x if x<0 { x x 2 = if x<0 ,0) (0, { 4 if x>0 2 if x<0 ). , we have x 2 if x>0 if x>0 = if x<0 x Note that g is not defined for x=0 . The domain is ( { 1 x 1 x ,0) (0, if x>0 if x<0 ). { x if x 0 x+1 if x>0 Domain is R , or ( , ). 37. f (x)= 8 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 38. f (x)= { 39. f (x)= { 2x+3 if x< 1 3 x if x 1 Domain is R , or ( , ). x+2 if x 1 2 if x> 1 x 2 Note that for x= 1 , both x+2 and x are equal to 1. Domain is R . 40. f (x)= { 1 if x 1 3x+2 if 1<x<1 7 2x if x 1 Domain is R . ( y y 41. Recall that the slope m of a line between the two points x ,y 1 1 2 ) and ( x ,y ) is m= x 2 2 2 1 x and an 1 equation of the line connecting those two points is y y =m(x x ) . The slope of this line segment is 1 1 6 1 7 7 7 4 = , so an equation is y 1= (x+2) . The function is f (x)= x , 2 x 4. 4 ( 2) 6 6 6 3 42. The slope of this line segment is f (x)= 3 ( 2) 5 5 = , so an equation is y+2= (x+3) . The function is 6 ( 3) 9 9 5 1 x , 3 x 6. 9 3 43. We need to solve the given equation for y . 9 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 2 2 x+ ( y 1 ) =0 ( y 1 ) = x y 1= x y=1 x . The expression with the positive radical represents the top half of the parabola, and the one with the negative radical represents the bottom x . Note that the domain is x 0 . half. Hence, we want f (x)=1 2 2 44. ( x 1 ) +y =1 0 x 2. 2 y= 1 (x 1) = 2 2 2x x . The top half is given by the function f (x)= 2x x , 45. For 1 x 2 , the graph is the line with slope 1 and y intercept 1 , that is, the line y=x+1 . For 3 3 3 2<x 4 , the graph is the line with slope and x intercept 4 , so y 0= (x 4)= x+6 . So the 2 2 2 x+1 if 1 x 2 3 function is f (x)= x+6 if 2<x 4 2 { 46. For x 0 , the graph is the line y=2 . For 0<x 1 , the graph is the line with slope 2 and y intercept 2 , that is, the line y= 2x+2 . For x>1 , the graph is the line with slope 1 and x intercept 1 , 2 if x 0 that is, the line y=1( x 1 ) =x 1 . So the function is f (x)= 2x+2 if 0<x 1 . x 1 if 1<x { 47. Let the length and width of the rectangle be L and W . Then the perimeter is 2L+2W =20 and the 20 2L area is A=LW . Solving the first equation for W in terms of L gives W = =10 L . Thus, 2 2 A(L)=L(10 L)=10L L . Since lengths are positive, the domain of A is 0<L<10 . If we further restrict L to be larger than W , then 5<L<10 would be the domain. 48. Let the length and width of the rectangle be L and W . Then the area is LW =16 , so that W =16/L . The perimeter is P=2L+2W , so P(L)=2L+2(16/L)=2L+32/L , and the domain of P is L>0 , since lengths must be positive quantities. If we further restrict L to be larger than W , then L>4 would be the domain. 49. Let the length of a side of the equilateral triangle be x . Then by the Pythagorean Theorem, the 2 2 3 1 2 2 2 1 2 3 2 x =x , so that y =x x = x and y= x . Using the height y of the triangle satisfies y + 2 4 4 2 3 3 2 1 1 formula for the area A of a triangle, A= ( base ) ( height ) , we obtain A(x)= (x) x = x , 2 2 2 4 with domain x>0 . 3 3 50. Let the volume of the cube be V and the length of an edge be L . Then V =L so L= V , and the 10 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function surface area is S(V )=6 ( 3 V ) 2=6V 2/3 , with domain V >0 . 51. Let each side of the base of the box have length x , and let the height of the box be h . Since the 2 2 2 volume is 2 , we know that 2=hx , so that h=2/x , and the surface area is S=x +4xh . Thus, 2 2 2 S(x)=x +4x(2/x )=x +(8/x) , with domain x>0 . 1 52. The area of the window is A=xh+ 2 2 2 x , where h is the height of the =xh+ 8 1 1 rectangular portion of the window. The perimeter is P=2h+x+ x=30 2h=30 x x 2 2 1 h= ( 60 2x x ) . Thus, 4 60 2x A(x)=x 4 x 1 x 2 2 x 1 2 4 2 2 2 2 2 + =15x x x + x =15x x x =15x x 8 2 4 8 8 8 +4 8 Since the lengths x and h must be positive quantities, we have x>0 and h>0 . For h>0 , we have 2h>0 1 60 60 30 x x>0 60>2x+ x x< . Hence, the domain of A is 0<x< . 2 2+ 2+ 53. The height of the box is x and the length and width are L=20 2x , W =12 2x . Then V =LWx and so ( 2 ) 3 2 V (x)= ( 20 2x ) ( 12 2x ) (x)=4(10 x)(6 x)(x)=4x 60 16x+x =4x 64x +240x . The sides L , W , and x must be positive. Thus, L>0 20 2x>0 x<10 ; W >0 12 2x>0 x>0 . Combining these restrictions gives us the domain 0<x<6 . 54. C(x)= { $2.00 2.20 2.40 2.60 2.80 3.00 3.20 3.40 3.60 3.80 4.00 x<6 ; and if 0.0<x 1.0 if 1.0<x 1.1 if 1.1<x 1.2 if 1.2<x 1.3 if 1.3<x 1.4 if 1.4<x 1.5 if 1.5<x 1.6 if 1.6<x 1.7 if 1.7<x 1.8 if 1.8<x 1.9 if 1.9<x<2.0 11 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function 55. (a) (b) On $14 , 000 , tax is assessed on $4000 , and 10%($4000)=$400 . On $26 , 000 , tax is assessed on $16 , 000 , and 10%($10 , 000)+15%($6000)=$1000+$900=$1900 . (c) As in part (b), there is $1000 tax assessed on $20,000 of income, so the graph of T is a line segment from (10 , 000,0) to (20 , 000,1000) . The tax on $30,000 is $2500, so the graph of T for x>20 , 000 is the ray with initial point (20 , 000,1000) that passes through (30 , 000,2500) . 56. One example is the amount paid for cable or telephone system repair in the home, usually measured to the nearest quarter hour. Another example is the amount paid by a student in tuition fees, if the fees vary according to the number of credits for which the student has registered. 57. f is an odd function because its graph is symmetric about the origin. g is an even function because its graph is symmetric with respect to the y axis. 58. f is not an even function since it is not symmetric with respect to the y axis. f is not an odd function since it is not symmetric about the origin. Hence, f is neither even nor odd. g is an even function because its graph is symmetric with respect to the y axis. 59. (a) Because an even function is symmetric with respect to the y axis, and the point ( 5,3) is on the graph of this even function, the point ( 5,3) must also be on its graph. (b) Because an odd function is symmetric with respect to the origin, and the point ( 5,3) is on the graph of this odd function, the point ( 5, 3) must also be on its graph. 60. (a) If f is even, we get the rest of the graph by reflecting about the y axis. 12 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function (b) If f is odd, we get the rest of the graph by rotating 180 about the origin. 61. f (x)=x 2 . 1 2 f ( x) = ( x ) = ( x) 2 1 = 2 x 2 =x = f (x) So f is an even function. 62. f (x)=x 3 . 3 f ( x) = ( x ) = = 1 x 3 = 1 ( x) 3 ( x 3) = = 1 x 3 f (x) So f is odd. 2 2 2 63. f (x)=x +x , so f ( x)= ( x ) + ( x ) =x x . Since this is neither f (x) nor f (x) , the function f is 13 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.1 Four Ways to Represent a Function neither even nor odd. 4 2 64. f (x)=x 4x . f ( x) = ( x ) 4 4 ( x ) 2 4 2 =x 4x = f (x) So f is even. 3 65. f (x)=x x . f ( x) = ( x ) 3 ( x ) = x3+x = ( x3 x ) = f (x) So f is odd. 3 2 3 2 3 2 66. f (x)=3x +2x +1 , so f ( x)=3( x) +2( x) +1= 3x +2x +1 . Since this is neither f (x) nor f (x) , the function f is neither even nor odd. 14 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions 5 1. (a) f (x)= x is a root function with n=5 . 2 (b) g(x)= 1 x is an algebraic function because it is a root of a polynomial. 9 4 (c) h(x)=x +x is a polynomial of degree 9 . 2 (d) r(x)= x +1 3 is a rational function because it is a ratio of polynomials. x +x (e) s(x)=tan 2x is a trigonometric function. (f) t(x)=log x is a logarithmic function. 10 2. (a) y= ( x 6 ) / ( x+6 ) is a rational function because it is a ratio of polynomials. 2 / (b) y=x+x x 1 is an algebraic function because it involves polynomials and roots of polynomials. x (c) y=10 is an exponential function (notice that x is the exponent ). 10 (d) y=x 6 is a power function (notice that x is the base ). 4 (e) y=2t +t is a polynomial of degree 6 . (f) y=cos +sin is a trigonometric function. 3. We notice from the figure that g and h are even functions (symmetric with respect to the y axis) and that f is an odd function (symmetric with respect to the origin). So (b) y=x g is flatter than h near the origin, we must have (c) y=x with h . 8 5 must be f . Since 2 matched with g and (a) y=x matched 4. (a) The graph of y=3x is a line (choice G ). x (b) y=3 is an exponential function (choice f ). 3 (c) y=x is an odd polynomial function or power function (choice F ). 3 1/3 (d) y= x =x is a root function (choice g ). 5. (a) An equation for the family of linear functions with slope 2 is y= f (x)=2x+b , where b is the y intercept. 1 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions (b) f (2)=1 means that the point ( 2,1 ) is on the graph of f . We can use the point slope form of a line to obtain an equation for the family of linear functions through the point ( 2,1 ) . y 1=m ( x 2 ) , which is equivalent to y=mx+ ( 1 2m ) in slope intercept form. (c) To belong to both families, an equation must have slope m=2 , so the equation in part (b), y=mx+ ( 1 2m ) , becomes y=2x 3 . It is the only function that belongs to both families. 6. All members of the family of linear functions f (x)=1+m(x+3) have graphs that are lines passing through the point ( 3,1) . 7. All members of the family of linear functions f (x)=c x have graphs that are lines with slope 1 . The y intercept is c . 8. (a) 2 Stewart Calculus ET 5e 0534393217;1. Functions and Models; 1.2 Mathematical Models: A Catalog of Essential Functions (b) The slope of 4 means that for each increase of 1 dollar for a rental space, the number of spaces rented decreases by 4 . The y intercept of 200 is the number of spaces that would be occupied if there were no charge for each space. The x intercept of 50 is the smallest rental fee that results in no spaces rented. 9. (a) 9 9 means that F increases degrees for each increase of 1 C. (Equivalently, F 5 5 increases by 9 when C increases by 5 and F decreases by 9 when C decreases by 5 .) The F intercept of 32 is the Fahrenheit temperature corresponding to a Celsius temperature of 0 . (b) The slope of 10. (a) Let d= distance traveled (in miles) and t= time elapsed (in hours). At t=0 , d=0 and at t=50 1 5 5 40 0 minutes =50 = h, d=40 . Thus we have two points: ( 0,0 ) and ,40 , so m= =48 and 60 6 6 5 0 6 so d=48t . (b) (c) Th...
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