MA119-A Applied Calculus for Business Fall Homework 4 Solutions Due 9/29/ :30AM

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1 MA9-A Applied Calculus for Business 006 Fall Homework Solutions Due 9/9/006 0:0AM. #0 Find te it #8 Find te it. #6 Find te it = (0) 5 0 (0) + (0) + =.!! r + +. r s r + + = () + 0 () + = 5 =. by given f () = and g () =. p 5f () + g (). #5 Find te it p q 5f () + g () = 5 f () + g () = p 5 + =. = ( ) = (0) =.

2 . #56 Find te it According to te grap, +! y te it does not eist (DNE).. #7 Find te it Since! + =! te it does not eist (DNE).. #76 Find te it 6 8 0! + + =!! + +!! +! = 0,

3 Te it is + + =!!. #78 Find te it Te it is! = + + =! + +!!!!!! =! + +!! = !! (0) + (0) = =.. #8 Concentration of a Drug in te Bloodstream Te concentration of a certain drug in a patient s bloodstream t r after injection is given by C (t) = 0:t t + mg/cm. Evaluate C (t) and interpret your result.! We ave C (t) =!! 0:t t + =! 0: t + t = = 0. So, te concentration of a certain drug in a patient s bloodstream goes to zero wen time goes in nity. Tis means tat te drug will eventually disappear from te patient s blood..5 # Find te one-side it Te it is.5 # Find te one-side it +! + + +! + + = + + =. + p 5 +! 5 + = 0.

4 p Since we approac 5 from te rigt side (all more ten 5), we are able to calculate 5 +. Tus, te it is + p 5 + = ( 5) + p 5 + ( 5) = 5.!5 +.5 #8 Find te one-side its f () and f (), were + + if 0 f () = + if > 0 We ave and f () = ( + ) = (0) + = + + f () = ( + ) = (0) + =. +.5 #5 Find te values of for wic te function f () = + If + = 0, ten f is not de ned. Oterwise, + continuous ecept = or. ( + = ( + ) ( ).) is continuous. is a real number. So, f is j j.5 #56 Find te values of for wic te function f () = is continuous. Note tat j j is continuous everytere (even = ) and is continuous everywere. So, f () = j j is continuous everywere ecept = 0, or, =..5 #58 Determine all values of at wic te function f () = is discontinuous. ( )( ) Since f is not de ned wen = or, f is not continuous tere..5 #7 Let For 6= and, ( ) ( ) is continuous. Tus, f () = ( )( ) ecept = or. So, f is discontinuous only at = or. f () = + if 6= k if =. is continuous For wat value of k will f be continuous on ( ; )? If a 6=, f (a) is de ned and f () = f (a). Tus, f is continuous at = a. Since f () =!! + = ( + ) ( ) =! +! =, we know tat te it eists at =.

5 If f is continuous at =, f () must be equal to f () = k. Tus, k =!. Terefore, if k =, ten f is continuous on ( ; )..5 #78 Let f () = 6 +. Also, let a = 0 and b =. (a) Sow tat te function f is continuous for all values of in te interval [a; b]. (b) Prove tat f must ave at least one zero in te interval (a; b) by sowing tat f (a) and f (b) ave opposite signs. (a) Since f is a polynomial, f is continuous in [0; ]. (b) Note tat f (0) = and f () =. Since f () = < 0 < = f (0), by te Intermediate Value Teorem, tere is a real number c (0; ) suc tat f (c) = 0. So, f must ave at least one zero in te interval (a; b)..5 #8 Use te Intermediate Value Teorem to nd te value of c suc tat f (c) = M, were f () = + on [ ; ] and M = 7. We ave tat f ( ) = and f () =. Since f ( ) = < M < = f (), by te Intermediate Value Teorem, tere is a real number c ( ; ) suc tat f (c) = M = 7. Tus, we ave 7 = f (c) = c c +, tat is, c c 6 = 0. Tis tells us tat c = or (since 0 = c c 6 = (c ) (c + ).).6 # Use te four-step process to nd te slope of te tangent line of te grap of te function 5 f () = at any point. By te four-step process, () f ( + ) = ( + ) = ( + + ) = () f ( + ) f () = = f(+) f() () = =. () f 0 f(+) f() () = Tus, te slope of te tangent line is. = =....6 #6 Use te four-step process to nd te slope of te tangent line of te grap of te function f () = + 5 at any point. By te four-step process, () f ( + ) = ( + ) +5 ( + ) = ( + + )+5 ( + ) = () f ( + ) f () = ( ) ( + 5) = f(+) f() = +5+ () () f 0 f(+) f() () = = = ( ) = + 5.

6 6 Tus, te slope of te tangent line is # Find te slope of te tangent line of te grap of te function f () = at te point ; and determine an equation of te tangent line. By te four-step process at =, () f ( + ) =. (+) () f ( + ) f () = () (+) f(+) f() = (+) =. (+) () = (+) () f 0 f(+) f() () = = =. (+) Tus, te slope of te tangent line at te point ; is of te tangent line is = (+) =. (+) (+). Terefore, te equation y = ( )..6 #6 Let f () =. (a) Find te derivative f 0 of f. (b) Find an equation of te tangent line to te curve at te point ;. (c) Sketc te grap of f. (a) By te four-step process, () f ( + ) = = (+) () f ( + ) f () = +. + f(+) f() (+ )( ) () = = (+. )( ) () f 0 f(+) f() () = = (b) Te slope at = is f 0 () = = ( ) (+ ) =. (+ )( ) (+ )( ) (+ )( ) = (( ) ) is y = ( ( )), or, + y + = 0. (c) Te grap of f is = =. ( )( ) ( ). Tus, te equation of te tangent

7 7 y #0 Velocity of a Ball Trown into te Air A ball is trown straigt up wit an initial velocity of 8 ft/sec, so tat its eigt (in feet) after t sec is given by s (t) = 8t 6t. (a) Wat is te average velocity of te ball over te time intervals [; ], [; :5], and [; :]. (b) Wat is te instantaneous velocity at time t =? (c) Wat is te instantaneous velocity at time t = 5? Is te ball rising or falling at tis time? (d) Wen will te ball it te ground? (a) First, we ave s () = 8 () 6 () = 9, s () = 0, s (:5) = 0, and s (:) = 98:. So, te average velocity over [; ] is f() f() 0 9 = = 8, te average velocity over [; :5] is f(:5) f() 0 9 = = 56, and te average velocity over [; :] is :5 :5 f(:) f() = : 98: 9 : = 6:. (b) s 0 () = s(+) s() = 8 () = 6. (c) s 0 s(5+) s(5) (5) = = 8 (5) =. So, te ball is falling. (d) Te ground is eigt 0. So, let s (t) = 0. We can solve t from 8t 6t = 0. Tis tells us tat t = 0 or 8. Terefore, we know tat after 8 seconds, te ball its te ground..6 # Cost of Producing Surfboards Te total cost C () (in dollars) incurred by Aloa Company in manufacturing surfboards a day is given by C () = (0 5) (a) Find C 0 (). (b) Wat is te rate of cange of te total cost wen te level of production is ten turfboards a day?

8 8 (a) By te four-step process, () C ( + ) = 0 ( + ) + 00 ( + ) + 0 = () C ( + ) C () = ( ) ( ) = C(+) C() () = = () C 0 C(+) C() () = = ( ) = (b) C 0 (0) = 0 (0) + 00 = #8 Te grap of a function is sown at page 5. Weter or not (a) f () as a it at = a, (b) f () is continuous at = a, (c) f () is di erentiable at = a. Justify your answer. (a) By te grap, te rigt-side it at = a is positive and te left-side it at = a is negative. Tey cannot be equal. Tus, f does not ave a it at = a. (b) Since f does not ave a it at = a, f is not continuous at = a. (c) Since f is not continuous at = a, f is not di erentiable at = a..6 #50 Te grap of a function is sown at page 5. Weter or not (a) f () as a it at = a, (b) f () is continuous at = a, (c) f () is di erentiable at = a. Justify your answer. (a) By te grap, te rigt-side it at = a is in nity and te left-side it at = a is negative in nity. Tey cannot be equal. Tus, f does not ave a it at = a. (b) Since f does not ave a it at = a, f is not continuous at = a. (c) Since f is not continuous at = a, f is not di erentiable at = a..6 #58 Sketc te grap of te function f () =. Is te function continuous at = 0? Does f 0 (0) eist? Wy or wy not? Te grap of f is

9 9 y 5 5 From te grap, we can see tat f () = 0 = f (0). Tus, f is continuous at = 0. f() f(0) For te derivative, we need to calculate. 0 First, let us see te left-side it p < 0. So te it is. f () = 0 and f () = 0. So, we ave + f() f(0) = + = p. Wen < 0, f() f(0) Similarly, te rigt-side it = = + p +. Wen > 0, p > 0. So te it is. f() f(0) Terefore, does not eists. Tis tells us tat f 0 (0) does not eist. 0