MATH 150: Answers to even numbered HW assignment problems

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1 MATH 5: Answers to even numbered HW assignment problems Section. # 4: (a) When $ is spent on advertising, the number of sales per month is 35; (b) I; (c) the number of sales if no money is spent on advertising. # : f(5) = 4.. # 2 (a) and (b): Expected return * Risk # 4: (a) At t = 3 minutes the temperature was C. (b) Initial temperature of the object was a C. At time t = b the object s temperature is C. # 8: (a) III; (b) Vertical intercept represents the temperature of potato at t =. # 2: Heart rate time Section.2 # 2: Slope is 3/2, vertical intercept is 4. #8: y = 7+3x. # 2: (a) P =3, t; (b) P () = 39, 2. (c) Population will reach 45, in about 6.82 years (t =6.82), or during the year 26.

2 Section.3 # 2: concave up. # 6: (a) intervals between D and E, between H and I; (b) intervals between A and B, between E and F ; (c) intervals between C and D, between G and H; (d) intervals between B and C, between F and G; # 8: (a) 6.4 billion $; (b) 3.28 billion $ per year; (c) 6. billion $ between 2 and 2. Section.5 # 2: (a) initial amount = ; growth; growth rate = 7 % =.7; (b) initial amount = 5.3; growth; growth rate = 5.4 % =.54; (c) initial amount = 35; decay; decay rate = - 7 % = -.7; (b) initial amount = 2; decay; decay rate = -2 % =.2. #6: (a)q =3 2t, 3 Q (grams) 5 t (days) (b) Q =3 (.88) t, 3 Q (grams) 5 t (days) 2

3 # : (a) II; (b) I; (a) III; (b) V; IV shows exponential decay; VI shows linear decay. # 8: (a) neither; (b) exponential: s(t) =3.2 (.6) t ; (c) linear: g(u) =.5u # 24: (a) The investment was worth $ after years; (b) it will take about years to get the investment back to $,. Section.6 #8: t =ln 2.3. # 22: (a) Town D is growing the fastest. (b) Town C is largest at t =. (c)townb is decreasing in size. # 24. P =2 (/ e) t 2 (.665) t. Section.8 #2: (a)f(t +) = (t +) 2 + = t 2 +2t +2, (b) f(t 2 +) = (t 2 +) 2 + = t 4 +2t 2 +2, (c) f(2) = 5, (d) 2f(t) =2t 2 +2, (e) [f(t)] 2 +=t 4 +2t #8: (a)f(g(x)) = 2x 2 +2x + 8, (b) g(f(x)) = 2x 2 +3, (c) f(f(x)) = 8x 4. # 2: (a) y = u 6,whereu =5t 2 2; (b) P =2e u,whereu =.6t; (c) C =2ln(u), where u = q 3 +. # 3: (a) f(g()) = f(2) = 3; (b) f(g()) = f(3) = 4; (c) f(g(2)) = f(5) = ; (d) g(f(2)) = g(3) = 8; (e) g(f(3)) = g(4) = 2. Section.9 #2: yes,y =3x 2, k =3,p = 2. #6: yes,y =(5/2)x /2, k =5/2, p = /2. # 4: E = kv 3,wherek is a constant. Review for Chapter # 6: (b) 2; (c) 8; (d) 56; (e) decreasing; (f) concave down. # 24: y =.4x +2. # 32: g(x) = xcould be linear; h(x) =5.6 x could be exponential; f(x) is neither. Section 2. #4: slope 3: point F ; slope : point C; slope : point E; slope /2: point A; slope : point B; slope 2: point D. 3

4 # 6: (a) The average rate of change between x =andx =3isgreater than the average rate of change between x =3andx =5sinceslopeofAB > slope of BC y B * * C A * 3 5 x (b) The function is increasing faster at x = than at x =4. Thus, instantaneous rate of change at x = is greater than that at x =4. y This slope is larger 4 x # 4: x : d b c a e f (x) : # 24: (a) f(4) >f(3); (b) f(2) f() >f(3) f(2); (c) f(2) f() 2 > f(3) f() ; 3 (d) f () >f (4). 4

5 Section 2.2 # : IV. # 4: (a) f (2) 3; (b) f (x) is positive for <x<4and is negative for 4 <x<2 Section 2.3 #4: (a)f(2) = 35 means that it costs $35 to produce 2 gallons of ice cream; (b) f (2) =.4 means that when the number of gallons produced is 2, it costs $.4 to produce an additional gallon. # 2: (a) f(4) = 2 means that a patient weighing 4 pounds should receive a dose of 2 mg; f (4) = 3 tells us that if the weight of a patient increases by pound (from 4 pounds), the dose should be increased by 3 mg; (b) f(45) = 5 mg. # 22: f(22) f(2) + f (2) 2 = = 357. Section 2.4 #2: f (x) >, f (x) >. #4: f (x) <, f (x) =. # : f (t) > on the intervals <t<.4 and.7 <t<3.4; f (t) < on the intervals.4 <t<.7 and3.4 <t<4; f (t) > on the interval <t<2.6; f (t) < on the intervals <t<and2.6 <t<4. # 2: (a) f(x) < atx 4 and x 5 ; (b) f (x) < atx 3 and x 4 ; (c) f(x) is decreasing at x 3 and x 4 ; (d) f (x) is decreasing at x 2 and x 3 ; (e) slope of f(x) is positive at x, x 2 and x 5 ; (f) slope of f(x) isincreasingat x, x 4 and x 5. Review for Chapter 2 # : f (x) x # 8: f(26) f(25)+f (25) =3.4; f(3) f(25)+f (25) 5 =

6 #2: (a) f(8) = 55: consuming 8 Calories per day results in a weight of 55 pounds; f (2) = : consuming 2 Calories per day causes neither weight gain nor loss; (b) units of dw/dc are pounds/(calories/day). # 26: B. Section: Focus on Theory (Chapter 2, pp. 35 4) # : x 2: NO, x.5: YES. # 6: YES. # 24: # 32: f (x) = lim h 5(x + h) 5x h = lim h 5=5. f (x) = lim h 2(x + h) 2 +(x + h) 2x 2 x h = lim h 4xh +2h 2 + h h Section 3. #4: y = 2x 3. # 4: y =24t 2 8t + 2. # 6: y = 2x 3 2x 2 6. # 24: y =2z 2z 2. = lim h (4x +2h +)=4x +. Section 3.2 #6: f =5 5 x ln(5) x ln(6). # 6: y = (ln ) x x. 2 # 8: D = /p. # 22: f = Ae t + B/t. # 26: f() = 4 megawatts; f(5) = 4 (.3) 5 = 53, 233 megawatts; f () = 4 ln(.3) = 273 megawatts/year; f (5) = 4 ln(.3) (.3) 5=3, 967 megawatts/year. Section 3.3 #6: w =5 (5r 6) 2. 6

7 # 6: f =3E 5x 2xe x2. # 22: f = 2t t 2 +. # 3: y =2(5+e x )e x. # 38: f(4) = 27e.4 (4) =5.4ng/ml; f (4) = 3.78e.4 (4) = 2.6 ng/ml per hour. Section 3.4 #8: y = e t (t 2 +2t +3). # 6: f = e z 2 z ze z. # 26: # 28: y (x) = e x ( + e x ) 2. y (z) = z ln z z z(ln z) 2. Section 3.5 #6: y =5cosx 5. #8: R =5cos(5t). # 4: y =2cos(2t) 4sin(4t). Review for Chapter 3 #8: s =5t 2 2t + 2. # 2: R =5(sint) 4 (cos t). # 36: h = 2p (3 + 2p 2 ) 2. Section 4. #8: f (x) =3x 2 6, and thus, f (x) =atx = ± 2. f (x) changes from positive to negative at x = 2, and so there is a local maximum at x = 2. f (x) changes from negative to positive at x =+ 2, and so there is a local minimum at x =+ 2. 7

8 # 2: f (x) =lnx +, so f (x) =whenlnx =, that is, for x = e.37. This is the only place where f changes sign. Since f () >, the function f increases for <x<e and increases for x>e.thus,we have local minimum at x = e. Section 4.2 # 6: critical points at x = (local min) and at x = (neither min nor max); inflection points at x =andatx =2/3. Section 4.3 # 22: Local and global maximum at x =2. # 3: de df =.25 2 (.7) F 3 =, thus, ( ) /3 2 (.7) F = =2.4 hours..25 This gives a local and global minimum. # 44: (a) At t =wehaveq() = ; (b) Maximum value occurs at t =ln2=.69 (where q (t) = ); maximum value is q(ln 2) = 5 mg; (c) as t we have q(t). Section 4.4 # 2: Profit function is positive for 5.5 <q<2.5 (whenr(q) >C(q)), and negative for <q<5.5, q>2.5(whenr(q) <C(q)). Profit maximized when R(q) >C(q) andr (q) =C (q) which occurs at about q =9.5. # 6: Profit is maximized at q = 75; profit = R(75) C(75) = $ Section 4.8 # 6: (a) At peak concentration C (t) = ; corresponding t =/.3 = 33.3 minutes;c(33.3) 245 ng/ml; (b) after 5 min C(5) 9 ng/ml; after one hour C(6) 98 ng/ml. Review for Chapter 4 # 8: (a) Increasing for all x. (b) No max or min. # : (a) Decreasing for x< ; <x<; increasing for x>; < x<. (b) f( ) and f() are local minima; f() is a local maximum. 8

9 Section 7. #6: t 8 /8+t 4 /4+C. # 4: # 8: ln z + C. # 46: # 5: 2x 4 +ln x + C. 2 3 z3/2 + C. e 2t 2 + C. Section 7.2 #4: e 5t+2 + C. # 2: 6 (x2 +3) 3 + C. # 2: 8 (cos θ +5)8 + C. # 34: # 36: 2e y + C. # 38: ln(2 + e x )+C. 2 ln(y2 +4)+C. Section 7.3 #6: # : # 22: 4 x dx =2. 5t 3 dt = x(x 2 +) 2 dx = Section 5.3 #2: Area=8. 9

10 y=x x # 4: Negative. #6: Zero. # 6: (a) 3; (b) 2; (c) ; (d) 5. Section 5.4 # 4: The change in velocity between times t =andt = 6 hours; it is measured in km/hr. # 2: 2627 acres. Section 5.5 # 6: The fixed cost is $5. Total variable cost is $ Total cost is $5 + $866.7 = $, Review for Chapter 5 # : 5 (x 2 +)dx = # 4: 3 ( ) x x ( ) z 2 (z +/z)dz = 2 +ln(z) # 32: Since f(x) = +x 3 is increasing for x 2, we have so that 2= f()dx dx f(x)dx f(x)dx # 36: Total cost is 4,25, riyals. f(2)dx, 3dx =6.

11 Section 6. #2: Average value = 3 f(x)dx =8. 3 # 4: Average value = 2. # 6: Average value # 2(a): Average inventory # 22: (c) < (a) < (b). Section 7.4 #6: F()= F()= x # 6: F () =, F (3) = 7, F () =, F (4) = 5. Review for Chapter 7 #6: ln x x 2x + C. 2 # 2: 3 sin(x)+7cos(x)+c. # 36: x2 +4+C. Section. # 2: (a) (III); (b) (V); (c) (I); (d) (II); (e) (IV). # 8: Rate of change of B =Ratein Rate out: db dt =.4B 2. Section.2 # 6: E.

12 # 8: A. # 22: (a) II; (b) I. Section.4 #2: w =3e 3r. #4: Q =5e t/5. #6: p = 64.87e.q. Section.5 #4: P = 4e t 4. #6: Q = 4 35e.3 t. 2