AMS 151 Old Final Exam Solutions

Solutions to AMS 151 Final Exam, Fall 2005

1. Derivative: f'(x) = 0 from 0 to 1 and from 2 to 3, = 1 from 1 to 2.
Integral: Decreasing (slope -1) from 0 to 1, turning (concave up) from 1 to 2 so that from 2 to 3 the slope is 1.
2. T(t) = 70 + 120(.95^t). t = ln(1/6)/ln(.95).
3. 25 - 3sin(pi*t/6).
4. a) -10sin(5x), b) 9x^2 + 1/x, c) -12e^(-4x).
5. y = 3x - 1,x = 1.1 => y = 2.3
6. a) 3(2x^2 + e^(-x^2) + x)^2 * (4x - 2xe^(-x^2) + 1), b) 7^(x^x)*ln(7)*e^(xln(x)*(ln(x) + 1), c) (5/2)[ln(x/cos(x))]^(3/2)*(cos(x)/x)*[[cos(x)+xcos(x)]/[cos(x)^2]], d) dy/dx = (ye^(-x) - 2xy^2)/(2yx^2 + e^(-x)).
7. a) 1/2 (after 2 rounds of l'Hopital's Rule), b) 1.
8. dx/dt = 10/(576pi).
9. local max x = 0, local min x = 2, global max x = 0 and 3, global min x = -1 and 2, pt of inflect x = 1, concave up [1,3], concave down [-1,1].
10. r = (300/pi)^(1/3).

Solutions to AMS 151 Final Exam, Fall 2004

1. Derivative: f'(x) = 0 from 0 to 1, =-1 from 1 to 3; Integral, I(x) = x, from 0 to 1, = x -x^2/2, from 1 to 3.
2. T(t) = 70 - 70(.94)^t. t = ln(3/7)/ln(.94).
3. 3000 -1000sin(pit/6).
4. a) 1/x + 2x, b) -15e^(-5x), c) -3sin(3x).
5, y = 4x - 3, y(1.1) = 1.4.
6 a) 3(e^(-x^2) + x)^2 * (-2xe^(-x^2) + 1), b) 5[cos{x/ln(x)^3}^4*(-sin{x/ln(x)^3}*[ln(x)^3 - 3ln(x)^2]/ln(x)^6
c) e^{ln(ln(x))*x^2]*[x/ln(x) + ln(ln(x))*2x], d) y' = [ye^(-x) - 3y^3]/[9xy^2 - e^(-x)].
7. a) 0/3 = 0, b) 3/0 - unbounded.
8. 20/(150pi).
9. Max x = -sqrt(5/3), Min x = sqrt(5/3), Inflection point x = 0.
10. Local max at 3/4. Local min at 3, global max at 4, global min at -1. Points of inflection at (1/56)[42 +/- sqrt(42^2 - 4*28*9)].
11. r = sqrt[100/(3pi)].

Solutions to AMS 151 Final Exam, Fall 2002

1. Derivative: f'(x)=0, x from 0 to 1 and from 2 to 3, f = -1, x 1 to 2.
2. T(t) = 80 + (150-80)*.95^t, t = ln(2/7)/ln(.95).
3. 3000 + 2000sin(pi*t/6).
4. a) -6e^(-3x), b) 2cos(x) + 2x/3, c) 1/x + 1.
5. y = 2x+1, y(.1) = 1.2,
6. a) 5(cos(3x) + x^2)^4 *(-3sin(3x)+2x), b) [(e^x + 1)(sqrt(ln(3x)+2) - (e^x + x)((1/2)(ln(3x)+2)^(-1/2)*(1/x)]/(ln(3x)+2),
c) e^(ln(4)xln(x))*ln(4)(ln(x)+1), d) y' = [-2xy -y/x]/[2yx^2 + ln(x)].
7. a) -1/2, b) 2.
8. 6/(5pi).
9. (3-sqrt(3))100/3.
10. Local & global max at x = 1, local and global min at x = 0, concave up from 0<= x <= p and p'<= x <= 5, otherwise concave down, where p = (2-sqrt(2))/8, p' = (2+sqrt(2))/2.
11. x = sqrt(1000/3).

Solutions to AMS 151 Final Exam, Fall, 2000

1. (See webwork set5: #24,25) Derivative: f'(x) = 0, 0 <= x <= 1, f'(x) decrease from 0 to -oo, 1 < x =< 2, f'(x) = approx 1, 2 < x =< 3. Integral looks like Fig 3.14 on page 156 of text.
2. (See webwork set 2: #14,15, set3: #4,9) 70 - 60(.97)^t, t = ln(.5)/ln(.97).
3. (See webwork set 3: #30,31, set 13: 6) 110 - 70cos(pi*t/6).
4. (See set 6) a) -2sin(x) + (3/5)x^2, b) -5e^(-5x), c) ln(3x) + 1,
5. y = -2x + 3, y(0.1) = 2.8.
6. (NOTE: This type of problem will be omitted on 2002 final exam) LHS = .2*(2/1 + 2/1.2 + 2/1.4 + 2/1.6 + 2/1.8); RHS = .2*(2/1.2 + 2/1.4 + 2/1.6 + 2/1.8 + 2/2).
7. (See webwork set 7: #14-22, set 8: #16-29, for implicit diff see set 8: #34-37) a) 5(3e^x +2x)^4 (3e^x + 2), b) [(1/x + 1)(sqrt(sin(3x)+2) - (ln(x) + x)* {3cos(3x)/2sqrt(sin(3x)+2)}] / (sin(3x)+2), c) cos(x)^x^2 {2x ln(cos(x)) - x^2 tan(x)}.
8. (See set 9: #4-10) a) -1/2, b) unbounded.
9. (See webwork set 7: #25, set 8: #32, set 9: #1-3) 5/(6pi).
10. (See webwork set 10: #4, set 11: #3,4) p'(x) = -6/x^3 = 2/(5-x)^3 = 0 == 3(5-x)^3 = x^3 == 3^(1/3) (5-x) = x == x = 3^(1/3)*5/[1 + 3^(1/3)].
11. (See webwork set 9: #16-21, set 10: #1-3) f'(x) = (2-9x)(2-x)^7 = 0 == x = 2/9 (or 2), x= 2/9 is local & global max; no local min, x = 0 is global min; f"(x) = (72x - 32)(2-x)^6 = 0 == x = 4/9 ( or 2), concave down [0,4/9], concave up [4/9,0].
12. (See webwork set 10: #7-11, set 11: #5-7, set 13: #10) Cost function = 3xy + 60x +100y = 600 + 60x + 20000/x (where y = 200/x), Min at x = sqrt(1000/3), y =sqrt(120).>i