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Math 112 (Calculus I)
Final Exam Form A KEY
Multiple Choice. Fill in the answer to each problem on your scantron. Make sure your name, section and instructor is on your scantron.
- Find the absolute minimum value for f (x) on the interval [− 4 , 3] when f (x) is given by
f (x) = x^2 − 4 x^2 + 4
a) − 1 b) 5 / 13 c) 4 / 5
d) − 2 e) 2 f) 1
Solution: a)
- For what value of c is the function g(x) below continuous?
g(x) =
cx^2 − 4 c x − 2 if x < 2 cx + 1 if x ≥ 2
a) c = 1 b) c = 2 c) c = 1/ 2
d) c = − 2 e) c = 0 f) None of the above
Solution: c)
- Evaluate: d dx
∫ (^) x 2
3
ln(t − 1) dt
a) 2 x ln(x^2 − 1) b)
x^2 − 1
c) ln(x^2 − 1) − ln(8)
d) 2 x ln(x^2 − 1) − 2 x ln(8) e) 2 x ln((x^2 − 1)^4 (x^2 − 1)) f) 2 x x^2 − 1
x 4
Solution: a)
- Use linear approximation to estimate sinh(0.1).
a) 0. 2 b) 0. 01 c) − 0. 2
d) 1. 1 e) 0. 9 f) 0. 1
Solution: f)
- A function s(t) is given by s(t) = t^1 /^3 + t^2 /^3. Find s′′(1).
a) 1 b) − 1 / 3 c) 0
d) − 4 / 9 e) 1 / 3 f) − 1
Solution: d)
- Find all vertical asymptotes.
y = 3(x − 2)(x + 1) ln(x + 5) (x + 4)(x − 2) i. x = − 4 ii. x = 2 iii. x = − 5 iv. x = 1
a) i. only b) i. and ii. c) i., ii., and iii.
d) i., ii., iii., and iv. e) i. and iii. only f) ii. and iii. only
Solution: e)
- For the definition of the limit
xlim→ 2 3 x^ = 6, find the largest value of δ that corresponds to = 0.06. a) 0. 06 b) 0. 02 c) 0. 2
d) 0. 1 e) 0. 03 f) 0. 01
Solution: b)
Short Answer. Fill in the blank with the appropriate answer.
- A rectangular box with a square base is to hold 4000 cubic centimeters. Material for the sides costs 2 cents per square centimeter, while material for the top and bottom costs 1 cent per square centimeter. (a) (4 points) If the base has side x, express the cost C of the box as a function of x. Solution: Form A x^2 y = 4000, y =
x^2 C = 2 · 4 xy + 1 · 2 x^2 =
x
- 2x^2. Form B x^2 y = 9000, y =
x^2 C = 6 · 4 xy + 2 · 2 x^2 =
x
- 4x^2. (b) (4 points) Find the dimensions of the most economical box. Solution: Form A C′^ = −
x^2
Thus, x^3 =
= 8000, x = 20, y = 10. Form B C′^ = −
x^2
Thus, x^3 =
= 27000, x = 30, y = 10
- (7 points) Find the equation of the tangent line to the curve given implicitly by
exy^ = x^2 + y^2
at the point (0, 1). Solution: By differentiating implicitly, we have
exy(y + xy′) = 2x + 2yy′
Plugging in x = 0 and y = 1, we have
e^0 (1) = 2y′, y′^ =
- (7 points) A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 2 m/s, at what rate is the angle between the rope and the horizontal increasing when that angle is π/6? Solution: Form A If s is the length of the rope and θ is the angle the rope makes from the horizontal,
s sin θ = 1,
and sin θ ds dt
When θ = π/6, s = 2. Thus,
dθ dt
or dθ dt
√^1
Form B When θ = π/4, s =
- Thus,
dθ dt
or dθ dt
√^3
- (7 points) Find the derivative of (arctan(x))^3 x. Don’t simplify your answer.
Solution:
Form A ln y = 3x ln(arctan(x)) 1 y
y′^ = 3 ln(arctan(x)) + 3 x arctan(x)(1 + x^2 )
y′^ = (arctan(x))^3 x
3 ln(arctan(x)) + 3 x arctan(x)(1 + x^2 )
Form B y′^ = (arctan(x))^4 x
4 ln(arctan(x)) + 4 x arctan(x)(1 + x^2 )
- Evaluate the integrals.
(a) (5 points)
0
v^2 cos(v^3 ) dv Solution: u = v^3 , du = 3v^2 dv. ∫ (^1)
0
v^2 cos(v^3 ) dv =
0
cos(u) du 3
(sin(1)).
(b) (5 points)
sec θ tan θ 1 + sec θ
dθ Solution: Form A u = 1 + sec θ, du = sec θ tan θ dθ. ∫ sec θ tan θ 1 + sec θ dθ =
du u = ln |u| + C
= ln |1 + sec θ| + C. Form B ln |3 + sec θ| + C.
(c) (5 points)
0
|x^2 − 4 | dx Solution: (^) ∫ (^3)
0
|x^2 − 4 | dx =
0
4 − x^2 dx +
2
x^2 − 4 dx
= (4x − x^3 3
)|^20 + (
x^3 2
− 4 x)|^42 = 8 −
END OF EXAM
