4/16/2015
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Practice Exam # 3 (3.10­4.7) (5680271)
Due:
Thu Apr 23 2015 11:59 PM PDT
Question
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Question Details
SCalcET7 3.11.023. [1644808]
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Use the definitions of the hyperbolic functions to find each of the following limits.
(a) lim tanh x
x → ∞ (b) lim tanh x
x → −∞ (c) lim sinh x
x → ∞ (d) lim sinh x
x → −∞ (e) lim sech x
x → ∞ (f) lim coth x
x → ∞ (g) lim coth x
x → 0+ (h) lim coth x
x → 0− (i) lim csch x
x → −∞ Solution or Explanation
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SCalcET7 4.1.041. [1639563]
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Find the critical numbers of the function. (Enter your answers as a comma­separated list. Use n to denote any arbitrary
integer values. If an answer does not exist, enter DNE.)
f(θ) = 18 cos θ + 9 sin2θ
θ =
Solution or Explanation
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3.
Question Details
SCalcET7 4.1.055. [1886712]
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Find the absolute maximum and absolute minimum values of f on the given interval.
f(t) = t
64 − t2 , [−1, 8]
absolute minimum value absolute maximum value Solution or Explanation
f(t) = t
64 − t2 , [−1, 8].
f '(t) = t · −t2
1
(64 − t2)−1/2(−2t) + (64 − t2)1/2 · 1 = 2
f '(t) = 0 64 − 2t2 = 0 t2 = 32 f '(t) does not exist if 64 − t2 = 0 f(−1) = −3
7 , f(
64 − t2
t = ±
+ 64 − t2 = 32 , but t = −
−t2 + (64 − t2)
64 − t2
= 64 − 2t2
64 − t2
.
32 is not in the given interval, [−1, 8].
t = ±8, but −8 is not in the given interval. 32 ) = 32, and f(8) = 0. So f(
32 ) = 32 is the absolute maximum value and f(−1) = −3
7 is the
absolute minimum value.
4.
Question Details
SCalcET7 4.2.011. [1912771]
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Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f(x) = ln x, [1, 3]
Yes, it does not matter if f is continuous or differentiable, every function satisfies the Mean Value Theorem.
Yes, f is continuous on [1, 3] and differentiable on (1, 3). No, f is not continuous on [1, 3].
No, f is continuous on [1, 3] but not differentiable on (1, 3).
There is not enough information to verify if this function satisfies the Mean Value Theorem.
If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your
answers as a comma­separated list. If it does not satisfy the hypotheses, enter DNE).
c = Solution or Explanation
f(x) = ln x, [1, 3]. f is continuous and differentiable on (0, ∞), so f is continuous on [1, 3] and differentiable on (1, 3). f(b) − f(a)
1
f(3) − f(1)
ln 3 − 0
ln 3
2
f '(c) = ⇔ = = = ⇔ c = , which is in (1, 3).
b − a
c
3 − 1
2
2
ln 3
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SCalcET7 4.3.025. [1777011]
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Sketch the graph of a function that satisfies all of the given conditions.
f '(0) = f '(2) = f '(4) = 0,
f '(x) > 0 if x < 0 or 2 < x < 4,
f '(x) < 0 if 0 < x < 2 or x > 4,
f ''(x) > 0 if 1 < x < 3,
f ''(x) < 0 if x < 1 or x > 3
Solution or Explanation
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f '(0) = f '(2) = f '(4) = 0 horizontal tangents at x = 0, 2, 4.
f '(x) > 0 if x < 0 or 2 < x < 4 f is increasing on (−∞, 0) and (2, 4).
f '(x) < 0 if 0 < x < 2 or x > 4 f is decreasing on (0, 2) and (4, ∞).
f ''(x) > 0 if 1 < x < 3 f is concave upward on (1, 3).
f ''(x) < 0 if x < 1 or x > 3 f is concave downward on (−∞, 1) and (3, ∞). there are inflection points where x = 1 and
3.
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SCalcET7 4.3.052. [1774965]
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Consider the function below.
f(x) = earctan 4x
(a) Find the vertical asymptote(s). (Enter your answers as a comma­separated list. If an answer does not exist,
enter DNE.)
x = Find the horizontal asymptote(s). (Enter your answers as a comma­separated list. If an answer does not exist,
enter DNE.)
y = (b) Find the interval where the function is increasing. (Enter your answer using interval notation.)
(c) Find the local maximum and minimum values. (If an answer does not exist, enter DNE.)
local maximum value local minimum value (d) Find the interval where the function is concave up. (Enter your answer using interval notation.)
Find the interval where the function is concave down. (Enter your answer using interval notation.)
Find the inflection point.
(x, y) = Solution or Explanation
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7.
Question Details
SCalcET7 4.4.017.MI. [1677948]
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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x → ∞ ln 3x
3x
Solution or Explanation
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SCalcET7 4.4.011. [1775011]
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Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it.
cos x
lim x → (π/2)+ 1 − sin x
Solution or Explanation
This limit has the form 9.
0
cos x
. lim 0
x → (π/2)+ 1 − sin x
lim −sin x
x → (π/2)+ −cos x
= lim x → (π/2)+ tan x = −∞.
Question Details
SCalcET7 4.4.044. [1641353]
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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim sin x ln 3x
x → 0+ Solution or Explanation
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Question Details
SCalcET7 4.4.053.MI. [1643905]
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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim (9x − ln x)
x → ∞ Solution or Explanation
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11.
Question Details
SCalcET7 4.4.063. [1836302]
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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim (5x + 1)cot x
x → 0+ Solution or Explanation
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SCalcET7 4.5.012. [1641321]
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SCalcET7 4.5.024. [1641082]
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Use the guidelines of this section to sketch the curve.
x
y = 2
x − 16
Solution or Explanation
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13.
Question Details
Use the guidelines of this section to sketch the curve.
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y = x2 + 4x − x
Solution or Explanation
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Question Details
SCalcET7 4.5.031. [1641208]
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SCalcET7 4.5.045. [1640965]
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Use the guidelines of this section to sketch the curve.
y = 3
x2 − 9
Solution or Explanation
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15.
Question Details
Use the guidelines of this section to sketch the curve.
y = x − ln x
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Solution or Explanation
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SCalcET7 4.7.016. [1642180]
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A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width.
Material for the base costs $15 per square meter. Material for the sides costs $9 per square meter. Find the cost of
materials for the cheapest such container. (Round your answer to the nearest cent.)
$ 245.31 Solution or Explanation
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17.
Question Details
SCalcET7 4.7.032.MI. [1642311]
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A Norman window has the shape of a rectangle surmounted by a semicircle. (Thus the diameter of the semicircle is equal
to the width of the rectangle. See the figure below.) If the perimeter of the window is 24 ft, find the value of x so that the
greatest possible amount of light is admitted.
x = ft
Solution or Explanation
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SCalcET7 4.3.016. [1777042]
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Consider the equation below.
f(x) = x7 ln x
(a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum value of f. (c) Find the inflection point. (x, y) = Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.) Solution or Explanation
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19.
Question Details
SCalcET7 4.5.001. [1775032]
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Use the guidelines of this section to sketch the curve.
y = x3 − 12x2 + 36x
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Solution or Explanation
y = f(x) = x3 − 12x2 + 36x = x(x2 − 12x + 36) = x(x − 6)2
A. f is a polynomial, so D = .
B. x­intercepts are 0 and 6, y­intercepts = f(0) = 0
C. No symmetry
D. No asymptote
E. f '(x) = 3x2 − 24x + 36 = 3(x2 − 8x + 12) = 3(x − 2)(x − 6) < 0 ⇔ 2 < x < 6, so f is decreasing on (2, 6) and
increasing on (−∞, 2) and (6, ∞).
F. Local maximum value f(2) = 32, local minimum value f(6) = 0
G. f ''(x) = 6x − 24 = 6(x − 4) > 0 ⇔ x > 4, so f is CU on (4, ∞) and CD on (−∞, 4). IP at (4, 16)
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SCalcET7 4.7.008. [1641629]
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Find the dimensions of a rectangle with area 1,728 m2 whose perimeter is as small as possible. (If both values are the
same number, enter it into both blanks.)
m (smaller value)
m (larger value)
Solution or Explanation
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Name (AID): Practice Exam # 3 (3.10­4.7) (5680271)
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