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These are the notes of Exam of Calculus which includes Traditional Problems, Symmetric Matrix, Property, Conditions, Constants, Matrix, Positive, Anti Symmetric etc. Key important points are: Undefined, Value, Limit, Critical Point, Function, Rock Shot Vertically, Moon, Velocity, Inflection Points, Absolute Minimum
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Math 1205 Common Final Exam Version A Spring 2006
Name: ID #
CRN: Pledge:
Instructions: please enter your NAME, ID NUMBER, FORM DESIGNATION, and your CRN on the op-scan sheet. The CRN should be written in the box labeled ‘COURSE’. Darken the appropriate circles below your ID number and below the Form designation letter. Use a number 2 pencil. Machine grading may ignore faintly marked circles.
Mark your answers to the test questions in row s 1 through 17 of the op-scan sheet. Your score on this test will be the number of correct answers. You have one hour to complete this portion of the exam.
C. Limit does not exist because f(-4) is undefined.
D. Limit does not exist because f is an oscillating function.
E. Limit does not exist because f does not approach the same value from both sides of x=a.
2. Determine the value of lim x → 1 4 x^ −^4 x^2 + x − 2
E. Limit does not.
3. Find the critical point(s) of f ( x ) = 2 x x^2 + 1
3 .
A. x = 0 B. x = 0 , 1
C. x = 0 , 2 , − 2 D. x = 0 , 1 , − 1
E. There are no critical points of this function.
-4.0 undefined -3.9999 -3. -3.999 -3. -3.99 -3.
4. A rock shot vertically upward from the surface of the moon at a velocity of 8 (^) sec m reaches a height of
s ( t ) = 8 t − 0.8 t^2 meters in t seconds. How high does the rock go?
A. 20 m B. 5 m
C. 50m D. 3.8 m
E. 36 m
5. If f ( x ) = sin x − cos x + x , then f(x) has inflection points at which of the following values on the interval [ 0 , 2 π ].
A. x = π 4
, 3 π 4
B. x = π 4
,^5 π 4
C. x = 5 π 4
, 7 π 4
D. x = 3 π 4
,^7 π 4
E. x = π 4
, 3 π 4
, 5 π 4
, 7 π 4
6. The absolute minimum of f ( x ) = x^3 − 12 x on [0,4] occurs at:
A. x=0 B. x=-
C. x=2 D. x=
E. x=
7. If f ( x ) = 3 x
(^2) − 12 x x^2 − 16
, which of the following is FALSE.
A. lim x → 1 f ( x ) = 3 5
B. lim x → 4 f ( x ) does not exist
C. (^) x lim→−∞ f ( x ) = 3 D. f(x) has a vertical asymptote at x=-
E. f(x) has a horizontal asymptote at y=
C. y ′ = sin(3 x ) cos^2 (3 x ) D. y ′= − 3 sin( 3 x ) cos(cos( 3 x ))
E. y ′ = − 3 sin x sin( 3 x ) + cos x cos( 3 x )
13. For f ( x ) = 2 x find all of the possible values on the interval [0,9] that satisfy the conclusion of the Mean Value Theorem?
A. x = 2 3
B. x = 4 9
C. x = 9 4
D. x = 3 2
E. The Mean Value Theorem does not apply to f(x) on this interval.
`
14. The following three graphs, labeled a, b, and c, are the graphs of the function and its first and second derivatives. Match the appropriate letter to f ( x ), f (^) ′( x ), f (^) ′′( x ).
A. a = f ( x ), b = f (^) ′( x ), c = f (^) ′′( x )
B. b = f ( x ), a = f (^) ′( x ), c = f (^) ′′( x )
C. b = f ( x ), c = f (^) ′( x ), a = f (^) ′′( x )
D. c = f ( x ), b = f (^) ′( x ), a = f (^) ′′( x )
E. c = f ( x ), a = f (^) ′( x ), b = f (^) ′′( x )
c a
b
15. An oil tank in the shape of a right circular cylinder of radius 8m is being filled at a constant rate of
10 m^
3 min
. How fast is the level of the oil rising? Recall: V = π r^2 h
A. Oil is rising at a rate of (^64) π m min
B. Oil is rising at a rate of 32 π 5
m min
C. Oil is rising at a rate of 5 32 π
m min
D. Oil is rising at a rate of 5 8 π
m min
E. Oil is rising at a rate of 1 64 π
m min
16. The Taylor Polynomial of degree three generated by (^) f ( x ) = 1 x
and centered at a=2 is:
A. (^) P 3 ( x ) = (^1) x − (^) x^1 2 ( x − 2) + (^) x^1 3 ( x − 2)^2 − (^) x^1 4 ( x − 2 )^3
B. (^) P 3 ( x ) = 12 − 14 ( x − 2) + 14 ( x − 2 )^2 − 83 ( x − 2 )^3
C. P 3 ( x ) = 12 − 14 ( x − 2) + ( x − 2 )^2 − ( x − 2 )^3
D. (^) P 3 ( x ) = 12 − 14 ( x − 2) + 12 ( x − 2 )^2 − 16 ( x − 2 )^3
E. (^) P 3 ( x ) = 12 − 14 ( x − 2) + 18 ( x − 2 )^2 − 161 ( x − 2 )^3
17. Determine g ′ ′( x ) if (^) g ( x ) = 12 x^3 − x + 1 x^2
A. (^) g "( x ) = 23 x^2 − 12 x −^2 1 + 1 x
B. (^) g "( x ) = 3 x + 14 x −^23 + 6 x^4
C. (^) g "( x ) = 23 x^2 − 12 x −^2 1 − 2 x
D. (^) g "( x ) = 3 x + 14 x −^23 + 1
E. (^) g "( x ) = 3 x + 14 x −^23 + (^12)