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This lecture is from AP Calculus. Key important points are: Interval Review, Work Excercise, Revison, Concave Upward, Concave Downward
Typology: Exams
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Calculus 3rd Quarter Exam Review Name___________________
x 1 f x x 4
2
x f x
x 2
2
3
x 3x f x x 9x
Horizontal asymptote Vertical asymptote(s)
A) y = 0 x = 0 and x = 3
B) y = 0 x = 0, x = − 3, and x = 3
C) y = 0 x = 3
D) y = 1 x = 0 and x = 3
E) y = 1 x = 3
2 x 3x 1 f x x 2
Oblique asymptote Vertical asymptote(s)
A) y = x − 1 x = 2
B) y = x + 5 x = − 2
C) y = x − 1 none
D) y = x + 1 x = − 2
E) y = x + 5 x = 2
D) not enough information E) no relative extrema
A) 4 ft by 9 ft B) 3 ft by 12 ft C) 6 ft by 6 ft
D) 1 ft by 36 ft E) 4ft by 12ft
second number.
the following dimensions should be used so that the enclosed area will be a maximum?
A) 45’ by 30’ B) 40’ by 30’ C) 50’ by 15’ D) 60’ by 20’ E) 1’ by 180’
3 f x = 2 x on the interval (^) [ −1,1]
A) f is not continuous at x = 0 B) f is not differentiable at x = 0 C) f is not continuous at x = 1
D) f is not differentiable at x = 1 E) Mean Value Theorem does apply
2 f x = x^3 on the interval [ 0,1]^.
A) f is not continuous at x = 0 B) f is not differentiable at x = 0 C) f is not continuous at x = 1
D) f is not differentiable at x = 1 E) Mean Value Theorem does apply
2 sec x − sin x −2x dx
A) (^) tan x − cos x − x + C B)
2 tan x − cos x − x + C C)^
2 tan x − cos x − 2x +C
D) tan x + cos x − x + C E)
2 tan x + cos x − x +C
2 tan x +1 dx
A) sec x + x + C B) sec x tan x + C C) tan x + C D)
2 sec x + C E) sec x +C
1 2
0
x − 2x +1 dx
2
1 2
x dx x
6
2
3 dx
6
0
6 −2x dx
(^9 2 )
0
3x x dx 3
9 2
0
6x −x dx
(^6 2 )
0
3x x dx 3
6 2
0
6x −x dx
2 (^9) y = 6x −x
6
7
1
2x + 1 dx
7 2
1
x +x dx
3
0
2x + 1 dx
7 2
1
x +x dx
3
0
2 dx
3x
2
x
3
dx
3
3 (^3 2 ) x 1 C 2
3 x + 1 +C
x 1 C 2
x 1 C 3
dy (^2) x cos 3x , then y dx
sin 3x C 6
sin 3x C 6
cos 3x C 6
2
2 cos 3x +C
2 x f (x) , 2
′ = where f (0) = 0 then 3f (4)=
y = 2x + 1 7
3
1
4
Answers to Calculus 3
rd quarter exam review: