# Calculus

Please check my answers! My teacher never gave us the answer sheet so I have no idea what to study TT
(also please give the correct answers to the ones I got wrong ^0^)

1. lim as x-> inf. of (9x-x^2-7x^4)/(x^3+12x)
9
-7/12
-7
DNE <----

2. If f(x) = cos x, then f'(pi/2) =
-1 <----
1
1/2
-(sqrt 3)/(2)

3. If y^5 + (3x^2)(y^2) + 5x^4 = 49, then dy/dx at the point (-1,2) is:
-1 <----
11/23
-23/11
-10/3

4. The graph of y=g(x) is shown below. At which point is g''<g'<g?
(gyazo.com/8100a6237282d58c81f7a6b760daf32c)
A
B <----
C
D

5. The maximum value of f(x) = -e^x + x is:
0 <----
-1
1
ln 2

6. An equation of the line tangent to the graph of y= (5x - 1)/(3x+1)
7x + y = 0
9x + 7y = 53
2x - 3y = 13
X - 2y = -1 <----

7. The graph of y = g(x) is shown below, where the curve below the axis is a semi circle. The value of ∫ (0 on bottom, 8 on top) g(x)dx =
(gyazo.com/05b57b20ed35de45d08716c9e72f5605)
6 - 2pi
4pi - 15 <----
15/2 - 2pi
15/2 + 2pi

8. d/dx (3^x)=
3^(x-1)
(3^(x-1)) ln3
3^x ln3 <----
(1/ln3)3^x

9. Lim as x->inf - (ln x+x^3)/(e^-x)=
Inf <----
1
0
-inf

10. For which value of x does the function f(x) = (x+2)(x+3)^2 have a relative maximum?
-2
-3 <----
Both -2 and -3
-5/2

11. Y = 2cotx - sqrt(x)sec x, then dy/dx
-2csc^2 x - 1/(2sqrt(x)) sec x tan x
-2csc^2 x - 1/(2sqrt(x))sec x - sqrt(x) sec x tan x <----
-2csc x cot x - 1/(2sqrt(x)) sec^2 x
-2csc^2 x - sqrt(x) sec x tan x

12. The area of the region enclosed by the curves y=2x and y=x^2 - 4x is:
36
18 <----
12
143/12

13. A particle moves along the x axis so that at any time t >= 0, it’s position is given by x(t) = t^3 - 12t^2 + 36t. For what values of t is the particle at rest?
No values
3 only
6 only <----
2 and 6

14. What is the average value of y for the part of the curve y = 4x - x^3 that is the first quadrant?
2/3
3/8 <----
3/2
2

15. How many critical points does the function f(x) = (x-1)^6 (x+5)^7 have?
13
7
6
3 <----

16. The region enclosed by the curve y=e^x, the x axis, and the lines x=0 and x=1 is revolved about the xaxis. Find the volume of the resulting solid formed.
Pi e^2
(e^2 - 1)/(2)
pi( (e^2/2) -1) <----
pi( (e^2 - 1)/(2) )

17. Consider the function f(x), whose graph is composed of two intersecting line segments shown below:
Which of the following are true for f(x) on the open domain (a,c)?
I. f(x) is continuous on (a,c)
II. f’(x) is continuous on (a,c)
III. f(x) is differentiable on (a,b)

I only <----

18. Lim as x->a (sin x - sin a)/(x-a) =
-sin a
cos a <----
-sin x
Undefined

19. ∫ (-3 on bottom, 3 on top) |x+1|dx =
-3.2
6
19/2
10 <----

20. ∫(3x^2 - 2)^2 dx=
9/5 x^5 - 4x^3 + 4x + C <----
((3x^2 - 2)^3)/(3) + C
((3x^2 - 2)^3)/(18x) +C
(6x(3x^2 - 2)^3)/(3) + C

21. The absolute minimum value of f(x) = x^3 - 3x + 12 on the closed interval [-2,4] occurs at x=
4
2
1 <----
-2

22. The graph of which of the following equations has y=-1 as an asymptote?
Y = lnx
Y = (x)/(x+1)
Y = (3 - 2x^2)/(2x^2 - 13x + 7) <----
Y = e^-x

23. ∫ (3x^2 - 1)/((x^3 - x)^2) dx =
-2/((x^3 - x)^3) +C
-1/(x^3 - x) + C <----
In(x^3 - x)^2 + C
(-6x^2)/((x^3-x)^3) + C

1. 👍
2. 👎
3. 👁
1. #1 ok
#2 ok
#3 nope. 11/23 use implicit differentiation
#4 ok
#5 f = -e^x+x
f' = 1-e^x
f'=0 at x = 0
f(0) = -1
#6 ok
#7 Nope
The area is triangle+rectangle+triangle-semicircle
3 + 3 + 3/2 - 2pi = 15/2 - 2pi
#8 ok
#9 maybe. If that leading "-" sign belongs there, then the limit is -inf
#10 ok
#11 ok
#12 I get 36
#13 since v(t) = 3t^2 - 24t + 36 = 3(t-2)(t-6)
v=0 at t=2,6
#14 No. The average value is (∫[0,2] 4x - x^3 dx) / (2-0) = 2
#15 ok
#16 I get pi/2 (e^2-1)
#17 I get I and III
check again the rule for differentiability on an open interval
#18 ok
#19 ok
#20 ok
#21 Hmmm. f(-2) = f(1) = 10
#22 ok
#23 ok

1. 👍
2. 👎
👤
oobleck

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