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Sp 2010 MLC Final Review
Answers
p. 1
Disclaimer: This answer sheet may contain errors, hopefully there are not many.
1.
a. 1
x+2 = pos # which means that
x  2  ( x  2) and you cancel
b. -1
x+2 = neg # which means that
x  2  ( x  2) and you cancel
2
c. 1
divide each term by the highest power in the denominator which is x (Note: x  x )
d. 0
divide each term by the highest power in the denominator which is x2
1
0
0
e.
you get , so use L’Hopital’s Rule repeated until you no longer get
6
0
0
f. DNE y  sin x does not “settle down” to a single value and continually oscillates
2.
a.
1
3
see problem 1c
5
5
and lim
and see that they are not the same value
x

4
x4
x4
2
lim q(h)  lim h  1  3 )
( lim q(h)  lim h  3  1
b. DNE Compute lim
x4
c. DNE
d. 5
e. 
h2
h2
h2
h2
Compute the left- and right-hand limits
1
4
Simplify by first using a common denominator
3.
a.
b.
4. Write the equation as cos x  x  0 , then show that for some value of x cos x  x = neg # and for a
different value of x cos x  x = pos #. (Hint: find an x-value that gives these two results). Then explain
how you know that cos x  x is continuous and use the Intermediate Value Theorem to claim that for some
of x, cos x  x  0 . (Or in terms of the original problem there is some x-value where cos x  x .)
5.
a.
b.
6. 0
f ( x)  2 x  2
1
f ( x)  2
x
 
y   cos x  ( sin 2 x)  2 and y    0  0 .
2
Sp 2010 MLC Final Review
Answers
p. 2
Disclaimer: This answer sheet may contain errors, hopefully there are not many.
7.  cm2/min
dA
 ? when r  50
dt
dr
 0.01
dt
Area of a circle: A  r
2
Be sure to take the derivative of the Area equation BEFORE plugging in any numbers.
8. 800
9.
3
mm
sec
dV
 ? when r  100
dt
dr
2
dt
Volume of a sphere: V 
4 3
r
3
Be sure to take the derivative of the Area equation BEFORE plugging in any numbers.
Note: Continuous if lim f ( x )  f ( a ) (or the limit = function value)
xa
a. c = 2
Note: lim f ( x ) 
b. No such c since the limit DNE
x 1
pos #

0
lim f ( x) 
x 1
pos #
 
0
10.
a. g (t )  4 sec t tan t  sec 2 t
b. y  
x2
1  x 
2
3 3
or y  

1
1  x3
3


2
3
3x 2
Chain Rule
c. h( x)  5sin 5 x  cos 5 x   cos5 x   5   sin 5 x   5
Chain Rule
1  cos x  cos x  sin x( sin x) and cos 2 x  sin 2 x  1
1
d. y  
Quotient Rule: y  
1  cos x 
1  cos x 2
3 1
5 7
e. y   x 2  x 2
Simplify first & Power Rule (or Product & Quotient Rule)
2
2
5
 8t  1
f. f (t ) 
or f (t )  4 t 2  2t  5 2t  2
Quotient Rule (or rewrite & chain rule)
5
2
t  2t  5
dy
11. Find
dx
a. If you simply first: y  tan x  1 and y   sec 2 x , if you use the product rule you would get:
y  (sin x  cos x)sec x tan x  sec x(cos x  sin x)
2
b. y   x 2 csc 2 x   2 x  cot x  3
Product Rule
x
dy
dy
dy
16 x
 0 for
c.
Solve 16 x  2 y 

dx
dx
dx
2y
4




dy
sin y 2
3
See problem 11c Note:  is a constant like (3.14)3

2
dx 1  x cos y  2 y
dy
dy 
dy
y cosxy

 cos xy   y  1  x  
e.
See problem 11c

dx
dx 
dx 1  x cosxy

2
f. 2 x ln x  ln x
We haven’t done ln x, but you can still do this using the FTOC and the chain rule
12. Find the equation of the tangent line to the graph of the equation at the given point.
1
y  y0  m( x  x0 ) where x0  1 , y 0  , and m  f (1)
a. 4 y  x  1
2
b. y  8  4( x  2)
See problem 11a
2
13. a 
Note: To find slope write 8 x  3 y  4 into y-intercept form: y  mx  b Then set the derivative
3
function y   m and solve for x. (This x will have a tangent line with the same slope as the line given.)
 
d.
 
Sp 2010 MLC Final Review
Answers
p. 3
Disclaimer: This answer sheet may contain errors, hopefully there are not many.
14.
3
994  9.98
3
x  10 
1
( x  1000) for x near 1000
300
15.
a. See problem 4 to show that it has “at least one solution”. Then give an explanation for why f ( x)  x 4  3x  1
does not cross the x-axis more than once on the interval [-2, 1].
b. See problem 15b
16. See example 7 in section 4.4 of the textbook
17. 72m of fence is needed
Draw a picture and label it:
x
y
x
y
Find the minimum fencing. Total fencing: F  4 x  3 y
(Note: Get F  4 x  3 y into one variable and determine which CP is min & inside the domain)
(Use information about the area of the patch to come up with another equation that relates x and y)
18. 18in by 9in
This is #11 in section 4.5 of the textbook.
Write two equations (the area of printing and the area of the paper) Use one to write the other in one
variable and find the CP that gives the min. (Be sure to test the CP to make sure that it is a min.)
19. y   2 x  1
a. x1  2
2
b. x1 
3
20. Note 1: Area of rectangle = (base) x (height). base 
length of the interval
b-a
, height = f(x0) where x0 is the

the number of rectangles
n
left- or right-endpoint.
Note 2: f ( x)  x is an increasing function on [0, ∞), so use left-endpoints for lower sums and right-endpoints for upper sums.
Note 3: Determine the area for each rectangle individually and add it all together
2
1
8
87
b.
32
a.
1 0
1
, heightRec #1 = f (0) , heightRec #2 = f  
2
2
2 1
 1
 2
 3
base 
, heightRec #1 = f 1   , heightRec #2 = f 1   , heightRec #3 = f 1   ,
4
 4
 4
 4
 4
heightRec #4 = f 1  
 4
3 5 3
Area   f     f ( 4)
2 2 2
base 
267
8
189
d.
See problems 20a-c
32
21. Evaluate the following definite and indefinite integrals
a. 2
The area of a semi-circle with radius 2.
2
1
2 3
F ( x)  x 3  x 2  x  C Then compute F (1)  F (0)
b.
3
3
3
1 1 2
1
F (t ) 
 t  C Then compute F (2)  F (1) Note: 2  t  2
c.  1
t
2
t
1
F ( x)   cos( 2 x)  C Then use the FTOC.
d. 1
Try u = angle
2
c.
Sp 2010 MLC Final Review
Answers
p. 4
Disclaimer: This answer sheet may contain errors, hopefully there are not many.
2
1
3
Try u = the thing to the power F ( x)  cos ( x)  C Then use the FTOC.
3
3
22. Evaluate the following definite and indefinite integrals
3
a.
See problem 21e
2
1
tan 4 x  C
b.
Try u = sec x or u = tan x and see which one will work and which one won’t.
4
1
c.  cos 2 x 2  1  C See problem 21d
2
e.

d. sin x  tan x
Simplify first
1
e.  csc   C
2
Try u = csc x
f.
23.
e4
4


 cos x  sec x dx then integrate
2
The integral them becomes:
1
 2 du
2
2
 3 C
2
t
3t 2
Use the FTOC and the chain rule to determine the formula for F (x ) , then evaluate at x = 4. (See problem 11f)
1
24. 4
The curves intersect at x = -1 and x = 1. Determine the top curve and evaluate
 top  bottomdx
1
25.
37
12
The curves intersect at y = -1, y = 0, and y = 2. Determine which curve is to the right curve on (-1, 0) and which is
to the right curve on (0, 2). Then evaluate
26. -1
27. r = 4
0
2
1
0
 right  left dy +  right  left dy
Hint: The limit exists if lim q ( h)  lim q ( h)
h2
h2
See problem 17 or 18. V  r h and by similar triangles
2
h
12

6r 6
3
5
cos 2 x  6 x 
2
2
29. s(t )   cos t  sin t  4t  4
3
cos 2 x  6 x  C , then use y (0)  1 to solve for C.
2
See problem 28. Note:  a (t )dt  v(t ) and  v(t ) dt  s (t )
30. You decide on the answer 
See problem 24.
28. y 
y