Solutions to MI 3 Problem set 6 F10
Introduction: At this point in the course students have studied polynomials and their graphs and
logarithms. They have not studied any trigonometry besides yet. They are also about to study
rational functions. Several problems in this problem set gets the students thinking about those
units.
1) Comment: Just a starting question to get student in a reasoning mode.
Soln: When we divided both sides by ( x  y ) we divided by 0 which is not allowed in
mathematics.
2) Comment: Students at IMSA will not formally study sequences and series until the first unit
of MI 4, so any questions like this are in anticipation of that unit. The  symbol probably
deserved a box to introduce, although most students figure it out. One may want to show an
example before giving this problem.
Soln: 14 17  20  23  26  29  32  161
3) Comment: A box indicates students are responsible for any information in the box and it is
important. Again there is spiraling of ideas through the problem sets well before the concept is
introduced in class.
Soln: a) 30, 37
b) 2.7, 3.8
c)
7 3
,
12 4
4) Comment: This problem anticipates calculus.
Soln:
1
1
1
1
 2  3  4    4  4  9    5  9  13   3 13  7   118
2
2
2
2
Note: Work is expected.
5) This problem reviews some basic analytic geometry. Part (c) the student needs to calculate
the distance to the origin and subtract the radius.
a) Line containing points ( 24, 7) and (0, 0) has a slope of m  
b) y  7 
24
 x  24 
7
c) Distance to the origin:
32.31  25  7.31
302  122  32.31
7
7
. y   x.
24
24
Solutions to MI 3 Problem set 6 F10
6 – 9 Comment: The next three problems are in anticipation of a rational function unit. For
problem 6 expect a student to use a graphing calculator.
3x  9  0
6) Soln: a) 3( x  3)  0
x3
x2  5x  6  0
b) ( x  3)( x  2)  0 Note: 3 is not in the domain.
x2
c)
7) Comment: This problem is to get the student to see an oblique asymptote and conjecture its
relationship to the function.
Soln: a) x  2,  2 but all other real numbers.
b) Quotient: x  4 remainder:
c)
2 x  22
2 x  22
or x  4  2
2
x 4
x 4
Solutions to MI 3 Problem set 6 F10
d) The line y  x  4 is an asymptote of the graph.
Justification: As x   the value of
2 x  22
 0 , so the curve approaches the quotient
x2  4
y  x4.
8) Comment: Hopefully the student thinks about these problems both algebraically and
geometrically.
Soln: a) 3
b) 0
9) Comment: No calculator. Students are expected to use the laws of logs to simplify and solve.
Soln:
1
1
1
1



3
log 3 x log 6 x log 8 x log12 x
log x 3  log x 6  log x 8  log x 12  3
log x  3  6  8 12   3
x3  1728
x  12
10) Comment: This problem is a variation on one from Phillips Exeter. This and the next
problem are to get students to think of circular motion in anticipation of the trigonometry units.
Soln:
Solutions to MI 3 Problem set 6 F10
sin  
1
   19.47 so   141.06
3
141.06
 30  11.8 sec.
360
11) Soln: Circumference  27
120
12  16.977 rev.
27
0.977  360  351.5
sin 81.5 
y
13.5
y  13.35
13.5 13.35  0.15 in.
Solutions to MI 3 Problem set 6 F10
12) Comment: Some beginning thinking of epsilon delta definition of a limit.
Soln: a)
1
8  3  7
2
b) g (5) 
11
2
g (11) 
17
 11 17 
Interval:  , 
2
2 2
g ( x)  7  3
13) Comment: The reference to Poly 5 is worksheet 5 IMSA student’s did in class. The
comment following this problem is to help students connect sign charts and the graphs of
polynomials. Of course later in calculus they will be interested in the sign chart in examining the
derivative.
Soln: y  a  x  2   x  1
3
5  a  23 1  a  
y
2
5
8
5
3
2
 x  2   x  1
8
14) Soln: a) ( x  7)( x  3)  0
_
+ + +
_
_
+ + +
7
-3
x  3 or x  7
b) ( x  5)( x  1)  0
x  1 or x  5
c)
( x  7)( x  3)
0
( x  5)( x  1)
3  x  1 or 5  x  7
--- +++++ ---- +++
++
-3
-1
5
7
Solutions to MI 3 Problem set 6 F10
15) Comment: The function graphed is periodic. With this problem students begin to think about
periodicity before they encounter the sinusoidal functions.
Soln: a)
1 1 1
, ,
b) 1, 0, 0
2 3 2
1
 2  3n
c) a  
 3  3n
 2
nZ