Algebra review problems:
a. The equation for a circle is
(x - h)2 + (y - k)2 = r2
where h and k are the x- and y-coordinates of the center of the circle and r is the radius.
So in this problem the equation for the circle centered at (3,4) and r = 5 is:
(x - 3)2 + (y - 4)2 = 25
b. Let y = ax + b. This line goes thru (-1,3) and (5,18) hence:
-a + b = 3 OR b = a +3
5a + b = 18
Substitute b = a + 3 into the above:
5a + a + 3 = 18
Hence 6a = 15 or a =15/6 =2.5
And therefore b = 5.5
The line has the form y = 2.5x + 5.5
c. Choose some point having coordinates (x,y). The distance between this point and (-5,4)
is given by:
x  52   y  42
The distance between point (x,y) and (3,16) is given by
x  32   y  162
Equating these distances, since the point is to be equidistant from the two given points,
we have:
x  52   y  42 = x  32   y  162
Squaring both sides, we have:
x  52   y  42  x  32   y  162
x 2  10 x  25  y 2  8 y  16  x 2  6 x  9  y 2  32 y  256
10 x  8 y  41  6 x  32 y  265
16 x  224  24 y
2
28
y  x
3
3
Reference:
http://www.tpub.com/math2/10.htm
Substitute one into the other:
 
x2  x 3
2
1
x 2  3x 2  1
4x 2  1
x2  1
4
1
1
x

4
2
1
y
3  0.87
2
Hence the points of intersection are (0.5,0.87) and (-0.5,-0.87)

 
x 3  y 3 x  y  x 2  xy  y 2
x 2  xy  y 2


x  y x  y 
x  y 
x2  y2

Reference:
http://mathworld.wolfram.com/PolynomialIdentity.html
a.
b.
c.
6
3
6
6
a
a
 12  2 3  2 3  4 3
 6
 a
1. Circumference = 2*pi*R =2*3.14*12 = 75.36
l(B,C) = Ra /180
2. The arc length for an angle a is (see reference):
l(B,C) = Ra /180
Hence when a = 30, l(A,B) = 6.28
3. l(B,C) = 12*80* /180 = 16.7
4. l(B,C) = 12*120* /180 = 25.12
5. Rearrange:
a = l(B,C)*180/(R )
Hence a = 3 *180/(12* ) = 45 (degrees)
6. a = 16 *180/(12* ) = 240 (degrees)
Reference:
http://math.rice.edu/~pcmi/sphere/drg_txt.html
Note that pages 136-137 are not attached in the problem.
Page 143, problem 1
Yes the function is periodic. The fundamental period of a function is the length of a
smallest continuous portion of the domain over which the function completes a cycle.
That is, it's the smallest length of domain that if you took the function over that length
and made an infinite number of copies of it, and laid them end to end, you would have the
original function.
In this case the fundamental period is 6.
The amplitude of a periodic function is the distance between the highest point and the
lowest point, divided by two. The amplitude is 1 in this case.
Note that for a periodic function g(x) with period p, the following is true:
f(p*k+a) = f(a) is true for all a real, k integer.
So f(1000) = f(166*6+4) = f(4) = -1
Also f(-1000) = f(-167*6+2) = f(2) = 1
Reference:
http://library.thinkquest.org/2647/algebra/ftperiod.htm
Page 143, problem 3
Period = 3
Amplitude = 0.5
f(1000) = f(333*3+1) = f(1) = 2
Also f(-1000) = f(-334*3+2) = f(2) = 3
Page 143, problem 7:
First re-arrange
a. y = x  2
b. y = x  3
c. y = x  5  4
d. y = 2 x  1
e. y =  x  1
f. y = 2 x  3
All of the graphs were provided in excel file attached.
Page 149, problem 1
a. f-1(6) = 2.
b. f-1(f(3)) = f-1(7) = 3.
c. f(f-1(7)) = f(3) = 7.
Reference:
http://www.analyzemath.com/inversefunction/inversefunction.html
Page 150, problem 11
Here's your original function: y = 3x - 5
y5
Try to solve for "x =": x 
3
Switch x and y; "y =" is the inverse. y 
x5
3
Hence an inverse exists as above.
 x 5  x 5
We see that f(f-1(x))= f 
  3
5  x
 3   3 
3x  5  5
x
Also f-1(f(x))= f 1 3x  5 
3
Hence f(f-1(x))= f-1(f(x))= x
Reference:
http://www.purplemath.com/modules/invrsfcn3.htm
Page 150, problem 21
The graphs were plotted in Excel
Using the above procedure to find a rule for g-1(x)
Here's your original function: y = 9 – x2 , x ≤ 0.
Try to solve for "x =": x   9  y
Switch x and y; "y =" is the inverse. y   9  x
Hence g-1(x) =  9  x , x ≤ 9.
Page 207, problem 32
8 x  87 x
a. x  7  x
x  3.5
0 .6 5 x  0 .6 x / 2
5x  x / 2
b.
4. 5 x  0
x0
e 3x  e x
3x  x
c.
2x  0
x0
x
1
6 x
  2
2
 
d. 2  x  2 6 x
 x  6 x
06
This is illegal so the inequality cannot be satisfied with any value of x.
Page 162, problem 9
We have the relationship due to geometry:
1. 8
s

3
sd
1.8s  1.8d  3s
1.8d  1.2 s
3
s d
2
a. Area of the base = w2
Cost for the base = 8*w2
Area of the 4 sides = 4*(8/w) = 32/w
Cost for sides = 6*32/w = 192/w
Hence total cost C = 8w2 + 192/w
b. I did this in Excel file. Plot C vs. w (w >0).
From graph, cost is minimum when w = 2.3 (m). The cost is about $127
350
300
250
200
C
150
100
50
0
0
1
2
3
4
5
6
7