REVIEW OF FUNDAMENTALS
R.R Review Exercises for Chapter
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1. Subtract 2 from both sides t o get 3x = 2 . Then divide by 3 to
get x = 2/3 .
5. Expansion gives (x + 1)^{2}  (x  1)^{2} = 2 = (x^{2} + 2x + 1)  (x^{2}  2x + 1) =
4x . Divide both sides by 4 to get x = 1/2 .
9. 8x + 2 > 0 is equivalent is 8x > 2 . Divide both sides by 8 to get
x > 1/4 .
13. Expand to get .
Add 1
to both sides to get 2x > 3 and divide by 2 to get x > 3/2 .
17. x^{2} < 1 is equivalent to x^{2}  1 < 0 or (x + l ) ( x  1) < 0 . One possible
solution is x < 1 and x > 1 , but this is not possible. Another
possibility is x > 1 and x < 1 . Thus, the solution is 1 <
x < 1 ,
i . e . , x ∈ (1,1) .
21. means
, i . e . ,
or .
For , we get
. For , we get
. Therefore, the solution is
or ,
25. x^{3} ∈ (8,27) implies x ∈ (2,3) , so x < 10 and x ∈ (2,3) means
2 < x < 3 ; i . e . , x ∈(2,3) .
29. is equivalent t o
or . Dividing
by 2 reverses the inequality and yields . Also, 3x  22 > 0
is equivalent t o 3x > 22 or x > 22/3 . Thus, the solution is
45. The distance between P_{1} and P_{2} is
.
In this case , it is .
49. The pointpoint form of a line is .
In t his case, the line is y = 1 + [ ( 3  (1))/(7  1 / 2 ) ] ( x 
1/2) =
53. The pointslope form of a line is . In t his case,
the
line is y = 13 + (3)(x  3/4) = 3x + 61/4 or 4y + 3x = 61 .
57. The slope of 5y + 8x = 3 is 8/5 . Thus, the slope of a perpendicular
line is 5/8 , and the line passing through ( 1 ,1) is y = 1 + (5/8) (x 
1) ,
using pointslope form of the line . The line is y = 5x/8 + 3/8 or
61. The equation of the circle with center (a,b) and radius r is
. In this case, the circle is
(y  5)^{2} = 8^{2} or (x^{2}  24x + 144) + (y^{2}  10y + 25) = 64 or
x^{2}  24x + y^{2}  10y + 105 = 0 .
Complete the square to get . This is a parabola opening upward with vertex at (2/3,2/3) . 
69. Solve the equations simultaneously. Substitute y = x in to x^{2} + y^{2} = 4
to get 2x^{2} = 4 , i . e . , x^{2} = 2 or
. Thus, the points of inter 
section are and
This is the graph of y = 3x if ; it is y = 3x if x < 0 . 

In addition t o the points in part (a), we have f(1.5) = 0.1875 ; f(1) = 0 ; f(0.5) = 0.0375 ; f(0.5) = 0.0375 ; 
81. If a vertical line passes through two
points of the graph, it is not a
function. Thus, (a) and (c) a r e functions.
TEST FOR CHAPTER R
1. True or false :
(a) I f a > b > 0 , then 1 / a > 1 / b .
(b) The interval (2,4) contains 3 integers.
(c) If a > b > 0 , then ac > bc for constant c .
(d) The domain of is x≥ 0 .
(e) The line x = 2 has slope zero.
2.Express the solution set of x^{2} + 3x + 2 ≥ 0 in terms of
absolute
values.
3.Write equations for the following lines:
(a) The line going through (1,3) and (2,4) .
(b) The line with slope 5 and passing through (3,2) .
(c) The line with slope 1 and yintercept 1/2 .
4.Sketch the graph of .
5.Do the following equations describe a circle or a parabola?
6.(a) Complete the square for the equation y^{2} + 4y + 3 = 0 .
(b) Solve the equation in part (a) .
7.Sketch the graph of y = Ix^{2}  1 I .
8.(a) Find the points of intersection of the graphs of y = x^{2} and
(b) Find the distance between the intersection points.
9.Factor the following expressions.
10.Dumb Donald had heard how wonderful chocolate mousse tasted, so he
decided to go on a hunting trip at Moose Valley. When he finally
spotted a moose, he chased the moose in circles around a tree located
at (4,2). Unfortunately for Dumb Donald, the moose ran much faster,
caught up with Dumb Donald, and trampled him. Their path was radius 7
from the tree. What is the equation of the circle?
ANSWERS TO CHAPTER TEST
1. (a) False; 1/a < 1/b .
(b) False; 2 and 4 are excluded from the interval.
(c) False; c must be positive for ac > bc .
(d) True.
(e) False; vertical lines do not have slopes.
5. (a) Circle
(b) Parabola
(c) Parabola