Solution Key: Homework due September 11, 2002

Page 9, Problem 10: Find all values of a such that g(a) = 5 where g(x) = 2x2 - x + 4.

Solution: Setting 2a2 - a + 4 = 5 leads to 2a2 a -1= 0 = (2a + 1)(a-1) so the solutions are a = 1 and a = -1/2.

Page 9, Problem 20: Let f(x) be the first class postage (in cents) for a letter mailed in the U.S. weighing x ounces, 0 < x < 12. As of January 7, 2001, the postage rate for such a letter was 34 cents for the first ounce and 21 cents for each additional ounce or fraction thereof. What is the range of f(x)?

Solution: The possible values, in cents, are 34, 55, 76, 97, 118, 139, 160, 181, 202, 223, 244, and 265. Note that 0 is not in the range.

As of June 30, 2002, the rates have gone up to 37 cents for the first ounce and 23 cents for each additional ounce or fraction thereof. How does the range change?

Page 9, Problem 30: Find the largest domain on which the function g(t) = (2/(3-t)) is defined.

Solution: In order for the function to be defined, the number under the radical sign must be non-negative, so 2/(3-t)≥0. This is correct if t < 3 sothe domain is all t less than 3.

Page 9, Problem 40: If a rectangle has side lengths x and y, then its area is A = xy and its perimeter is 100 = 2(x + y). Show that the area is given by the function A = x(50-x) for 0 < x < 50.

Solution: We get 50 = x + y so y = 50 - x and A = x(50 - x). Area can never be non-negative. So the domain of this function is all x between 0 and 50.

Page 11, Problem 47: An open-topped box is to be made from a square piece of cardboard of edge length 50 in. First, four small squares, each of edge length x inches, are cut from the corners of the cardboard. Then the four resulting flaps are turned up to form the four sides of the box, which will thus have a square base and a depth of x inches. Express its volume V as a function of x. (See Figures 1.1.21, 1.1.22 on page 11.)

Solution: The box will have height x when it is folded up, and the base will have area (50-2x)2. The volume will then be V = x((50-2x)2. Since volume must be non-negative the domain of V is all x with 0 < x < 25.

Section 1.2

Page 21, Problem 5: Write an equation of the line L that passes through (2,-3) and (5,3).

Solution: First, we calculate the slope of the line to be m = (3 - (-3))/(5 - 2) = 2. Using the point-slope equation with the point (5,3), the line L has equation y - 3 = 2(x-5), or y = 2x - 7.

Page 21, Problem 10: Write an equation of the line L that passes through (-2,4) and is perpendicular to the line with equation x + 2y = 17.

Solution: The slope of the line with equation x + 2y = 17 is -1/2, so the slope of L is 2. Using the point-slope equation again, the line L has equation y - 4 = 2(x + 2), or y = 2x + 8.

Page 21, Problem 15: Find the center and radius of the circle with equation 2x2 + 2y2 + 2x - 2y = 1.

Solution: First factor out the 2, giving 2[(x2 + x) + (y2 - y)] = 1. Now complete the square in x and in y, giving 2[(x + 1/2)2 - 1/4 + (y - 1/2)2 - 1/4] = 1. Dividing both sides by 2 and then simplifying this expression, (x + 1/2)2 + (y - 1/2)2 = 1/2 + 1/4 + 1/4 = 1. Thus, the center of the circle is (-1/2, 1/2) and the radius of the circle is 1.

Page 21, Problem 20: Find the vertex of the translated parabola 2y = x^2 - 4x + 8.

Solution: Complete the square in x, giving 2y = (x - 2)2 - 4 + 8 = (x - 2)2 + 4, so y - 2 = 1/2(x - 2)2. The vertex of this parabola is (2,2). For the rest of the problems, see this site.