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Material Type: Exam; Class: COLLEGE ALGEBRA; Subject: MATHEMATICS; University: La Sierra University; Term: Fall 2004;
Typology: Exams
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Math 121, Test 2, October 19, 2004
Name: Hints and Answers
Instructions. Part I: Do each of the following 10 problems. Each problem is worth 5pts. Show all appropriate details in your solutions. Calculators are not allowed. Good Luck! When you turn this in, please pick up Part II which has 2 problems.
Answer. (a) d =
(b) The midpoint is
Answer. Complete the square(s) to write the circle in standard form:
x^2 + 2x + 1 + y^2 โ 6 y + 9 + 1 = 1 + 9 and so (x + 1)^2 + (y โ 3)^2 = 9.
Therefore, the center is (โ 1 , 3) and the radius is 3.
x โ 2 and g(x) =
8 โ x. Find the domains of (a) f (b) g (c) f g (d) f /g. Write your answers in interval form.
Answer. (a) x โ 2 โฅ 0, thus x โฅ 2, i.e. [2, โ).
(b) 8 โ x โฅ 0, thus x โค 8, i.e. (โโ, 8].
(c) The intersection of the domains of f and g which is [2, 8].
(d) The intersection of the domains of f and g, except where g(x) = 0, which is [2, 8).
Answer. The orginal line in slope intercept form is y = 12 x โ 2, thus the slope of the original line is m 1 = โ 1 /2. The slope of the perpendicular line is m 2 = โ 1 /m 1 = โ2. Therefore, the equation of the perendicular line is y โ 3 = โ2(x โ 2) which simplifies to y = โ 2 x + 7 in slope-intercept form.
Answer. The x-coordinate of the vertex is โb/ 2 a = โ 6 / โ 4 = 3/2. The y-coordinate is y = โ2(3/2)^2 + 6(3/2) โ 5 = โ 1 /2. The parabola opens downward because a < 0. Therefore:
(a) The range is (โโ, โ 1 /2].
(b) There is no minimum because the graph opens downward.
(c) The maximum is f (3/2) = โ 1 /2.
f (x + h) โ f (x) h
for the function f (x) = 2x^2 โ 3 x.
Answer.
f (x + h) โ f (x) h
2(x + h)^2 โ 3(x + h) โ (2x^2 โ 3 x) h
=
2(x^2 + 2xh + h^2 ) โ 3 x โ 3 h โ 2 x^2 + 3x h
=
4 xh + 2h^2 โ 3 h h = 4 x + 2h โ 3.
Name:
Part II. You may use a calculator on this part of the test. Show all appropriate work in your solutions.
(a) Find a formula for P (x).
Answer. The formula for P is the formula of a line going through (100000, 300000) and (150000, 800000). Thus the slope of P is
m =
Thus P (x) = 10x + b, plug in one of the points to find b, i.e. 300, 000 = 10(100, 000) + b and so b = โ 700 , 000. Therefore, the equation is P (x) = 10x โ 700 , 000.
(b) How many subscribers are needed to make a profit of $1,000,000 per year?
Answer. Solve 1, 000 , 000 = 10x โ 700 , 000, to get that x = 170, 000 subscribers.
(c) What would the profit be if there were 300,000 subscribers?
Answer. The profit would be P (300, 000) = 10(300, 000) โ 700 , 000 = $2, 300 , 000.
Answer. Let x be length of the rectangle along the expensive side, and y be the length of the other dimension of the rectangle. Then the cost of the fence is 30x + 10x + 10y + 10y = 20, 000 and we wish to maximize area A = xy. Now solve the cost equation for x to get y = 1000 โ 2 x. Then A = xy = x(1000 โ 2 x) = โ 2 x^2 + 1000x.
This is a quadratic function whose maximum occurs at the vertex when
x = โb/ 2 a = โ 1000 / โ 2(2) = 250.
Thus y = 1000 โ 2(250) = 500. Therefore, the dimensions are 250 feet by 500 feet (where 250 feet is the length of the expensive side).