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Material Type: Exam; Class: COLLEGE ALGEBRA; Subject: MATHEMATICS; University: La Sierra University; Term: Unknown 1989;
Typology: Exams
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Math 121, Chapter 2 Practice Problems Hints and Answers
Answer. d =
(b) find the midpoint of the line segment with endpoints (3, 5) and (− 5 , 2).
Answer.
Answer. Complete the squares:
x^2 + 10x + 25 + y^2 + 4y + 4 = −20 + 25 + 4 (x + 5)^2 + (y + 2)^2 = 9
Therefore, the center is (− 5 , −2) and the radius is r = 3.
(b) Write the equation of a circle whose center is (− 5 , 1) and passes through the point (3, 1).
Answer. The radius is the distance from (− 5 , 1) to (3, 1) which is 8. Therefore, the equation is (x + 5)^2 + (y − 1)^2 = 64.
(c) Find the equation of a circle that has diametral endpoints of (0, 0) and (6, 8). (Hint: the center is the midpoint of the diametral endpoints).
Answer. The center is at (3, 4), the radius is r =
(3 − 0)^2 + (4 − 0)^2 = 5. Thus the equation is (x − 3)^2 + (y − 4)^2 = 25.
(a) (f ◦ g)(−5) (b) (g ◦ f )(x) (c) (f g)(0) (d) (f + g)(0)
Answer. (a) f (g(−5)) = f (6) = 2(6^2 ) + 7 = 72 + 7 = 79.
(b) g(f (x)) = |f (x) − 1 | = | 2 x^2 + 7 − 1 | = 2x^2 + 6 (we can drop the absolute values because 2 x^2 + 6 ≥ 0 for all values of x.)
(c) and (d): f (0) = 7 and g(0) = 1, therefore (f g)(0) = 7(1) = 7 and (f + g)(0) = 7 + 1 = 8.
f (x + h) − f (x) h
Answer.
f (x + h) − f (x) h
4(x + h)^2 − 3(x + h) − (4x^2 − 3 x) h
=
4 x^2 + 8xh + 4h^2 − 3 x − 3 h − 4 x^2 + 3x h =
8 xh + 4h^2 − 3 h h
=
h(8x + 4h − 3) h = 8 x + 4h − 3.
Answer. Notice that the graph of f (x) is the graph of y = |x| shifted left 3 units, and down 2 units. Thus, f is decreasing on (−∞, −3] and increasing on [− 3 , ∞). The sketch is left to the reader. Notice that f is not one-to-one because it fails the horizontal line test.
(a) f (x) =
x + 3 (x + 2)
16 − x^2
(b) g(x) =
x − 4
(c) h(x) =
4 − x
(d) k(x) =
3(x − 1) (x + 2)(x − 11)
Answer. (a) We need 16 − x^2 > 0 and x + 2 6 = 0. Therefore, − 4 < x < 4 and x 6 = 2 which in interval form is (− 4 , −2) ∪ (− 2 , 4).
(b) x ≥ 4; (c) x ≤ 4 (d) x 6 = −2, x 6 = 11.
Answer. The slope is m = 4 −−^7 (−−^3 1) = −2. Using the point-slope form of a line, we get
y − 3 = −2(x + 1) and so y = − 2 x + 1 is the slope-intercept equation of the line. One should plug both points in to make sure they work.
(b) Find the slope-intercept form of the line that passes through the point (− 3 , −7) and is perpendicular to the line 2x + 5y = 10.
Answer. First, the line 2x + 5y = 10 has slope m 1 = −^25. So the perpendicular line should have slope m 2 = − 1 /m 1 = 52. Thus the desired line has point-slope form y + 7 = 52 (x + 3) which leads to the slope-intercept equation y = 52 x + 12.
(b) the profit function;
Answer. P (x) = 12x − (. 75 x + 875) = 11. 25 x − 875.
(c) the minimum number of parcels the company must ship on a flight to break even.
Answer. Solve P (x) ≥ 0, i.e., 11. 25 x − 875 ≥ 0 implies x ≥
≈ 77 .8. Thus the company
must ship at least 78 parcels to break even or make a profit.
h(t) = − 16 t^2 + 64t + 80.
(a) Find the maximum height of the projectile.
Answer. Using the answer in (b), we compute h(2) = −64 + 128 + 80 = 144 feet.
(b) Find the time t when the projectile reaches its maximum height.
Answer. The maximum height occurs at t =
= 2; that is 2 seconds into its flight.
(c) Find the time t when the projectile hits the ground (has a height of 0 feet).
Answer. h(t) = 0 implies −16(t^2 − 4 t − 5) = 0, and so (t − 5)(t + 1) = 0. Therefore, the projectile lands after 5 seconds.
(d) The difference quotient
h(1.01) − h(.99)
gives the average velocity of the projectile for
. 99 ≤ t ≤ 1 .01. Compute this difference quotient. Do you think it would provide a good estimate of the instantaneous velocity of the projectile when t = 1?
Answer. Using a calculator,
h(1.01) − h(.99)
= 32 feet/second. It should be a good estimate,
because the average velocity of a very small time interval should be close to the instantaneous velocity at the time in the midpoint of the interval.
Answer. See answer in text.
(b) Determine whether the graph of y = x^3 − 4 x is symmetric about the (i) x-axis, (ii) y-axis, (iii) origin.
Answer. Symmetric to the origin since the equation is unchanged if x is replaced with −x and y is replaced with −y. Please check this.
(c) Determine whether the function g(x) = x^5 − x^3 is even, odd or neither.
Answer. Odd, because g(−x) = −x^5 + x^3 = −(x^5 − x^3 ) = −g(x).
(d) In terms of shifts or translations, how does the graph of y = f (x + 5) − 10 compare to the graph of y = f (x)?
Answer. The graph of y = f (x + 5) − 10 is the graph of y = f (x) shifted 5 units to the left, and 10 units down.
(e) In terms of shifts or translations, how does the graph of y = f (x + 5) − 10 compare to the graph of y = f (x − 3) + 2?
Answer. Shift the graph of y = f (x − 3) + 2 to the left 8 units and down 12 units to get the graph of y = f (x + 5) − 10.
Answer. Let the two numbers be x and y. Then y − x = 10 and so y = x + 10. Now we minimize x^2 + y^2 = x^2 + (x + 10)^2. In other word, we minimize the quadratic function
f (x) = 2x^2 + 20x + 100. The minimum occurs when x = −
b 2 a
= −5. Thus x = − 5
and y = −5 + 10 = 5. So the two numbers are −5 and 5, and the sum of their squares is 50.
5 − x and g(x) =
x + 7. Find the domain of (i) f + g, (ii) f − g, (iii) f g, (iv) f /g.
Answer. Remember, the domains of f + g, f − g and f g are all the same, and they are the intersection of the domains of f and g. The domain of f is (−∞, 5] and the domain of g is [− 7 , ∞). The intersection of these domains is [− 7 , 5] which is the answer for (i), (ii), and (iii).
For (iv), the domain is all x ∈ [− 7 , 5] such that g(x) 6 = 0. Now g(x) = 0 if x = −7. Therefore, the domain of f /g is (− 7 , 5].
Answer. Let the dimensions be x and y, with the y being the length of the expensive side. Then 2x+2x+2y+6y = 1000. Therefore, 4x+8y = 1000 and so x = 250− 2 y. Now we maximize the area xy = y(250 − 2 y). So we maximize the quadratic function f (y) = − 2 y^2 + 250y. This maximum occurs when y = − −^2504 = 62.5 feet and so x = 250 − 125 = 125 feet. Thus the dimensions are 62.5 feet by 125 feet, where the expensive side is 62.5 feet long.
C(x) = 180 + 2. 50 x
A full tour consists of 60 people. The ticket price per person is $15 plus $0.25 for each unsold ticket. Determine
(a) Find the profit function.
(b) How many cups of lemonade does Julie need to sell to break even on a given day?
(c) How many cups of lemonade does Julie need to sell to make $100 in a day?
(d) How much would she make on a day when she sells 1000 cups of lemonade?
Answer. (a) Let x be the number of cups sold, then P (x) = mx + b where m =
= 0.3. Thus P (x) =. 3 x + b, and so 40 = .3(300) + b which means b = −50. Hence
P (x) =. 3 x − 50.
(b) To break even, she must have. 3 x − 50 = 0, and so x = 50/.3 = 166.67. That is, she must sell 167 cups of lemonade.
(c) To make $100, we solve. 3 x − 50 = 100, and so x = 150/.3 = 500 cups.
(d) She will make P (1000) = .3(1000) − 50 = 250 dollars.