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Math 121, Chapter 2 Practice Problems Hints and Answers
- (a) Find the distance between the points (− 3 , 2) and (9, −3).
Answer. d =
(9 − (−3))^2 + (− 3 − 2)^2 =
(b) find the midpoint of the line segment with endpoints (3, 5) and (− 5 , 2).
Answer.
- (a) Determine the center and radius of the circle whose equation is x^2 +y^2 +10x+4y+20 = 0.
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.
- Let f (x) = 2x^2 + 7 and g(x) = |x − 1 |, find
(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.
- Let f (x) = 4x^2 − 3 x, find the difference quotient
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.
- Sketch the graph of f (x) = |x + 3| − 2 and find intervals where f is (a) increasing; (b) decreasing. Is f one-to-one?
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.
- Determine the domains of the following functions.
(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.
- (a) Find the slope-intercept form of the line through the points (− 1 , 3) and (4, −7).
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.
- The height in feet of a projectile with an initial velocity of 64 feet per second and an initial height of 80 feet is a function of time t in seconds, given by
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)
- 01 −. 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)
- 01 −. 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.
- (a) Do # 41, p. 273.
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.
- Find two numbers whose difference is 10 and the sum of whose squares is a minimum.
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.
- Let f (x) =
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].
- A farmer has $1000 to spend to fence a rectangular corral. Because extra reinforcement is needed on one side, the corral costs $6 per foot along that side. It costs $2 per foot to fence the remaining sides. What dimensions of the corral will maximize the area of the corral?
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.
- A Hollywood charter bus company that provides tours through the movie star neighbor- hoods in Beverly Hills has determined that the cost of providing x people a tour is
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
- Julie opened a lemonade stand and found that daily her profit is a linear function of the number of cups of lemonade sold. When she sells 300 cups of lemonade, she makes $40 and when she sells 600 cups of lemonade, she makes $130.
(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.
