10. 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) = −16t2 + 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
;
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(t2 − 4t − 5) = 0, and so (t − 5)(t + 1) = 0.
Therefore, the
projectile lands after 5 seconds.
(d) The difference quotient
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 ,
.
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.
11. (a) Do # 41, p. 273.
Answer. See answer in text.
(b) De termine whether the graph of y = x3 − 4x is symmetric about the (i)
x-axis, (ii) y-axis,
(iii) origin.
Answer. Symmetric to the origin since the equation is un changed if x is replaced
with −x
and y is replaced with −y. Please check this.
(c) Determine whether the function g(x) = x5 − x3 is even, odd or neither.
Answer. Odd, because g(−x) = −x5 + x3 = −(x5 − x3) = −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.
12. 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 x2 + y2 = x2 + (x + 10)2. In other
word, we minimize the quadratic function
f(x) = 2x2 + 20x + 100. The minimum occurs when
. Thus x = −5
and y = −5 + 10 = 5. So the two numbers are −5 and 5, and the sum of their
squares is 50.
13. Let 
and
. Find the domain of (i) f + g, (ii) f − g, (iii) fg,
(iv) f/g.
Answer. Remember, the domains of f + g, f − g and fg 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) ≠ 0. Now g(x) = 0 if x =
−7. Therefore,
the domain of f/g is (−7, 5].
14. 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 = 250−2y. Now we
maximize
the area xy = y(250 − 2y). So we maximize the quadratic function f(y) = −2y2
+ 250y. This
maximum occurs when 
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.
15. A Hollywood charter bus company that provides tours through the movie star
neighborhoods
in Beverly Hills has determined that the cost of providing x people a tour is
C(x) = 180 + 2.50x
A full tour consists of 60 people. The ticket price per person is $15 plus $0.25
for each unsold
ticket. Determine
(a) The revenue function.
(b) The profit function.
(c) The company’s maximum profit.
(d) The number of ticket sales that yields the maximum profit.
Answer. (a) R(x) = x(15 + .25(60 − x)) = −.25x2 + 30x.
(b) P(x) = R(x) − C(x) = −.25x2 + 27.5x − 180.
(d) The number of tickets is −27.5/(2(−.25)) = 55.
(c) The maximum profit is P(55) = −.25(552) + 27.5(55) − 180 = $576.25.
16. Answer the fol lowing in terms of shifts, reflections, stretching or
shrinking.
(a) How does the graph of y = f(−x) relate to the graph of y = f(x)?
(b) How does the graph of y = −f(x) relate to the graph of y = f(x)?
(c) How does the graph of y = −f(x + 2) relate to the graph of y = f(x)?
(d) How does the graph of y = f(5x) relate to the graph of y = f(x)?
(e) How does the graph of 
relate to the graph y = f(x)?
(f) How does the graph of y = 10f(x) relate to the graph of y = f(x)?
Answer. (a) It is a reflection about y-axis. (b) It is a reflection about
x-axis. (c) Shift graph
of f two units left and then reflect about x-axis. (d) Horizontally com pressed
by factor of 1/5
towards y-axis. (e) Stretched horizontally by factor of 12 away from the y-axis.
(f) Vertically
streched by factor of 10 away from the x-axis.
17. For conceptual graphing questions, see 57,59,63,73 on page 239.
18. (a) The function I(x) = 12x converts feet to inches and the function F(x) =
5280x
converts miles to feet. Compute (I o F)(x) and explain its meaning.
(b) Let f(x) = x2 + 4x − 1 and g(x) = x + 2. Find f o g and g o f.
(c) Let f(x) = x2 + 1 and 
. Compute f o g and g o f. What are their domains?
Are f o g and g o f equal?
Answer. (a) (I o F)(x) = I(F(x)) = I(5280x) = 12(5280x) = 63360x. Converts miles
to
inches.
(b) (f o g)(x) = f(g(x)) = (x + 2)2 + 4(x + 2) − 1 = x2 + 4x + 4 + 4x +
8 − 1 = x2
+ 8x + 11.
On the other hand, (g o f)(x) = (x2 + 4x − 1) + 2 = x2 + 4x
+ 1.
(c)
(note: domain is x ≥ 1) and so (f o g)(x) = x − 1 + 1 = x for
x ≥1. On the other hand, 
and the domain is (−∞,∞).
19. 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
. Thus P(x) = .3x + b, and so 40 = .3(300) + b which means b = −50. Hence
P(x) = .3x − 50.
(b) To break even, she must have .3x − 50 = 0, and so x = 50/.3 = 166.67. That
is, she must
sell 167 cups of lemonade.
(c) To make $100, we solve .3x − 50 = 100, and so x = 150/.3 = 500 cups.
(d) She will make P(1000) = .3(1000) − 50 = 250 dollars.
