Math 113 HW #2 Solutions
§ 1.3
12: Graph the function by hand, not by plotting points, but by starting with the graph of one of the
standard functions given in Section 1.2, and then applying the appropriate transformations:
y = x2 − 4x + 3.
Answer: We expect the graph of this function to be a parabola, translated from the origin so
that the base of the parabola is centered on some point (a, b). The model for such a parabola
is
y = (x − a)2 + b,
which expands as
y = x2 − 2a + (a2 + b).
If the given equation fits this model, then
2a = 4
2
a + b = 3.
Therefore, a = 2 and b = −1. Hence, the given equation is equivalent to
y = (x − 2)2 − 1,
which is easy to graph:
4
3
2
1
-5
-4
-3
-2
-1
0
1
-1
-2
Figure 1: y = x2 − 4x + 3
1
2
3
4
5
34: Find the functions (a) f ◦ g, (b) g ◦ f , (c) f ◦ f , and (d) g ◦ g and their domains, given
√
√
f (x) = x, g(x) = 3 1 − x.
Answer:
(a)
f ◦ g(x) = f (g(x)) = f
√
3
1−x =
q
√
3
1−x=
√
6
1−x
has the domain {x ∈ R : x ≤ 1}.
(b)
g ◦ f (x) = g(f (x)) = g
q
√ √
3
x = 1− x
has the domain {x ∈ R : x ≥ 0}.
(c)
q
√ √
√
f ◦ f (x) = f (f (x)) = f
x =
x= 4x
has the domain {x ∈ R : x ≥ 0}.
(d)
√
3
g ◦ g(x) = g(g(x)) = g
1−x =
q
3
1−
√
3
1−x
has the domain R (i.e. there are no restrictions on the domain).
44: Express the function
r
G(x) =
3
x
1+x
in the form f ◦ g.
Answer: Define the functions
f (x) =
√
3
x
x
g(x) =
.
1+x
Then
f ◦ g(x) = f (g(x)) = f
x
1+x
r
=
3
x
= G(x),
1+x
as desired.
56: An airplane is flying at a speed of 350 mi/h at an altitude of one mile and passes directly over
a radar station at time t = 0.
(a) Express the horizontal distance d (in miles) that the plane has flown as a function of t.
Answer: Since the plane is traveling at a constant speed of 350 mi/h and since distance
equals rate times time, we have that
d(t) = 350t
is the horizontal distance function.
2
d
1
s
Figure 2: Schematic picture of airplane and radar station
(b) Express the distance s between the plane and the radar station as a function of d.
Answer: Using the Pythagorean Theorem, the distance s from the plane to the radar
station is given by
s2 = 1 + d2 ,
so we have that
s(d) =
p
1 + d2 .
(c) Use composition to express s as a function of t.
Answer: If we want to express the distance from the radar station as a function of t,
we simply take the composition
p
p
s ◦ d(t) = s(d(t)) = s(350t) = 1 + (350t)2 = 1 + 122, 500t2 .
§ 1.5
18: Find the exponential function f (x) = Cax whose graph is given
2
2, 29
0
Answer: We know that the function f is of the form Cax and that its graph passes through
the two points (0, 2) and (2, 2/9). Since the graph passes through (0, 2), we know that
2 = f (0) = Ca0 = C,
3
so we know C = 2.
Since the graph passes through (2, 2/9), we know that
2
= f (2) = 2a2 .
9
Therefore, a2 = 1/9, meaning that a = 1/3. Therefore, the function f is given by
x
1
.
f (x) = 2
3
20: Suppose you are offered a job that lasts one month. Which of the following methods of payment
do you prefer?
I. One million dollars at the end of the month.
II. One cent on the first day of the month, two cents on the second day, four cents on the
third day, and, in general, 2n−1 cents on the nth day.
Answer: Option II. To see why, let’s just figure out how much you would get paid on the
last day of the month...and to make things as even as possible, let’s assume the month is
February (and not in a Leap Year). Since February has 28 days, on the last day of the month
you would get paid
227 = 134,217,728 cents,
which equals $1,342,177.28. Therefore, under Option II, you get paid more on just the last
day than you get paid for the whole month under Option I.
26: A bacterial culture starts with 500 bacteria and doubles in size every half-hour.
(a) How many bacteria are there after 3 hours?
Answer: Since 3 hours equals 6 half-hours, the culture will have doubled 6 times.
Therefore, there will be
500 · 26 = 32,000
bacteria.
(b) How many bacteria are there after t hours?
Answer: Since t hours is the same as 2t half-hours, the culture will have doubled 2t
times. Therefore, there will be
500 · 22t
bacteria.
(c) How many bacteria are there after 40 minutes?
Answer: There are two possible answers depending on how you interpret the set-up
to the problem. If each bacterium in the culture doubles once every half-hour on the
half-hour, then each one will double after exactly 30 minutes, and then not again until
60 minutes have passed. In that case, there will be
500 · 2 = 1000
4
bacteria after 40 minutes.
On the other hand, if each bacterium doubles exactly once per half-hour, but at some
random time within that half-hour, then it makes sense to think of the population
function P (t) = 500 · 22t as continuous. In that case, since 40 minutes is
40
2
=
60
3
of an hour, the population will be
4
2
500 · 22 3 = 500 · 2 3 ≈ 1259
after 40 minutes.
(d) Graph the population function and estimate the time for the population to reach 100,000.
120000
100000
80000
60000
40000
20000
-4
-3
-2
-1
0
1
2
3
4
Answer: Using Mathematica or a graphing calculator, it’s not too hard to check that
the function
P (t) = 500 · 22t
passes 100,000 at approximately t = 3.82.
29: If you graph the function
1 − e1/x
1 + e1/x
you’ll see that f appears to be an odd function. Prove it
f (x) =
Proof. We will prove the function f is odd by showing that
f (−x) = −f (x).
5
To see this, consider
1−
1 − e−1/x
=
f (−x) =
1 + e−1/x
1+
Pull out a factor of
1
e1/x
1
e1/x
1
e1/x
.
from both the numerator and denominator to get
f (−x) =
1
e1/x
1
e1/x
e1/x − 1
e1/x − 1
= 1/x
.
e +1
e1/x + 1
Therefore,
1 − e1/x
= −f (x),
1 + e1/x
completing the proof that f is an odd function.
f (−x) = −
6