Math 113 HW #3 Solutions
1. Exercise 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,
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
2. Exercise 1.5.20. 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.
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3. Exercise 1.5.26. A bacterial culture starts with 500 bacteria and doubles in size every halfhour.
(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
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
2
4
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.
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.
4. Exercise 1.6.26. Find a formula for the inverse of the function
y=
2
ex
.
1 + 2ex
120000
100000
80000
60000
40000
20000
-4
-3
-2
-1
0
1
2
Answer: To find the inverse, we first swap the roles of x and y:
x=
ey
.
1 + 2ey
Now, the goal is to solve for y, so multiply both sides by 1 + 2ey :
x(1 + 2ey ) = ey .
Subtracting ey from both sides yields:
x + 2xey − ey = 0.
To isolate the y’s, first subtract x from both sides:
ey (2x − 1) = −x.
Now, dividing by 2x − 1, we see that
ey =
−x
.
2x − 1
Finally, taking the natural logarithm of both sides yields
−x
y = ln
,
2x − 1
which is an expression for f −1 (x).
5. Exercise 1.6.36. Find the exact value of each expression:
3
3
4
(a) e−2 ln 5
Answer: First, notice that
−2 ln 5 = ln 5
−2
= ln
1
52
= ln
1
25
.
Therefore, since eln x = x for any x > 0, we have that
1
e−2 ln 5 = eln( 25 ) =
1
.
25
10
(b) ln ln ee
Answer: Since ln (ex ) = x for any x, we have that
10
ln ee
= e10 .
Therefore,
10
ln ln ee
= ln e10 = 10.
6. Exercise 1.6.38. Express the quantity
ln(a + b) + ln(a − b) − 2 ln c
as a single logarithm.
Answer: Using the properties of logarithms, we know that
ln(a + b) + ln(a − b) = ln [(a + b)(a − b)] = ln a2 − b2
and that
2 ln c = ln c2 .
Hence,
ln(a + b) + ln(a − b) − 2 ln c = ln a2 − b2 − ln c2 ;
in turn, this is equal to
ln
a2 − b2
c2
.
7. Exercise 1.6.50. Solve each equation for x.
(a) ln(ln x) = 1
Answer: Since eln x = x, we can exponentiate both sides to see that
ln x = e1 = e.
Exponentiating both sides again yields
x = ee .
4
(b) eax = Cebx , where a 6= b.
Answer: Dividing both sides by ebx , we have that
eax
= C.
ebx
However, the left side can be re-written, using the properties of exponentials, as eax−bx =
e(a−b)x , so we have that
e(a−b)x = C.
Now, taking the natural logarithm of both sides, we have that
(a − b)x = ln C.
Dividing both sides by a − b gives the expression for x:
x=
ln C
.
a−b
8. Exercise 2.2.14. Sketch the graph of a function f that satisfies all of the following conditions:
lim f (x) = 1,
x→0−
lim f (x) = 1,
x→2+
lim f (x) = −1,
x→0+
f (2) = 1,
lim f (x) = 0,
x→2−
f (0) is undefined
Answer:
1.6
1.2
0.8
0.4
-1.5
-1
-0.5
0
0.5
1
-0.4
-0.8
-1.2
-1.6
5
1.5
2
2.5
3
9. Exercise 2.2.28. Determine the infinite limit
lim
x→5−
ex
.
(x − 5)3
Answer: Whenever x < 5, the expression x − 5 will be a negative number. Therefore, as
x → 5− , we see that x − 5 becomes a very small negative number. Since taking the cube
preserves sign and since ex > 0 for all x, this means that
lim
x→5−
ex
= −∞.
(x − 5)3
10. Exercise 2.2.40. In the theory of relativity, the mass of a particle with velocity v is given by
m0
m= p
1 − v 2 /c2
where m0 is the mass of the particle at rest and c is the speed of light. What happens as
v → c− ?
Answer: Whenever v < c, the fraction
v
c
< 1. Therefore, as v → c− , the fraction
v2
c2
gets closer and closer to 1, but is always less than 1. Hence, the quantity
1−
v2
c2
is always positive but approaches zero as v → c− . Therefore, as v → c− , the mass
m0
m= p
1 − v 2 /c2
approaches +∞.
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