Example Problems
September 23, 2015
1
Linear Functions
Example 1.1. King Mukla is planning to open a new banana factory. It’ll cost 320 thousand dollars in seed
money to get the factory set up and every ton of bananas after that will cost 2 thousand dollars to ”produce”.
Create a function C(b) which gives the cost of producing b bananas. If bananas are going for 6 thousands
dollars per ton, then how many bananas need to be sold before the business venture will be profitable?
Proof. The equation for the cost, C, in thousands of dollars of producing b tons of bananas is given by
C(b) = 2b + 320
(1.1)
The revenue, R, from selling b tons of bananas at 5 thousands dollars per ton is given by
R(b) = 6b
giving an expected profitability of
P (b) = R(b) − C(b)
= 6b − (2b + 320)
= 4b − 320
1
(1.2)
Since profitability is a strictly increasing function, the break point is when the profit is zero
P (b) = 0
4b − 320 = 0
4b = 320
b = 80
Example 1.2. Market research at Mukla Inc. has shown the demand for bananas, D, in tons that will be
sold at the price point of p dollars is given by the function.
D(p) = 100 − 4p
(1.5)
At what price is a single bananas prohibitively expensive?
Also, based on the number of firms that decide to enter the market, the expected supply of bananas, S, in
tons to be produced at a price point of p thousand dollars per ton
S(p) = 6p
(1.6)
Find the equilibrium price.
Proof. The bananas will be bought when the demands is zero, D(p0 ) = 0.
100 − 4p0 = 0
4p0 = 100
p0 = 25
In terms of the cost of one banana this is p0 = 25, 000
Dollar
Ton
1/2000
Ton
Pound
1/3
Pound
Banana
= 4.02 Dollars
The equilibrium price occurs when the supply equals the demand S(p) = D(p). This gives the equation
2
for p
6p = 100 − 4p
10p = 100
p = 10
Example 1.3. At max capacity Kind Mukla’s factory will only be able to produce 50 tons of bananas. Should
he decide to go through with his plan to build the factory?
Proof. At a price point of 10 thousand dollars per ton it the max possible revenue is 500 thousand dollars
at a cost of C(50) = 2 ∗ 50 + 320 = 420 thousand dollars. Thus, the venture would be profitable for King
Mukla.
2
Quadratic Equations
Example 2.1. Plot the parabola
y = x2 + 2x − 8
(2.1)
Proof. Completing the square is accomplished by putting the equation of the parabola in the form
y = a(x − x0 )2 + y0
(2.2)
where (x0 , y0 ) is the location of the parabola’s peak. Expanding this out into a polynomial we get
y = ax2 − 2a(x0 )x + a(x0 )2 + (y0 )
(2.3)
So, for our case, matching the x2 terms gives a = 1, matching the x terms gives −2a(x0 ) = 2, and matching
the constant terms gives a(x0 )2 + (y0 ) = −8. Thus, a = 1, x0 = −1 and y0 = −9.
3
From here finding x intercepts (y = 0) is fairly simple.
0 = (x + 1)2 − 9
9 = (x + 1)2
±3 = (x + 1)
x = −1 ± 3
x = −4 or 2
This method is equivalent to using the quadratic formula on y = ax2 + bx + c
x=
Where x0 =
−b ±
√
b2 − 4ac
2a
(2.5)
−b
b2 − 4ac
and y0 =
2a
4a
Figure 1: Parabolas.
y = − 12 x 2 + 4x − 6
y = x 2 + 2x − 8
5
5
(4, 2)
y
0
y
0
−5
−5
(−1, −9)
−10
−6
−4
−10
−2
0
2
4
x
0
2
4
6
x
Example 2.2. Plot the parabola
1
y = − x2 + 4x − 6
2
(2.6)
Proof. An alternative approach is find the roots by factoring the polynomial and then using the fact that
4
the peak of the parabola will be found halfway between the two roots.
1
1
− x2 + 4x − 6 = − (x2 − 8x + 12)
2
2
1
= − (x − ?)(x − ?)
2
1
= − (x − 2)(x − 6)
2
At x = 4 halfway between the roots, y = − 21 ∗ 42 + 4 ∗ 4 − 6 = −8 + 16 − 6 = 2. Thus,
1
y = − (x − 4)2 + 2
2
3
(2.8)
Polynomials
The general form for a polynomial is
p(x) = an xn + an−1 xn−1 + . . . + a1 x + a0
This is said to be a an order n polynomial. Unlike linear and quadratic polynomials, polynomials of higher
order are too complicated for a general approach (though cubic and quadrics do have complicated formulas
for their roots).
Example 3.1. Plot y = x4 − 5x2 − 4.
Proof. Notice that this can be rewritten as
y = (x2 )2 − 5(x2 ) − 4
Now when looking for roots we can solve for x2 using whatever method we like to use on quadratic equations.
x2 = 1, 4
Which gives 4 roots x = ±1, ± 2.
Example 3.2. Plot y = x3 + 2x2 − 5x − 6
5
Proof. This equation can be factored. The sum of the factors should be 2 and the product should be −6
x3 + 2x2 − 5x − 6 = (x − ?)(x + ?)(x + ?)
= (x + 3)(x + 1)(x − 2)
Once we have what we think the factors are, we can plug these back in to check.
4
Rational Functions
Rational functions have the form
f (x) =
p(x)
q(x)
where both p(x) and q(x) are both polynomials.
Things to check
• What happens as x gets large (positive or negative).
• When the denominator is zero.
• Can the numerator or denominator be factored.
Example 4.1. Plot y =
x2 −2x−3
x−3
Proof. The numerator here can be factored
x2 − 2x − 3 = (x + 1)(x − 3)
So
x2 − 2x − 3
x−3
(x + 1)(x − 3)
=
x−3
x−3
= (x + 1)
x−3


 x + 1 x 6= 3
=

 DNE x = 3
y=
So, this should look like the line y = x + 1 with a hole at x = 3.
6
Example 4.2. Plot y =
x2 −2x−3
x−2
Proof. In this case factoring the numerator does not help. Instead we use a simplified version of long division
for polynomials.
x2 − 2x − 3
x−2
x(x − 2) − 3
=
x−2
x(x − 2)
3
=
−
x−2
x−2
3
=x−
x−2
y=
From this, it’s clear that for large x the function looks just like y = x and then near x = 2 there’s a vertical
asymptote.
Figure 2: Rational Functions.
x 2 − 2x − 3
y=
x−3
8
y=
20
x 2 − 2x − 3
x−2
7
10
6
y
y
5
0
4
3
−10
2
1
1
2
3
4
−20
−10
5
x
5
−5
0
5
10
x
Exponential Functions
Where polynomials are functions in which a variable x is raised to a constant power n, exponential functions
consider the case where the base a is fixed, however the exponent depends the variable on x.
f (x) = ax
There are several natural laws that follow for a, b, n and m positive integers.
• an+m = an · am
7
• an·m = (an )
m
n
• (a · b) = an · bn
Example 5.1. Plot y = 2x and y = 2−x and y = 2x+1 + 1.
Example 5.2. Solve 4x = 32 ∗ 8x
Example 5.3. King Mukla, after making a killing the banana manufacturing business, now has to go out
and invest his money. The Second Nations Bank of Two’s is willing to give him 2% interest compounded 2
times a year, while the InfiniBank is only able to give 1.9 percent interest but will compound continuously.
Write down functions that give the amount of money Mukla will have after x years. Which deal is better
for the King? How much of a difference will this make after 20 years if the initial investment is 2 million
dollars.
6
Logarithms
Here we see the first concrete example of an inverse function
y = loga x,
means
x = ay
So, if f (x) = loga x and g(y) = ay , then g(f (x)) = x. When the base is e this is called the natural log
written as
f (x) = ln x
Example 6.1. Plot y = log2 x.
Rules for logarithms
• loga x + loga y = loga xy
• y loga x = loga xy
•
ln x
ln a
= loga x
Example 6.2. Reduce log2 12 − log2 3.
8
Proof.
log2 12 − log2 3 = log2 4 ∗ 3 − log2 3
= log2 4 + log2 3 − log2 3
= log2 4 = log2 22
= 2 log2 2 = 2 ∗ 1 = 2
Example 6.3. Solve 2 ln x + ln 4 = 0.
Proof.
2 ln x + ln 4 = 0
ln x2 + ln 4 = 0
ln 4x2 = 0
4x2 = e0 = 1
x2 =
1
4
x=±
1
2
Example 6.4. How many digits does the number 102016 have in hexadecimal. Assume you only know
ln 2 ≈ 0.69 and ln 5 ≈ 1.61 as far as taking logs goes. What about in binary?
Proof. Hexadecimal is a base 16 numbering system, so to find the number of digits we take
log16 102016
and round up to the nearest whole number. Another way of saying this, is that we seek x such that
9
16x = 102016 .
log16 102016 = 2016 log16 10
ln 10
ln 16
ln 2 ∗ 5
= 2016
ln 24
ln 2 + ln 5
= 2016
4 ln 2 ln 5
= 504 1 +
ln 2
= 2016
= 504 ∗ 3.3̄
= 1680
If instead we use exact values for ln 10 and ln 16 we find the actual number of digits is 1675, giving a 0.2%
error.
To go from hexadecimal to binary, we have that our number is around the size of 161680 . Thus,
log2 161680 = 1680
ln 16
ln 2
= 1680 ∗ 4
= 6720
ln 2
ln 2
Or, if we do an exact transform from 102016 , we get 6698. This is a 0.3% error.
7
Word Problems
Example 7.1. Half Life: Half-life is the amount of time required for the amount of something to fall to
half its initial value. The term was originally coined (half-life period) by Earnest Rutherford in 1907 with
respect to the radioactive decay of radium to lead-206. In this case, the half-life was the time it took for there
to be half as much radium. But, the principle can be applied to any problem exhibiting exponential decay.
Imagine we have a pile of dice. Each hour we’ll roll all the dice and then get rid of the all the unlucky
dice that came up 1. What’s the half-life of dice in this example?
Proof. Let t be measured in hours and lets assume we start our with D0 dice. So, since 1 in 6 dice will role
a 1 on average, the number of dice after 1 hour has passed should be.
D(1) = D0
10
5
6
and after two hours the number of dice should be
D(2) =
D0
5
6
5
25
= D0
6
36
and after t hours we’ll have
D(t) = D0
t
5
6
The half-life is then the time t 21 at which the number of dice left is D(λ) = 12 D0 . As an equation this looks
like
1
D(t 12 ) = D0
2
t 1
1
5 2
= D0
D0
6
2
t 1
5 2
1
=
6
2
t 1
5 2
1
ln
= ln
6
2
5
1
t 12 ln = ln
6
2
ln 2
t 12 =
ln 6 − ln 5
t 12 ≈ 3.8h
Another way to do this is to write
D(t) = D0
t
6
5
= D0 e− ln 5
6
in this case λ = ln 65 is the decay constant and t 21 =
t
= D0 e−λt
ln 2
λ .
Example 7.2. Present Value: Wimpy often says that he’ll pay a dollar tomorrow for a hamburger today.
If Wimpy is a loan shark who gets an exorbitant 24% interest per day compounded hourly, then how much
would he be willing to pay for that hamburger today. (assuming he’d being paying for it exactly 24 hours in
the future).
Proof. Let t be measured in days. If Wimpy has W0 dollars ”invested” today, then the amount of money
he’ll have tomorrow is.
24t
0.24
W (t) = W0 1 +
24
11
and we want to find the value of W0 that would result in 1 dollar tomorrow, i.e., W (1) = 1. Thus, the
equation we must solve is
1 = W (1)
24
1 = W0 (1 + 0.01)
−24
W0 = (1.01)
≈ W0 1.27
≈ 0.79
This is how much money a dollar tomorrow is worth to Wimpy today.
12