Download CHAPTER 7 ELEMENTARY FUNCTIONS I

CHAPTER 7
ELEMENTARY FUNCTIONS I
1.
2.
3.
4.
5.
Polynomial Functions
Rational and Algebraic Functions
Exponential and Log Functions
Exponential and Log Applications
Introduction to Financial Mathematics
7.1 : Polynomial Functions
7.1 POLYNOMIAL FUNCTIONS
You have previously examined linear and quadratic equations which are also called
first and second degree polynomial equations, respectively. Polynomials can be easily
created using addition, subtraction, multiplication and taking powers. Consequently
values are easy to calculate, and so polynomials are sometimes used to approximate other
more complicated functions. Examples of this are seen in Infinite Series in calculus.
Polynomial function equations take the form:
y = P(x) = aoxn + a1xn-1 + a2xn-2 + . . . + an-1x + an
where all the exponents are whole numbers. The number n is called the degree of the
polynomial. The ai 's are called the coefficients of the polynomial (an is often referred to
as the constant term). The term aoxn is called the leading term and the number ao is
called the leading coefficient. The name polynomial means many terms. A polynomial
with just one term is called a monomial, with two terms - a binomial, with three terms - a
trinomial, etc. You might observe the nth degree polynomial seen above has n+1 terms
(some of which could have a zero for a coefficient and hence not be written).
GRAPHS OF POLYNOMIALS
Polynomial function graphs are continuous smooth curves which means that they
have no holes or gaps or corners or cusps. (See the figures below.)
o
Gap
Hole
Corner
Cusp
Polynomial
Not a polynomial
Examine the five monomial graphs below.
y=x
2
y=x
y = x3
y = x4
y = x5
Notice that for the odd degree graphs the ends go in opposite (up-down) directions,
whereas for the even degree graphs the ends go the same direction (up in this case). Now
think about transformations and realize that if you negated the functions above, the
graphs would be reflected about the x-axis and the right ends (as well as the left ends for
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the even degree functions) would go down. This pattern is seen in all polynomial graphs
and is stated below.
If a polynomial's leading coefficient is positive, then its graph goes up
to the right (and to the left as well for even degree polynomials), and
conversely if its leading coefficient is negative, then its graph goes
down to the right (and to the left as well for even degree polynomials).
Now look at the five graphs again. Notice that the first degree graph has 0 bends, the
second degree graph has 1 bend, and the third degree graph has 2 bends. In general the
following is true for polynomial graphs.
An nth degree polynomial graph has up to n-1 bends.
Exercise: Use your grapher to produce pictures of various polynomials
and convince yourself of the truth of the preceding statement.
Now that you know how to identify the shape of a polynomial graph, it is time to
develop techniques for determining some details concerning the graph. What we would
be interested in is (1) Where are the intercepts? and (2) Where are the turning points of
the graph? With regard to turning points, except for the turning point of a parabola
(which you recall is called the vertex), finding turning points requires calculus or the use
of a computer algebra system (CAS). Consequently, at this time we will not be
concerned with finding turning points for polynomials with degree greater than two.
Now with regard to intercepts, you first should realize that the y-intercept occurs when x
= 0 which produces (0,an) as the y-intercept. Secondly, as you have seen previously, the
x-intercepts are real zeros, often called roots. Although methods exist to find zeros of
cubic and quartic equations they are generally cumbersome and not used today. If
finding exact roots that exist for a polynomial equation is desired, it can be done by the
methods discussed in the next subsection.
ROOTS OF POLYNOMIAL FUNCTIONS
A root of an equation, y = f(x), is a real zero of the equation, or graphically speaking, an
x-intercept of the graph.
Consider the polynomial equation y = 6x4 - 7x3 - 9x2 + 7x + 3
which factors into y = (2x - 3) (3x + 1) ( x + 1) ( x - 1)
3
1
and thus has x-intercepts at 2 , - 3 , - 1, 1 ( the roots or zeros).
Observe that for each zero, the numerator is a factor of the constant term, 3, and the
denominator is a factor of the leading coefficient, 6. This makes sense since when you
multiply the factors together 2 • 3 • 1 • 1 = 6 and - 3 • 1 • 1 • - 1 = 3. The generalization
of this notion is called the Rational Root Theorem.
7.1 : Polynomial Functions
If f(x) = aoxn + a1xn - 1 + . . . + an-2x2 + an-1x + an is a polynomial with integer
p
coefficients, then each of its rational roots is of the form q where p and q have no
common factors other than 1 and such that p is a factor of an and q is a factor of ao.
p
Proof: Let q be a root.
Then f(q
p
) = ao(pq )n + a1(pq )n - 1 + . . . + an-2(pq )2 + an-1(pq ) + an = 0
Multiplying by qn produces aopn + a1pn-1q + . . . + an-2p2qn-2 + an-1pqn-1 + anqn = 0.
Now since the sum is 0 and p is a factor of each of the first n terms, it must also be a
factor of the n+1st term. P is not a factor of qn and hence must be a factor of an. Similar
analysis shows that q must be a factor of ao.
Now let us pretend that we do not know the zeros of y = 6x4 - 7x3 - 9x2 + 7x + 3.
The Rational Zeros Test tells us that any possible rational zeros
are
factors of 3
(1,3)
± factors of 6 = ± (1,2,3,6) = ±
{ 1, 3, 12 , 32
1
1
, 3 , 6
}
Now recall that if r is a zero, then x - r is a factor. We can use our grapher to graph the
polynomial, estimate which of our possible rational zeros is likely to be a true zero, and
then use long division to check and see if it is a zero. For example, x = 1 appears to be a
zero.
x-1
6 x3 - x2 - 10 x - 3
6 x4 - 7 x3 - 9 x2 + 7 x + 3
6 x4 - 6x3
- x 3 - 9 x2 + 7 x + 3
- x 3 + x2
- 10 x2 + 7 x + 3
- 10 x2 + 10 x
- 3x+3
- 3x+3
0
Now it would be nice if there was an easier method other than long division to find/verify
roots of polynomial equations. Happily, there is and it is called synthetic division.
Review the preceding long division and observe that after each division, the first term
always cancels out upon subtraction and also that writing the x's is a waste of time and
effort. Eliminating the unnecessary work produces the following procedure.
1
6
-7
6
-9
-1
7
-10
3
-3
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6
-1
- 10
-3
0
(1) Write the coefficients in order of descending powers of x.
(if any powers of x are missing use a coefficient of 0)
(2) Divide by the root possibility
(3) Bring down the first coefficient
(4) Multiply and add
Observe that the numbers in the bottom row; 6, - 1, - 10, and - 3 are the coefficients of
the quotient polynomial as seen earlier in the long division and that as before, for this
problem, 0 is the final remainder.
To generalize this development, let a polynomial P(x) be divided by x - r. Call the
quotient polynomial Q(x) and the remainder R.
P(x) = (x - r)Q(x) + R
If R = 0, then x - r is a factor and r is a root (This is often called the Factor Theorem)
Suppose R ≠ 0. Observe that P(r) = (r - r) Q(r) + R = R.
In other words, the remainder is the functional value of the polynomial when x = r.
(This result is often called the Remainder Theorem)
Example:
2
6
6
-7
12
5
-9
10
1
7
2
9
3
18
21
Thus P(2) = 21
Because of this development, synthetic division is sometimes called synthetic
substitution. Ordered pairs generated by y = f(x) can be found by synthetically dividing
by the x value. The remainder is the corresponding y-value.
Now return to our initial example, y = 6x4 - 7x3 - 9x2 + 7x + 3.
We previously found that the possible rational roots are ± { 1, 3,
3
1
The following work shows that the roots are 1, - 1 , 2 and - 3 .
1
6
-1
6
3/2 6
-1/3
6
6
-7
6
-1
-6
-7
9
2
-2
0
-9
-1
- 10
7
-3
3
0
7
- 10
-3
3
0
1 3
1
1
2 , 2 , 3 , 6
}
3
-3
0
Of course this process can be stopped at any point when you can find the roots by some
other method. For example, after finding roots of 1 and - 1, the quotient polynomial
6x2 - 7x - 3 is quadratic and thus the other roots can be found by quadratic methods.
7.1 : Polynomial Functions
Now for the fourth degree polynomial equation we have been working with, we
found four roots. In general, an nth. degree polynomial equation will have n zeros
and up to n roots (real zeros). This result comes from repeated application of The
Fundamental Theorem of Algebra:
If y = f(x) is a polynomial equation of degree n > 0,
then it has at least one complex zero.
Now if r1 is a zero then x - r1 is a factor and P(x) = (x - r1) Q1 (x)
Q1(x) will be degree n - 1.
If n - 1 > 0, apply the Fundamental Theorem again and
P(x) = (x - r1)(x - r2)Q2(x) where Q2(x) is degree n - 2.
Continuing in this manner
P(x) = (x - r1)(x - r2)(x - r3) . . . (x - rn)a0.
For our initial example
1
3
6x4 - 7x3 - 9x2 + 7x + 3 = (x - 1) ( x + 1) ( x - 2 ) (x + 3 ) 6
and thus the zeros set is {1, -1, 3/2, -1/3}.
Now some roots of polynomial equations may be real numbers which are not
rational. These numbers, which are called irrational, may be either found exactly using
quadratic methods or approximately by numerical methods or by using your grapher.
For example, consider the equation
y = f(x) = x2 + x - 1.
The only possible rational roots are ±1, neither of which work.
Thus any real roots must be irrational.
Using the quadratic equation produces roots of x =
-1±
12 - 4 (1) (-1)
2
1± 5
2
≈ 1.62 or - .62
=
With regard to numerical methods, observe that for the example above, f (- 1) = -1
and f(2) = 5. Since the graph is continuous and below the x-axis at -1 and above the
x-axis at 2, there is at least one real root between x = -1 and x = 2. This root set can be
found by bracketing, wherein you keep narrowing down the interval for which there is a
function value sign change to approximate the real roots to any degree of accuracy. This
technique lends itself nicely to employing a computer program. We will not be
concerned with this process in this book.
NON REAL ZEROS
Now some of the zeros of a polynomial may be non-real numbers of the form a + bi
where a and b are real numbers and I = √(-1). Observe that if b = 0, then a+bi is a real
number. If a = 0 and b ≠ 0, then a+bi is called a pure imaginary number. Two complex
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numbers a+bi and c+di are equal if and only if a = b. We will shortly show that if a +
bi is a zero of a polynomial equation with real coefficients, then its conjugate a - bi is
also a zero. In other words:
For a polynomial equation with real coefficients,
complex zeros always occur in conjugate pairs.
For example consider y = x2 + x + 1.
The zeros are x =
-1± 1-4
1
3
= -2 ± 2 i
2
So the notion of conjugate pairs seems reasonable. To justify this notion necessitates
some arithmetic with complex numbers. First of all, the arithmetical operations of
addition, subtraction and multiplication are performed as if complex numbers were
normal algebraic binomials.
Addition: (a + bi) + (c + di) = (a + c) + (b + d) i
Subtraction: (a + bi) - (c + di) = (a - c) + (b - d) i
Multiplication: (a + bi) • (c + di) = ac + adi + bci + bdi2 = (ac - bd) + (ad + bc)i
Now to see how to divide complex numbers, recall that division is defined as the inverse
x
of multiplication, i.e. y = z means that there exists z such that yz = x.
a + bi
So first suppose c + di = x + yi
Then a + bi = (c + di)•(x + yi) = (cx - dy) + (dx + cy)i
cx - dy = a
For this to be true, dx + cy = b
Solving this system of equations for x and y yields
bc - ad
ac + bd
+
i
x + yi = 2
c2 + d2
c + d2
[You should verify this.]
Examining the result above reveals that the expression c2+ d2 occurs in each
denominator. Also observe that (c + di)(c - di) = c2+ d2. This suggests that:
To divide complex numbers, multiply both the numerator and the denominator by
the conjugate of the denominator.
Example:
7.1 : Polynomial Functions
-4 + 5i
-4 + 5i
2 + 3i -8 - 12i + 10i - 15 -23 - 2i
23 2
=
•
=
=
=
2 - 3i
2 - 3i
2 + 3i
4 + 6i - 6i + 9
13
13 - 13 i
Now you probably realize that powers of complex numbers can be found by repeated
multiplication. For example:
(a + bi)2 = a2 + 2abi + b2i2 = (a2 - b2) - 2bi.
In the next chapter you will see how to use trig to find powers and roots of complex
numbers.
This subsection concludes with an explanation, which uses powers of
complex numbers, of why complex zeros of polynomial equations with real coefficients
occur in conjugate pairs.
First, we need Fact A: for a complex number,
the conjugate of a sum equals the sum of the conjugates.
(a + bi) + (c + di) = (a + c) + (b + d)i and the conjugate is (a + c) - (b + d)i.
Now observe that the sum of the congugates is (a - bi) + (c - di) = (a + c) - (b + d)i
which is the same complex number as the conjugate of the sum.
Second, we need Fact B: for a complex number,
the power of the conjugate equals the conjugate of the power.
Observe that (a + bi)2 = (a2 - b2) + 2abi and the conjugate is (a2 - b2) – 2abi.
Now observe that (a - bi)2 = (a2 - b2) + 2abi and the conjugate is (a2 - b2) – 2abi.
So this notion seems plausible. In general, using the Binomial Theorem you can show
this notion is true for any power.
Now let z = a + bi be a zero of the polynomial equation with real coefficients
p(x) = a0xn + a1xn-1 + . . . + an-2x2 + an-1x + an. Then
p(z) = a0zn + a1zn-1 + . . . + an-2z2 + an-1z + an = 0.
To facilitate the rest of this discussion, we will adopt the convention that placing a bar
_______
over a complex number denotes the conjugate of that complex number, i.e. a + bi = a
- bi.
Now since the conjugate of a real number is the same real number, the conjugate of
p(z) = 0.
____ ____________________________________
Thus p(z) = a0zn + a1zn-1 + . . . + an-2z2 + an-1z + an
= 0.
By Fact A,
____
____
____ __
____
a0zn + a1zn-1 + . . . + an-2z2 + an-1z + an = 0.
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By Fact B,
_
_
_
_
a0 z n + a1 z n-1 + . . . + an-2 z 2 + an-1 z + an = 0.
_
This means that p( z ) = 0 and hence the conjugate is also a zero of p(x).
INEQUALITIES
You previously have worked with linear and quadratic inequalities. Although
strictly algebraic techniques can be applied to solve polynomial inequalities, generally it
is easier to employ a graphical approach. By examining the graph (including the xintercepts) of some polynomial function, you can see where points are above, on, or
below the x-axis and thus solve a related inequality.
Example: Solve x3 - 3x2 -10x ≤ 0 for x.
The graph of y = x3 - 3x2 -10x is shown below.
Clearly y ≤ 0 when x ≤ -2 or 0 ≤ x ≤ 5.
Hence the solution set for the inequality is {x | x ≤ -2 or 0 ≤ x ≤ 5}.
In interval notation this is (- ∞,-2] ∪ [0,5] .
Problem Set 7.1
1.
Describe the end behavior and identify the possible number of bends in the graph for
each of the following polynomial equations.
7.1 : Polynomial Functions
(a) y = 2x - 3
(b) y = 3x2 + 2x - 1
(d) y = -5x4 + 6x3 - 2x + 1
(c) y = -4x3 - 5x + 6
(e) y = 6x5 + 3
2.
Can the graph of a polynomial equation have no y-intercepts? Why or why not?
Can it have no x-intercepts? Why or why not?
3.
Given that 2, -1, and -3 are the roots of a cubic polynomial equation whose graph
passes through (0,12), write its equation in standard form.
4.
Given that x = 3 is a root of the polynomial 2x3 - 5x2 - 4x + 3 = 0 find the other
roots.
5.
List the possible rational roots for y = 6x4 + 5x3 - 65 x2 + 50x + 24 and then use
synthetic division to assist in finding all the roots for this equation.
6.
Prove that the equation x3 + x2 - 3 = 0 has no rational roots. Then show that it
does have an irrational root between x = 1 and x = 2.
7.
Given that i is a zero, find all the zeros for y = x4 + 2x3 - x2 + 2x -2 = 0.
8.
Examine the graphs of y = (x - 1)2, (x - 1)3, (x - 1)4, (x - 1)5, etc. Observe that the
roots for these equations are x = 1 with increasing multiplicity. Make a general
conjecture concerning root multiplicity and what the graph is doing at the intercept.
9.
Discuss the difference between a root and a zero of a polynomial equation.
10. Write two polynomials (of differing degrees) which each has the following
characteristics: Crosses the x-axis at –2. Crosses the y-axis at 1. Is above the x-axis
between –2 and 1
11. Find a possibility for my equation using the following clues.
A.
B.
C.
D.
I am a 4th degree polynomial.
I am tangent to the x-axis at x = 2.
My y-intercept is y = 4.
I go through (1,8) and (-1,18).
12. An n cm by (n +3) cm by (n+9) cm rectangular wooden prism is painted and then
cut into 1 cm cubes through cuts parallel to the faces. Following this, half of the
cubes have at least one face painted. What are the dimensions of the original block?
13. Solve the following inequalities. Write your answers in interval notation.
(a) x3 - x ≤ 0
(b) x2 + 12 < 4x
(c) x5 + x2 ≥ 2x3 + 2
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7.2 : Rational and Algebraic Functions
7.2 RATIONAL AND ALGEBRAIC FUNCTIONS
Just as a rational number is a positive or negative ratio of two whole numbers,
a rational function is the ratio of two polynomials.
P(x)
R(x) = Q(x)
ASYMPTOTES
1
Possibly the simplest example of a rational function is y = x which as you previously
saw, graphs as a rotated hyperbola.
Note that the x and y axes are the asymptotes for this hyperbola. Now more complicated
rational functions can be created by employing the operations seen earlier in the Algebra
1
3x - 5
of Functions Section (Section 3.5) . For example y = 3 + x - 2 = x - 2 translates the
1
graph of y = x right 2 and up 3. Thus the new asymptotes are x = 2 and y = 3. The
new graph is shown below.
y
y=3
x
x=2
Now examination of the equation and the graph may prompt you to conjecture two
notions. First, you might believe that a vertical asymptote occurs where the denominator
equals 0. This conjecture is true provided the numerator polynomial does not also equal
0 at the x value that makes the denominator 0.
P(x)
For a rational function R(x) = Q(x) ,
vertical asymptotes occur where Q(x) = 0 and P(x) ≠ 0.
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Chapter 7 : Elementary Functions I
Now what if both P(x) and Q(x) are both 0 for some x value. Then the rational function
can be simplified.
Example: Consider y =
x2 - 3x -10
which simplifies to y = x + 2 where x ≠ 5.
x-5
o (5,7)
The graph is a line containing a hole at (5,7)!
A second conjecture you might make is that a horizontal asymptote occurs at the
ratio of the leading coefficients of the polynomials P(x) and Q(x). This notion is true
provided the degrees of P and Q are equal. To understand why this is so, consider the
work below.
1 5
2-x+ 3
x
2x3 - x2 + 5
y= 3
=
where x ≠ 0.
3x + 5x +2 3 + 5 + 2
x2 x3
As x →± ∞ , all the terms with "x" in the denominator ⇒ 0.
2
2
Thus y → 3 , so y = 3 is a horizontal asymptote.
The preceding discussion takes care of the case where P and Q have the same degree.
There are two other cases. If the degree of P is less than that of Q, then the x-axis (y = 0)
is a horizontal asymptote.
Exercise: Using algebra or graphs, convince yourself the preceding
statement is true.
Finally if the degree of P is greater than that of Q, then there is no horizontal asymptote.
This should be obvious because as x →± ∞ , the numerator grows more rapidly than the
denominator and hence y →± ∞. It should be noted however, that rational functions of
this type do exhibit asymptotic behavior. This class of graphs asymptotically approach a
polynomial graph of degree equal to the difference in degree between P and Q. To
3x2 - 2x + 4
. Using long division this equation can
illustrate this notion, consider y =
x+2
20
20
be written as y = 3x - 8 + x + 2 . Now as x →± ∞, x + 2 → 0 and thus the graph
approaches the line y = 3x - 8 which is frequently referred to as a slant asymptote.
7.2 : Rational and Algebraic Functions
The preceding development of the relationship between the degrees of P and Q and
axn + • • •
can be summarized as follows.
the horizontal asymptote for R(x) = m
bx + • • •
a
n = m ⇒ the horizontal asymptote is y = b
n < m ⇒ the horizontal asymptote is y = 0
n > m ⇒ there is no horizontal asymptote (but there is asymptotic behavior)
GRAPHING RATIONAL FUNCTIONS
The first step in graphing rational functions is to graph the asymptotes (as dotted
lines) and plot the intercepts. If the graph has a horizontal asymptote, the next step is to
see if the graph crosses that asymptote. You probably realize that this task is
accomplished by setting the function equal to the asymptote value and solving this
resulting equation. The final step is to test values in the equation to determine how the
graph at that point relates to the asymptotes and axes. An example problem will serve to
demonstrate the process.
Example: Graph y =
x2 - 3x - 10
x-2
First, observe that this equation can be written as y =
(x - 5)(x + 2)
,
x-2
and hence there is a vertical asymptote at x = 2 and no horizontal asymptote
12
Second, observe that this equation can be written as y = x - 1 - x - 2
and hence y = x - 1 is a slant asymptote.
Third, observe that the intercepts are (5,0), (-2,0) and (0,5).
Fourth, graph the asymptotes and the intercepts.
Y
Y=X-1
X
X=2
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Now we need to know where y is positive or negative. You notice that there are
three factors in the equation, x -5, x + 2, and x -2. When an odd number of these
factors is/are negative then y < 0 and when an even number of these factors is/are
negative then y > 0. A number line is useful in determining the sign of y.
+
-2
_2
+
5
We can now complete the graph as shown below.
x=2
(0,5)
(-2,0)
(5,0)
y=x-1
RATIONAL INEQUALITIES
The number line work used at the end of last subsection is the easiest technique with
(x - 5)(x + 2)
≥ 0. As seen
which to solve rational inequalities. Consider the inequality
x-2
earlier the break points are -2, 2 and 5. An even number of the factors are positive when
x > 5 or when -2 < x <2. Also the expression is zero when x = -2 or 5. Thus the solution
set for the inequality is [x | -2 ≤ x < 2 or x ≥ 5], which in interval notation is
[-2,2) ∪ [5,∞ ).
ALGEBRAIC FUNCTIONS
Recall that in arithmetic you studied whole numbers, then rational numbers and then
irrational numbers to complete the study of real numbers. You also recall that a whole
number is a rational number and that a rational number is a real number. In other words
there is a subset relationship in the progression. We will now complete the analogous
study with functions. Perhaps you have previously realized that a polynomial function is
a rational function (with denominator = 1). Similarly a rational function is a special type
of what is classified as an algebraic function. Loosely described, an Algebraic Function
involves only the operations of addition, subtraction, multiplication, division and taking
7.2 : Rational and Algebraic Functions
roots (rational functions do not have roots). Almost all of the functions we have studied,
including the Square Root Function, are algebraic. As with numbers, it is customary to
label functions with the most restrictive name. For example, even though it is correct to
call y = 2x - 3 an algebraic function we call it a linear function. On the other hand, the
function y = 2 - 2x- 3 is properly called algebraic. Now if you recognize that the
algebraic function mentioned in the last sentence is a transformed square root function,
you can easily sketch the graph. Many algebraic functions can be graphed in this
manner, but many more are extremely difficult to graph. Consequently it is nice to rely
heavily on a grapher to graph most algebraic functions.
Problem Set 7.2
1.
Graph each of the following, showing exact values for coordinates of the intercepts
and the equations of the vertical, horizontal and/or slant asymptotes.
x
2x2 - 8
x3 - 1
2x2 + 2x - 4
(a) y = 2
(b) y = 2
(c) y = 2
(d) y =
x - 16
x + 5x
x -9
x2
2.
Find a possibility for my equation using the following clues.
A. I am rational.
B. I have vertical asymptotes at x = -2 and x = 3.
C. I have a horizontal asymptote at y = 4.
D. I have intercepts at (2,0) and (-3,0).
3. The temperature T (in oF), of a person during an illness is given by the equation
4t
T= 2
+ 98.6 where time t, is given in hours since the onset of the illness.
t +1
(a) Find the horizontal asymptote.
(b) Find the maximum temperature during the illness.
(c) Find the interval of time in which temperature is greater than 100o.
4. UPS is contracting you to design a new closed box for shipping that is made with a
square base and has a volume of 6 cubic feet.
(a) Find, in terms of the base dimension x, a function for the surface area of the box.
(b) What are the dimensions (to 0.1 inch) of the box that minimize surface area?
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5. The acceleration, in m/sec2, of an object due to gravity, g, can be measured by the
equation shown below, where height, h, represents meters above sea level.
3.99 • 1014
g(h) =
(6.374•106 + h)2
(a) Find the horizontal asymptote of g(h).
(b) If the Sears Tower in Chicago, Illinois, is 443 m above sea level, what is the
acceleration due to gravity from the top of the tower?
(c) Set g(h) = 0 and solve for h. How do you interpret the solution in the context of
this problem?
6.
Solve the following inequalities. Write your answers in interval notation.
x
x2 - 9
(a) 2
> 0 (b) x - 6 ≤ 0
x - 16
7.
Use transformations to sketch the algebraic function y = -2 + 3 x - 2 .
8.
Plot points and/or use a grapher to compare the graphs of y = x2, y = x3/2, y = x,
y = x2/3, and y = x1/2 for x ≥ 0. Discuss the "bending" nature of the graphs and then
make a general conjecture concerning the graphs of y = xp/q for x ≥ 0, where p/q is a
rational number.
7.3 : Exponential and Log Functions
7.3 EXPONENTIAL AND LOG FUNCTIONS
INTRODUCTION
An Amoebae can do an unusual trick. It multiplies by dividing. After about a day a
new amoebae can divide and form 2 amoeba. The next day the 2 amoebas divide and
form 4. The next day the 4 amoeba divide and form 8. And so on. The number of
amoeba is a function of the time that has passed.
Time in days 0 1 2
# of amoeba 1 2 4
3 4 5 6 7
8 9
10
8 16 32 64 128 256 512 1024
Note: The first row is an arithmetic sequence and the second row is a geometric
sequence, both of which were studied in an earlier chapter.
John Napier, the Scottish mathematician who lived from 1550 to 1617, noticed a
remarkable feature concerning a table such as the one above. Observe that 16 x 32 = 512
and that 4 + 5= 9. Multiplication in the second row corresponds to addition in the first
row. So just like Amoeba, we can multiply without really multiplying.
Upon further investigation of his observations, Napier developed logarithms.
Logarithms were the third of 4 remarkable labor saving inventions. Many of the fields in
which numerical calculations are important, such as astronomy, navigation, trade,
engineering and war. made ever increasing demands that these computations be
performed more quickly and accurately. These demands were met successively by the
invention of:
(1) the Hindu-Arabic numeration system
(2) decimals
(3) logarithms
(4) modern computing machinery (Calculators and computers).
Referring back to the amoeba table, Napier would call the numbers in the first
sequence the logarithms of the numbers in the second row. Although at the time of
Napier's invention of logarithms, exponents were not yet known, today we would call the
numbers in the first sequence the exponents of the base 2 that produces the numbers in
the second sequence. Hence we say a logarithm is an exponent.
After 25 years of intense effort, in 1614, Napier published A Description of the
Wonderful Law of Logarithms. Three hundred years later Lord Mouton wrote:
The invention of logarithms came upon the world as a bolt from the blue.
No previous work had led up to it, foreshadowed it or heralded its arrival.
It stands isolated, breaking in upon human thought abruptly without
borrowing from the work of other intellectuals or following known lines
of mathematical thought.
Napier's book contained log tables and Laws of Logarithms. Henry Briggs (1561-1631),
a professor of geometry at Greshan College in London, read Napier's book and was so
inspired that he traveled to Edinburgh to visit Napier. Together they decided that 10
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Chapter 7 : Elementary Functions I
would be the most useful base for log tables. Ten years later, in 1624, Briggs published a
book of base 10 logarithms. This table, coupled with Napier’s Laws of Logarithms and
Scientific Notation, facilitated in Briggs own words, "the more easy working of questions
in arithmetic and geometry. By them all troublesome multiplication's are avoided and
performed only by addition."
Shown below is a table for numbers from 1-10 and their base 10 log values rounded
to two decimal places.
Numbers
1
2
3
4
5
Logs
0 0.30 0.48 0.60
0.70
6
7
8
9
0.78 0.85
0.90
0.95
10
1
Observe that to multiply 2 x 3 you could add 0.30 + 0.48 = 0.78 and read the answer
2x3 as 6. This is actually how a slide rule multiplies using logarithmic scales.
Originally logarithms were utilized to reduce the level of arithmetical computation.
Powers and roots were converted to products and quotients and multiplication and
division were performed by additions and subtractions. Today, with our modern high
speed computing technology, the labor-saving aspect of logarithms has been greatly
reduced, but log functions are still utilized for many applications related to sight, sound
and the other senses, including c-e-n-t-s (finance). For example, both the decibel scale
for measuring the loudness of sound and the Richter Scale for measuring the intensity of
earthquakes are logarithmic scales. Logarithmic functions will be studied in more detail
later in this section.
Now recall that as mentioned earlier, exponents were not known at the time Napier
invented logarithms. Today it is customary to study exponential functions before
logarithmic functions and then study log functions as the inverses of exponential
functions. Then all the log work can be based on the previous exponential work.
Exponential functions have great utility in modeling population growth, radioactive
decay and compound interest (finance) problems which will be studied later in this
chapter.
EXPONENTS
Once upon a time an English couple attended their first baseball game in America.
Having never seen baseball played before, they were very confused by the play but
enjoyed the antics of the "performers". After some time the couple went to the snack bar
to try some "American cuisine". The snack bar attendant asked them, "How's the game
going?" The English gentleman recalled seeing the scoreboard as shown below,
1 0 3
1 0 2
0 0 0
0 0 0
0 0 0
0 0 0
and replied, I don't know what's going on, but the score is one hundred three million to
one hundred two million.
Recall that large numbers, such as the ones the English couple thought they were
seeing, as well as small numbers are frequently written in scientific notation as a decimal
7.3 : Exponential and Log Functions
number between 1 and 10 multiplied times ten to an integer power. For example
103,000,000 is written as 1.03 x 108. Not only is scientific notation a more compact
representation of such "unhandy" numbers, but it also facilitates computing with such
numbers. For example, consider the problem of multiplying 103,000,000 x 0.0000025.
Written in scientific notation this problem is (1.03 x 108) x (2.5 x 10-6 ) which, by the
commutative property of multiplication, is equal to 1.03 x 2.5 x 108 x 10-6 which
equals 2.575 x 102 or 257.5. This computation makes use of the familiar Laws of
Exponents which are stated below.
Laws of Exponents for Real Numbers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
ao=1
an=a•a•a• ... •a
1
a -m = m
a
m
(ab) = a m b m
⎛ a ⎞ m = am
⎜b⎟
⎝ ⎠
bm
⎛a ⎞ -m = ⎛b⎞ m
⎜b⎟
⎜a⎟
⎝ ⎠
⎝ ⎠
m
n
m+n
a a =a
am
= am - n
an
n
(am) = amn
p
q
ap/q = ( a )
(for n factors)
EXPONENTIAL FUNCTIONS
Casey Stengel, a former coach of the New York Yankees baseball team, once
remarked, "If you come to a fork in the road, take it." Consider the diagram shown
below.
•
start •
3•
•
•
2
•
3
•
•1
•
3 • 2 • 3
Observe that after Casey's first fork in the road, he could have gone 2 different ways.
After his second fork, he could have gone 4 different ways. After 3 forks - 8 different
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Chapter 7 : Elementary Functions I
ways. A table showing the number of different paths as a function of the number of
forks is shown below.
Table I
# forks 1
# paths 2
2
4
3
8
4
16
...
...
10
...
1024 . . .
n
2n
The expression 2n is called the generator of the sequence of the number of paths.
Recall that the sequence seen above is an example of a geometric sequence and that a
geometric sequence is a sequence of numbers for which there is a common ratio between
succeeding terms. In this case the common ratio is 2.
Exercise: Suppose Casey came to successive crossroads where there were
3 ways to go each time. Make a table, similar to the one shown earlier,
showing the number of crossroads encountered and the number of
different paths Casey could have gone.
# forks
# paths
What is the generator of this sequence?
Writing the function represented by your table as an equation produces P = f(n) = 2n.
An equation of this type where the independent variable (n in this case) occurs as an
exponent is called an exponential equation. In general, a real valued exponential
equation is of the form: y = ax where a is any positive number not equal to 1 and x is
any real number.
Consider the function y = 2x. Use your grapher and confirm that the picture of this
function looks as follows.
y = 2x
(0,1)
y=0
The point (0,1) is the y-intercept and the line y = 0 is the horizontal asymptote
(Recall that an asymptote is a line that a graph approaches as x ∅ ± ∞ .)
In general all graphs of y = ax, a > 1 resemble the picture shown above.
7.3 : Exponential and Log Functions
1 x
Now consider y =( 2 ) . Use your grapher to confirm that the picture of this
function looks as follows.
(0,1)
y=0
Observe that the y intercept and the asymptote remain the same but now the curve
approaches the asymptote to the right. This result agrees with our transformational work
seen previously because
graph over the y-axis.
x
1 x
y = ( 2 ) = (2-1) = 2-x
and we recall that negating x reflects the
As an aid to graphing exponential functions in general, the standard transformational
rules are reviewed below.
Let y = f(x) and let a and b be positive numbers.
y = f(x - h) translates left h units if h < 0 and right h units if h > 0 .
y = f(x) + k translates up k units if k > 0 and down k units if k < 0.
y = f(bx) horizontally shrinks if b > 1 and stretches if b < 1.
y = af(x) vertically shrinks if a < 1 and stretches if a > 1.
y = f(-x) reflects the graph over the y-axis.
y = - f(x) reflects the graph over the x-axis.
Remember: Reflection and dilation take priority over translation.
THE NATURAL EXPONENTIAL FUNCTION
In Section 7.5 we will develop the "Pert" formula A = Pert, used in compound
interest problems employing continuous compounding. This exponential function has the
irrational number e ≈ 2.718 as its base and is useful in a wide range of applications
including the financial problems you will see later in this chapter.
LOGARITHMIC FUNCTIONS
Consider the previously studied function y = f(x) = ax , a > 0, a ≠ 1
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Chapter 7 : Elementary Functions I
Recall from the introduction that a logarithm is an exponent. In particular it is the
exponent of the base that produces the number. Thus in log form we write Loga y = x.
As examples:
1
1
log2 8 = 3 because 23 = 8 ; log2 16 = -4 because 2-4 = 16 ;
a3 = a3 ; . . .
loga a3 =3 because
Now recall from your study of inverse functions that the inverse is obtained by
interchanging x and y. Thus y = loga x and y = ax are inverses and their graphs are
symmetric about the line y = x.
Example:
y = 2x
y = log x
2
y=x
Consequently to graph a log function, just think of the related exponential function and
then flip it over the line y = x.
Now every exponential function graph has a horizontal asymptote so every log
function graph has a vertical asymptote. The transformations that we have used before
apply equally to log graphs. However transformations can be described differently
because of the conversions that can be made on log functions according to the Laws of
Logarithms which are stated below
For a, x, y > 0 and a ≠ 1.
2.
loga xy = loga x + loga y (The log of a product equals the sum of the logs of the factors.)
x
loga y = loga x - loga y
(The log of a quotient equals the difference of the logs of the
3.
loga xy = y loga x
1.
factors.)
(The log of a quantity raised to a power equals the power times the log
of the quantity.)
4.
loga b =
log b
c
log a
c
(This law often is called the change of base formula.)
These laws can be proved using the Laws of Exponents. For example to prove Law 1:
7.3 : Exponential and Log Functions
Let m = loga x , n = loga y
Then am = x, an = y and xy = aman= am+n
Thus loga xy = loga am+n = m + n = loga x + log a y
Some other useful properties that can be deduced from the Log Laws are listed
below.
loga x
a
=x
loga ax = x
loga 1/a = -1
THE NATURAL LOG FUNCTION
Log and exponential functions have substantial applications in a large number of real
world problems. Of particular use is the natural log function y = loge x which is
denoted by ln x. Ln stands for log natural. You will see work with this function as well
as other log and exponential functions in the problem set for this section and also in the
next chapter with applications.
Note: Similarly, there is a shorthand notation for Base 10 logs which are indicated,
omitting the base as log x.
Problem Set 7.3
1.
Sketch the graphs of each of the following functions showing exact values for
coordinates of the intercepts and the equations of the asymptotes.
(a) y = -1 + 2x
2.
(b) y = 3x-2 (c) y = - 4x
(d) y = 2 - 3(4x+3)
Find a possibility for my equation using the following clues.
A.
B.
C.
D.
I am exponential with base 2.
I have a horizontal asymptote at y =-2.
My y-intercept is y = 3.
I am an increasing function.
3.
Write the inverse, using the same base, for each of the equations in problem 1 and
then sketch these graphs showing exact values for coordinates of the intercepts and
the equations of the asymptotes.
4.
Find a possibility for my equation using the following clues.
A. I am log with base 3.
B. I have a vertical asymptote at x = 3.
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Chapter 7 : Elementary Functions I
C. My x-intercept is x = 1.
D. I am a decreasing function.
5.
Log Law I was proved in this section. Prove the other Log Laws.
6.
Given logb 2 = K and logb 3 = L, find in terms of K and L:
(a) logb 6 (b) logb 8
7.
1
(c) logb 12
Evaluate:
(a) log9 (log5(log2 32))
(b) log2 56 - log4 49
(c) 10
log100 9
8.
Discuss why logs base 10 are called computational logs whereas logs base e are
called natural logs.
9.
If the number 9 9 were typed on a piece of paper using one centimeter for each
group of 3 digits, how many kilometers long would the paper have to be?
(Hint: Use the base 10 logarithm to find how many digits are in the number.)
9
7.4 : Exponential and Log Applications
7.4 EXPONENTIAL AND LOG APPLICATIONS
You studied Log and Exponential Functions and their graphs in the last section. You
also learned the Laws of Logarithms listed again below.
For a, x, y > 0 and a ≠ 1.
1.
2.
loga xy = loga x + loga y (The log of a product equals the sum of the logs of the factors.)
x
loga y = loga x - loga y (The log of a quotient equals the difference of the logs of the
factors.)
3.
loga xy = y loga x
4.
logc b
loga b = log a (This law often is called the change of base formula.)
c
(The log of a quantity raised to a power equals the power times the log
of the quantity.)
These Log Laws are used to simplify log expressions that occur in equations that
model many real world problems. For example consider the equation shown below.
log 3x - log 4 = 2 (remember this is base 10)
3x
Using Law 2 produces
log 4 = 2
3x
2
Changing to exponential form yields
4 = 10 = 100
400
And so
x= 3
Now in the preceding example, notice that we solved a log equation by changing it
to exponential form. In an analogous manner, an exponential equation is solved by
changing it to log form.
Consider the equation
Changing to log form yields
23x - 2 = 5
3x - 2 = log2 5
And thus
x=
2 + log2 5
3
Now if we wanted a decimal approximation for the answer, we can use the Change of
Base Formula (Law 4) and write
x=
log 5
2+ log 2
3
≈ 1.44
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Chapter 7 : Elementary Functions I
Another way to solve this equation is to perform the operation of taking the log of both
sides of the equation as shown below.
23x-2 = 5
log 23x-2 = log 5
(3x-2) log 2 = log 5
log 5
3x-2 = log 2
log 5
2+ log 2
x=
≈ 1.44
3
Some log and exponential equations are solved by employing techniques that were used
to solve algebraic equations, as illustrated in the next two examples.
Example 1:
ln2 x - ln x5 + 4 = 0
ln2 x - 5 ln x + 4 = 0
(ln x - 4)(ln x -1) = 0
ln x = 4 ln x = 1
Example 2: 32x - 7(3x) - 8 = 0
(3x - 8)(3x + 1) = 0
3x = 8 3x = -1
x = log3 8
x = e4 x = e
(Note that 3x > 0)
EARTHQUAKES
In 1935, geologist Charles Richter defined the magnitude M (now, in his honor,
I
called the Richter Number, R ) of an earthquake as M = log I where I is the intensity
0
of the quake and I0 is the smallest measurable intensity. If we take Richter's formula and
solve for I, we find that I = 10M I0. Thus the Richter number for an earthquake measures
relative intensity. This idea is used to measure the relative intensity of one earthquake to
another. For example, the 1906 San Francisco quake is believed to have been a 8.25
quake and the 1989 San Francisco quake was measured to be a 7.1 quake. To find the
relative intensity between the two quakes observe that :
I1906
I
and 7.1 = log 1989
I0
I0
I
I
I
Hence 8.25 - 7.1 = 1.15 = log 1906 - log 1989 = log 1906
I0
I0
I1989
I
Thus 1906 = 101.15 ≈ 14
I1989
8.25 = log
which means the 1906 quake was 14 times as intense as the 1989 quake.
7.4 : Exponential and Log Applications
In general, the relative intensity between two quakes is 10difference of the Richter Numbers.
Other relative measures such as sound intensities in Decibels are measured in a similar
manner
EXPONENTIAL GROWTH
In calculus, you will see that the rate of growth of some amount A, symbolized by
dA
the derivative dt , is directly proportional to the amount A, currently present. This
dA
statement is symbolized by the differential equation dt = kt. Solving this equation for
A produces the exponential growth equation A = A0ekt. In a typical application, the
variables A0, k, t and A are found one by one (not necessarily in that order), using given
information. An example problem will illustrate the process.
Example: At the start of a study (t = 0) the population of a colony of Zairs is 50. Two
months later the population had increased to 65. What will the population be
at the end of a year?
50 = A0ek(0) = A0
65 = 50 ek(2)
A = 50 e0.1312 t
So A = 50 ekt 1.3 = e2k
ln 1.3 = 2k
1
k = 2 ln 1.3
A(t=12) = 50 e0.1312(12)
≈ 241
k ≈ 0.1312
The exponential growth equation seen above, is applicable in any problems wherein
a rate of growth of a particular substance is directly proportional to the current amount of
the substance. This includes money problems involving compounding continuously.
The "Pert" formula for computing future amounts A = Pert, has been mentioned several
times previously in this book. Observe that this formula is the same as the exponential
growth equation discussed above. Consequently these "money problems" can be solved
using the same technique, seen in the example above.
RADIOACTIVE DECAY
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Chapter 7 : Elementary Functions I
Radioactive isotopes are produced in a nuclear reaction. Radioactive substances
decay (decrease) by emitting radiation. Just as for rate of growth, the rate of decay is
directly proportional to the amount present. A common isotope produced in an atomic
bomb is strontium-90 which has a half-life of 28 years. Half-life is how long it takes for
a substance to decay to half its original amount. A typical problem asks how long will it
take for some amount of a radioactive substance to decay to some specified lesser
amount. Useful, though not absolutely necessary to solve these types of problems, a
special half-life formula can be derived starting with the same exponential equation used
for growth.
Let T be the half-life of some substance.
Then 2 A0 = A0ekT ⇒ 2 = ekT ⇒ kT = ln 2 ⇒ k = T ln 2 = ln [ 2
1 t/T
t ln(1/2)1/T
ln(1/2)t /T
Thus A = A0ekt = A0e
= A0e
= A0(2 )
1
1
1
1
1
1
]1/T
Example: How long will it take a mass of 200 mg of strontium-90 to decay to 20 mg?
1
1
t
1
1 t/28
1 t/28
20 = 200 (2 ) ⇒ 10 = (2 ) ⇒ log 10 = 28 log 2 ⇒
28 log(1/10)
t = log(1/2)
≈ 93 years
An important application of radioactive decay is seen in Carbon Dating. The
radioactive isotope, Carbon-14 is produced by nuclear reactions in the atmosphere,
caused by cosmic-ray bombardment. Living organisms absorb C-14 and have the same
proportion (which is very small) as the atmosphere. When an organism dies, it stops
absorbing C-14 and the C-14 present begins to decay with a half life of about 5750 years.
Consequently we can measure the age of some fossil by measuring the proportion of C14 in the remains. For example, suppose a skeleton is found containing 10% of its
original proportion of C-14. Then its age would be found as shown below.
t
5750 ln 0.1
1 t/5750
⇒ ln 0.1 = 5750 ln 1/2 ⇒ t = ln 1/2
≈ 19101 years
0.1A0 =A0(2 )
NEWTON'S LAW OF COOLING
Newton's Law of Cooling is a formula that gives the temperature of an object,
placed in a cooler environment, as a function of time. It is similar to the formulas seen
previously, because the rate of cooling is proportional to the temperature difference
between the object and the surroundings, in which it is placed. If T0 is the initial
temperature of the object, Ts is the temperature of the surrounding medium, and k is a
7.4 : Exponential and Log Applications
constant that depends on the type of object, then the formula for the new temperature of
the immersed object is
T = Ts + (T0 - Ts)ekt
Now if you rewrite this formula as T - Ts = (T0 - Ts)ekt, you can observe that it says the
temperature differential equals the original temperature differential times ekt. Thus the
law of cooling is just our old friend the standard growth/decay formula A = A0ekt where
A represents temperature differential. Further, this law should be called the change in
temperature differential formula because it can be applied to objects both cooling down
when placed in cooler surroundings and warming up when placed in warmer
surroundings. One additional thing concerning applying the growth/decay formula to this
type of problem should be noted here. Regardless of whether an object is cooling down
or warming up, the temperature differential is decreasing. Consequently for this
application, k is always negative.
Exercise: A really hot cup of coffee, at 210o F, is placed on a counter in a
room which has a temperature of 75o F. After 10 minutes the coffee had
cooled to 150o F. How much longer will it take for the coffee to cool
down to 120o?
In this section you have seen only a few of the many types of problems that involve
solving log and exponential equations. More problems similar to these types, plus others
will be seen in the problem set for this section.
Problem Set 7.4
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Chapter 7 : Elementary Functions I
1.
Solve for the exact values of x without using a calculator.
(a) log 3x = 2
(d) 2x - 22-x = 3
2.
(b) 3x = 5x-2 (c) ln x + ln(2x - 1) = 2 ln(x + 2)
(e) 2 log6 x = 4 log3 x - 1
2
(f) x2 log 8 - x log 5 = log 10 - x
2
Determine the exact coordinates, without using a calculator, for the point(s) of
8
intersection of the graphs for log9 x + logy 8 = 2 and logx 9 + log8 y = 3 .
3.
When a certain drug enters the blood stream, it dilutes exponentially with a half-life
of 3 days. If some initial amount of drug A0 is injected into the blood stream, what
percentage of the drug will still be in the blood stream 30 days later?
4.
Cars arrive at the drive-thru of McDonald’s at a rate of 90 cars per hour between
12:00 pm and 1:00 pm. The equation F(t) = 1 - e-1.5t can be used to determine the
probability that a car will arrive t minutes after 12:00 PM
Determine how many minutes are needed for the probability to reach 50%.
5.
The risk, R(%), of having an accident is given by the equation R = 3ekx where x is
the variable concentration of alcohol in a person’s blood and k is a constant.
(a) Solve for k if the concentration of alcohol in a person’s blood,0.06, results in a
10% risk of an accident.
(b) Using this k value, what is the concentration of alcohol in a person’s blood that
results in a 100% risk?
(c) The law asserts that anyone with a risk of having an accident of 15% or more
should not have driving privileges. In this case, at what concentration of alcohol
should a driver be arrested?
6.
The 9.5 Chilean Earthquake, which occurred on May 22, 1960, is the strongest
quake ever recorded. How does it compare in intensity to the 8.25 San Francisco
quake?
7.
The loudness or intensity of sound is measured simarily as earthquakes but in
1
decibels ( 10 of a bel named in honor of Alexander Graham Bell). The number of
7.4 : Exponential and Log Applications
I
decibels is given by the formula dB = 10 log I where Io is the minimum audible
o
sound and I is the intensity of the sound being measured. Ordinary conversation is
about 50 dB. The threshold of pain for sound is about 120 dB. How many times
more intense is a 120 dB sound than a 50 dB sound?
8.
When I was a student in 1960, the US population was 180 million. By the next
census in 1970, it had increased to 204 million. Assuming exponential growth,
predict what the population would have been in 2000 when you were a student. The
US Census in 2000 is listed as 281.422 million. Compare this value with your
predicted population. What do you conclude? Create a new growth model using the
populations in 1970 and 2000 and then predict the 2010 population when you will be
a working adult.
9.
If you invest $10,000 today (in 2003) at 5% interest compounded continuously, how
many years will it take to double? What will be the value of this account in 2050
when you retire?
10. Atmospheric pressure decreases exponentially with altitude above the surface of the
Earth. One atmosphere ( 1 atm.) which is the pressure at sea level, is 14.7 pounds
1
per square inch. At the top of Mount Everest (29,000 feet) the pressure is 3 atm.
What is the pressure. in Mexico City (7500 feet altitude)? What is the pressure at
the edge of space (defined by NASA as 50 miles up)?
11. The Rhind papyrus contains much of what we know about Egyptian mathematics
(some of which you saw in Chapter 1). Chemical analysis showed that it contained
about 75% of its original C-14. What is the age of this document?
12. A dead body was found at 8 am in a hotel room having a constant 72o temperature.
When the coroner arrived at 8:30, she measured the body temperature to be 80o. At
10 am the temperature was measured again and found to have decreased to 78o.
Assuming a normal temperature of 98.6o, when did the person die? [Use Newton's
Law of Cooling.]
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Chapter 7 : Elementary Functions I
7.5 : Introduction to Financial Mathematics
7.5 INTRODUCTION TO FINANCIAL MATHEMATICS
SIMPLE AND COMPOUND INTEREST
Money placed in a bank account earns interest at a certain yearly rate. Borrowing
money to pay for a car, a house, college, etc. involves calculating amounts to be repaid
1
using a specified interest rate such as 6%, 62 %, 7%, etc.. The simplest financial
transaction involves having an investment or borrowing money at simple interest. If
$100 is invested in an account which accrues 4% interest, then:
In 1 year the account value is $100 + $100(4%) = $104.
In 2 years the account value is $100 + $100(4%)(2) = $108.
In 3 years the account value is $100 + $100(4%)(3)=$112.
In banking (and Math) terminology, the invested (or borrowed) amount is called the
principal and is denoted by P. The interest rate, which is yearly unless stated otherwise,
is denoted by r and the number of years is denoted by t. If the letter, I, denotes the
interest, then I = Prt. The amount, or value, of an account at the end of the time period is
usually denoted by A, and thus for simple interest problems:
A = P + I = P + Prt = P (1 + rt ).
Most banks pay compound interest on accounts. With compound interest, new
interest is paid on the old interest that accumulates as well as on the principal. The
interest compounds. Consider $100 invested at 4% interest compounded annually for 3
years. After the first year the account value will be $104 as in the previous example. But
then $104 becomes the principal for the next year and thus in two years, A = $104 +
4%($104) = $108.16, and after three years A = $108.16 + 4%($108.16) = $112.49 (to the
nearest cent). Notice the compounding effect. This account has $.49 more than the
account that earned simple interest seen previously.
Now it would be tedious to use the procedure just shown to calculate the final
amount of an investment at compound interest over a significant number of years.
Fortunately, there exists a formula which calculates the amount. Observe the pattern
below for the $100 principal at 4% interest. After 1, 2, 3, . . ., t years, A =
100 + 100 (4%) = 100 (1 + 4% ) = 100(1.04)
[100(1.04)] + [100(1.04)]4% = [100(1.04)](1+4%) = 100(1.04)(1.04)= 100(1.04)2
[100(1.04)2]+[100(1.04)2]4%=[100(1.04)2](1+4%) =[100(1.04)2](1.04)= 100(1.04)3
.
.
.
100 (1.04)t
In 10 years, the amount will be 100 (1.04)10 = $148.02.
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Chapter 7 : Elementary Functions I
In 100 years, the amount will be 100(1.04)100 = $5050.49.
In 1000 years, the amount will be 100 ( 1.04)1000= 1.079789994 • 1019 dollars!
By comparison, after 1000 years at simple interest ( of 4% ), the account would only have
a value of $4,100 which is significantly less than the over ten billion billion in the
compounding account!
The general formula for compound interest, when compounding once a year , is
A = P(1+r)t.
The statement above suggests that interest can compound more frequently than once a
year. This is true and the preceding formula can be modified to reflect the compounding
n times per year as shown below:
r ⎞ nt
⎛
A=P⎜1+n⎟
⎝
⎠
For example, suppose the $100 at 4% is in an account compounding quarterly for 10
years.
Then n = 4 and the amount is
.04 4(10)
= $148.89
A = 100 ⎛⎜1 + 4 ⎞⎟
⎝
⎠
If n = 12 (monthly), then
.04 12(10)
A = 100 ⎛⎜1 + 12 ⎞⎟
= $149.08
⎝
⎠
If n = 365 (daily), then
.04 365(10)
A = 100 ⎛⎜1 + 365 ⎞⎟
= $149.18
⎝
⎠
r nt
Now when using the compound interest formula A = P ⎛⎜1 + n⎞⎟ , you might observe
⎝
⎠
that as n increases, A also increases. For example, compounding daily produces a larger
amount than compounding monthly, monthly more than quarterly, etc.. A natural
question might be, "Is there an ultimate or maximum amount that can accrue for a fixed
set of P, r, and t ?" Another way to express this question is to inquire as to what happens
as n increases without bound (approaches infinity). Answering this question necessitates
1
r
a little algebra. In the formula for compound interest above, let x = n . Then n = xr.
This produces:
1 xrt
1 x ⎞ rt
⎛
A = P ⎛⎜1 + x⎞⎟
= P ⎜ ⎛⎜1 + x⎞⎟ ⎟
⎝
⎠
⎠ ⎠
⎝⎝
7.5 : Introduction to Financial Mathematics
577
Now as n approaches infinity, x must also approach infinity.
1 x
Exercise: Using your calculator find the value of ⎛⎜1 + x⎞⎟
⎝
⎠
for x = 10; 100; 1000; 10000; etc. and write them below.
These numbers approach the irrational number which is denoted by e. To see a calculator
approximation for e , enter one and press the ex key. To the nearest thousandth, e = 2.718.
Written in terms of e, the formula for what is called compounding continuously is
A = Pert
and is often called the "PERT" formula (think of the shampoo!).
Exercise: Determine the amount of money that would accrue if $1000 is
deposited in an account at 4% compounded continuously for 5 years.
Now with regard to typical problems involving compound interest, notice that in the
nt
r
compound interest formula, A = P ⎛⎝1 + n⎞⎠ , there are five variables and in the "pert"
formula there are four variables. If you know all but one of the variables, you can find
the last one. For example, suppose some amount P is deposited at 4% compounded
quarterly. If after twenty years the account value is $350, how much was the initial
deposit, P?
04
Solution:
350 = P ⎛⎜1 + 4 ⎞⎟ 4(20)
⎝
⎠
= P ( 2.2167. . .)
350
2.2167 = P
$157.89 = P
Question:
ANNUITIES
The "Rule of 72" states that the approximate number of years it takes for
an account, in which interest compounds annually, to "double" in value
equals 72 divided by the interest rate number. How long will it take an
account at 8% to double? Verify with a specific example.
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Chapter 7 : Elementary Functions I
An account into which money is periodically deposited or paid is called an Annuity.
If the same amount is deposited at the end of each interest period, the annuity is called an
ordinary or simple annuity. This is the only type of annuity with which we will be
concerned in this book.
The Future Value of a simple annuity is the sum of all the payments plus the
interest. Suppose that $100 is deposited at the end of each month into an account that
pays 6% interest compounded at the end of each month for a period of 2 years. Then in
nt
r
the compound interest formula, A = P ⎛⎝1 + n⎞⎠
r
, P = $100, n =0.06/12 = 0.005 and nt =
the number of months a deposit accumulates interest (the first payment accumulates for
23 months, the second for 22 months, etc.). Thus the sum of all the deposits and interest
for our example is: S = 100 + 100(1 + 0.005)1 + 100(1.005)2 + . . . + 100 (1.005)23.
Observe that this is a Geometric Series with a = 100, r = 1.005 and n = 24 terms. Thus,
a(rn - 1)
(1.005)24 - 1
using the sum formula for a geometric series, S = r - 1 , S = 100
=
0.005
$2543.20.
(
(1 + i)n - 1
In general, S = P
i
[
)
] where P = the deposit or payment
i = the interest rate per period
n = the total number of payments.
A Sinking Fund is some future amount, S, to be accumulated by making periodic
payments, P. Generally, you know how much you want to have at the end and want to
know how much you must deposit at the end of each period. Although it is not necessary
to have another formula, you could take the Future Value formula seen above and solve
for P.
Example: What monthly payment is necessary to accumulate $5000 in 2 years
in an account paying 6% (yearly rate) cp (compounded) monthly?
0.005
[ 1.005
24 - 1 ] = $196.60
P = $5000
The Present Value of an annuity is simply, as the name suggests, how much the
accumulated value of a future annuity is worth now. Another way to think of this is,
"What is the lump sum necessary to finance a future stream of annuity payments?" The
formula for computing present value is best understood via an example. Suppose you are
setting up an annuity that will pay $100 at the end of each quarter for 3 years and the
bank interest rate is 8%. To determine how much you must deposit to fund this annuity,
consider how much is necessary to finance each of the future payments; i.e. what amount
will grow to $100 in 12 quarters; in 11 quarters; etc.
(
)
(
)
.08 12
.08 -12
⇒ P12 = 100 1 + 4
100 = P12 1 + 4
7.5 : Introduction to Financial Mathematics
(
)
(
)
(
)
(
)
(
)
.08 11
.08 -11
100 = P11 1 + 4
⇒ P11 = 100 1 + 4
.08 10
.08 -10
100 = P10 1 + 4
⇒ P10 = 100 1 + 4
.
.
.
(
)
.08 1
.08 -1
100 = P1 1 + 4
⇒ P1 = 100 1 + 4
Observe that the sum of these 12 necessary funding deposits is a geometric series.
The sum, S12 =
(
.08 -12
) ((1 + .084)12 - 1)
1 - (1 + 4 )
= $100
.08
.08
(1 + 4 )- 1
4
.08 -12
100 1 + 4
In general P = R
1 - (1 + i)-n
is called the present value of a future annuity R.
i
Amortization is the process of paying off a loan. The loan payments are simply an
annuity stream financed by the loan amount. Typically for amortization problems, in the
present value formula seen above, P, i and n are known and you are looking for R. Thus
R=
iP
.
1 - (1 + i)-n
Example: Suppose you borrow $1000 for 2 years at 8% with monthly payments.
.08
12 1000
The monthly payment, R =
.08 -24
1 - 1 + 12
(
Problem Set 7.5
)
= $45.23
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Chapter 7 : Elementary Functions I
1. Mrs. Jackson took out a loan of $42,000 at 13% annual simple interest to buy her
house. If she won the LOTTO exactly one year later and was able to repay the loan
without penalty, how much interest did she owe?
2. A man collected $28,500 on a loan of $25,000 he made 4 years ago. If he charged
simple interest, what was the yearly interest rate?
3. Burger Queen will need $50,000 in 5 years for a new addition. To meet this goal,
money is deposited today in an account that pays 7% annual interest compounded
quarterly. Find the amount that should be invested to total $50,000 in 5 years.
4. A money-market fund pays 4% annual interest compounded daily. What is the value
of $10,000 invested in this fund after 7 years?
5. Rework problem 3 if interest is compounded continuously.
6. Rework problem 4 if interest is compounded continuously.
7.
The parents of a newborn child decide to save for college. If they deposited $1 in a
dresser drawer on their child's first birthday and then twice as much on each
subsequent birthday as on the previous one, up through the 18th birthday, how much
would they have saved?
8.
If instead, the parents described in the preceding problem, made constant, fixed
yearly deposits in a college account paying 6% interest cp annually, and
accumulated the same amount in the college account as in the dresser drawer, how
much would they need to contribute each year?
9.
Ima Professor has deposited $250 each month for 19 years into an ordinary annuity
account which pays 6% cp monthly. What is the value of this account? What future
20 year monthly annuity back to her would this sum finance assuming the same
interest rate?
10. Liz T. Reen deposits $2400 each year for 9 years at 8% cp annually. After 9 years
she stops making deposits but leaves the money in the account (still drawing interest
at 8% cp annually) for another 26 years. Meanwhile, Hal I. Tosis decides to try to
catch up with Liz, so he makes deposits of $2400 each year for these same 26 years
that Liz's account is accumulating, also at 8% cp annually. Who has more money at
the end? Who contributed more money? What does this result tell you about saving
for your "Golden Years"?
11. John Q. Student purchases some property for $14,000. Five years later, he sells the
property for a profit of $9,000. Find the annual rate of return on his investment,
expressed as a % cp annually.
12. If you invest money at 6.84% cp monthly, how long will it take your money to
double; to triple?
7.5 : Introduction to Financial Mathematics
13. If you need $22,000 in 9 years, how much should you deposit each quarter at 7.1%
cp quarterly?
How much interest will you earn in the first year; in total?
14 Sally Savvy buys a car for $20,000. She pays $2000 down and finances the balance
at 4% cp monthly for 5 years. Determine her monthly payment and the total cost
(including interest) of her car.
15. Sam Smart is paying $332.14 per month on a car note at 12% interest. He just made
his 21st payment and has 15 payments to go. He comes into some money and
decides to pay off the note. Calculate his payoff amount and the amount he would
save by paying off the loan now.
581
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Chapter 7 : Elementary Functions I