Polynomials and Rational Functions (2.1)

The shape of the graph of a polynomial function is related to the degree of the polynomial

Shapes of Polynomials

Look at the shape of the odd degree polynomials 3( ) 27f x x x= −

5 3( ) 5 4 1f x x x x= − + +

Graph of Odd polynomial

y=x^5-5x^3+4x+1

-3

-2

-1

0

1

2

3

4

5

-3 -2 -1 0 1 2 3

Graph of Odd Polynomial

y=x^3-27x

-60

-40

-20

0

20

40

60

0 10 20 30 40 50 60

3( ) 27f x x x= −

Graphs of even degree polynomials

Now, look at the shape of the even degree polynomial

y=x^4-6x^2

-15

-10

-5

0

5

10

15

20

25

30

-4 -3 -2 -1 0 1 2 3 4

4 2( ) 6f x x x= −

Graph of even degree polynomial

Here is another example of an even degree polynomial :

2( ) 3 6 1f x x x= + −

f(x)=3x^2+6x-1

-10

0

10

20

30

40

50

-6 -4 -2 0 2 4

Generalization:

The graphs of odd-degree polynomials start negative, end positive and cross the x-axis at least once. The even-degree polynomial graphs start positive, end positive, and may not cross the x axis at all

Characteristics of polynomials:

Graphs of polynomials are continuous. One can sketch the graph without lifting up the pencil. 2. Graphs of polynomials have no sharp corners. 3. Graphs of polynomials usually have turning points, which is a point that separates an increasing portion of the graph from a decreasing portion.

Turning points and x intercepts

Theorem 1 : Turning points and x Intercepts of Polynomials The graph of a polynomial function of positive degree n can have at most n-1 turning points and can cross the x axis at most n times.

Largest value of the roots of a polynomial

Theorem 2: Maximum value of an x-intercept of a polynomial. If r is a zero of the polynomial P(x) this means that P(r) = 0. For example,

is a second degree polynomial . and

, so r = 4 is a zero of the polynomial as well as being an x-intercept of the graph of p(x).

2( ) 4p x x x= −

2(4) 4 4(4) 0p = − =

Cauchy’s Theorem

A theorem by a French mathematician named Cauchy allows one to determine the maximum value of a zero of a polynomial (maximum value of the x-intercept). Let’s take an example: the polynomial

2( ) 4p x x x= −

Cauchy’s TheoremAccording to this theorem

The numbers within the absolute value symbols are the coefficients of the polynomial p(x).

r < 1 + maximum value of { }1 , 4−= 1 + 4 = 5

2( ) 4p x x x= −

Result of application of Cauchy’s theorem

From this result we have , which means -5 < r < 5 . This tells us that we should look for any potential x intercepts within the range of -5 and 5 on the x –axis. In other words, no intercepts (roots) will be found that are greater than 5 nor less than -5.

Conclusion

From the graph of 2( ) 4p x x x= −,

we find that the other zero is located at (0,0). Thus, the two zeros , 0 , -4, are within the range of -5 to 5 on the x-axis. Now, let’s try another example:

An Example:

Example: Approximate the real zeros of

First step: Coefficient of cubic term must equal one, so divide each term by three to get a new polynomial Q(x)=

3 2( ) 3 12 9 4P x x x x= + + +

3 2 44 33

x x x+ + +

Roots of new polynomial are the same as the roots of P(x).

Example, continued

Step 2: Use the theorem: 11 max 4 , 3 ,3

r ⎧ ⎫< + ⎨ ⎬

⎩ ⎭

1 4 5

5

r

r

< + =

<

Example, continuedStep3: We know that all possible x intercepts (roots) are found along the x-axis between -5 and 5. So we set our viewing rectangle on our calculator to this window and graph the polynomial function.

Step 4. Use the zero command on our calculator to determine that the root is approximately -3.19 (there is only one root).

Rational Functions

Definition: Rational function: a quotient of two polynomials, P(x) and Q(x), is a rational function for all x such that Q(x) is not equal to zero.

Example: Let P(x) = x + 5 and Q(x) = x – 2 then

R(x)=

is a rational function that is defined for all real values of x with the exception of 2 (Why?)

52

xx+−

Domain of rational functionsDomain :

and x is a real number. This is read as “the set of all numbers, x , such that x is not equal to 2. X intercepts of a rational function: To determine the x-intercepts of the graph of any function, we find the values of x for which y = 0 . In our case y = 0 implies that 0 =

This implies that x + 5 = 0 or x = -5 .

{ }2x x ≠

52

xx+−

Y-intercept of a rational function

Y intercept: The y intercept of a function is the value of y for which x = 0 . Setting x = 0 in the equation we have y = , or -5/2. So, the y-intercept is located at ( 0, -2.5). Notice that the y-intercept is a point described by an ordered pair, not just a single number. Also, remember that a function can have only one y intercept but more than one x-intercept ( Why?)

Graph of a Rational function:1. Plot points near the value at which the function is

undefined. In our case, that would be near x = 2. Plot values such as 1.5, 1.7. 1.9 and 2.1, 2.3, 2.5. Use your calculator to evaluate function values and make a table.

2. Determine what happens to the graph of f(x) if x increases or decreases without bound. That is, for x approaching positive infinity or x approaching negative infinity.

3. Sketch a graph of a function through these points.

4. Confirm the results using a calculator and a proper viewing rectangle.

Graph of rational function

-50

-40

-30

-20

-10

0

10

20

30

40

50

-2 -1 0 1 2 3 4 5 6 7 8 9 10 11Series1

y=(x+5)/(x-2)

undefined at x = 2

approaches zero as x gets largeapproaches zero as x gets small

Conclusions: From the graph we see that there is a vertical asymptote at x = 2 because the graph approaches extremely large numbers as x approaches 2 from either side. We also see that y = 0 is a horizontal asymptote of the function since y tends to go to zero as x tends to either a very large positive number or very small negative number.

Exponential functions

The equation

defines the exponential function with base b . The domain is the set of all real numbers, while the range is the set of all positive real numbers ( y > 0). Note y cannot equal to zero.

( ) xf x b=

RiddleHere is a problem related to exponential functions: Suppose you received a penny on the first day of December, two pennies on the second day of December, four pennies on the third day, eight pennies on the fourth day and so on. How many pennies would you receive on December 31 if this pattern continues? 2) Would you rather take this amount of money or receive a lump sum payment of $10,000,000?

Solution (Complete the table)

2^3

2^22^1

64732616584

432211

No. pennies

Day

GeneralizationNow, if this pattern continued, how many pennies would you have on Dec. 31? Your answer should be 2^30 ( two raised to the thirtieth power). The exponent on two is one less than the day of the month. See the preceding slide. What is 2^30? 1,073,741,824 pennies!!! Move the decimal point two places to the left to find the amount in dollars. You should get: $10,737,418.24

Solution, continued The obvious answer to question two is to take the number of pennies on December 31 and not a lump sum payment of $10,000,000(although, I would not mind having either amount!) This example shows how an exponential

function grows extremely rapidly. In this case, the exponential function

is used to model this problem.

( ) 2xf x =

Graph of

Use a table to graph the exponential function above. Note: x is a real number and can be replaced with numbers such as as well as other irrational numbers. We will use integer values for x in the table:

( ) 2xf x =

2

Table of values

1 122

− =

44

1 122 16

− = =3 12

8− =

2 124

− =

02 1=12 2=

22 4=2

1

0

-1

-2

-3

-4

yx

Graph of y = ( ) 2 xf x =

Characteristics of the graphs of

where b> 1

1. all graphs will approach the x-axis as x gets large.

2. all graphs will pass through (0,1) (y-intercept)

3. There are no x – intercepts. 4. Domain is all real numbers 5. Range is all positive real numbers. 6. The graph is always increasing on its domain. 7. All graphs are continuous curves.

( ) xf x b=

Graphs of if 0 < b < 1

1. all graphs will approach the x-axis as x gets large.

2. all graphs will pass through (0,1) (y-intercept)

3. There are no x – intercepts. 4. Domain is all real numbers 5. Range is all positive real numbers. 6. The graph is always decreasing on its domain. 7. All graphs are continuous curves.

( ) xf x b=

Graph of Using a table of values once again, you will obtain the following graph. The graphs of and will be symmetrical with respect to the y-axis, in general.

1( ) 22

xxf x −= =

0

2

4

6

8

10

12

-4 -2 0 2 4

graph of y = 2 (̂-x)

approaches the positive x-axis as x gets large

passes through (0,1)

( ) xf x b= ( ) xf x b −=

Graphing other exponential functions

Now, let’s graph

Proceeding as before, we construct a table of values and plot a few points.Be careful not to assume that the graph crosses the negative x-axis. Remember, it gets close to the x-axis, but never intersects it.

( ) 3xf x =

Preliminary graph of ( ) 3xf x =

Complete graph

0

5

10

15

20

25

30

-4 -2 0 2 4

Series1

y = 3 x̂

Other exponential graphs This is the graph of

It is symmetric to the graph ofwith respect to the y-axisNotice that it is always decreasing. It also passes through (0,1).

( ) 4 xf x −=

( ) 4xf x =

Exponential function with base e

The table to the left illustrates what happens to the expression

as x gets increasingly larger. As we can see from the table, the values approach a number whose approximation is 2.718

(1+1/x)^x2.7182804691000000

2.71814592710000

2.7169239321000

2.704813829100

2.5937424610

21

11x

x⎛ ⎞+⎜ ⎟⎝ ⎠

Leonard EulerLeonard Euler first demonstrated that

will approach a fixed constant we now call “e”.So much of our mathematical notation is due to Eulerthat it will come as no surprise to find that the notation e for this number is due to him. The claim which has sometimes been made, however, that Eulerused the letter e because it was the first letter of his name is ridiculous. It is probably not even the case that the e comes from "exponential", but it may have just be the next vowel after "a" and Euler was already using the notation "a" in his work. Whatever the reason, the notation e made its first appearance in a letter Euler wrote to Goldbach in 1731. (http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/e.html#s19)

11x

x⎛ ⎞+⎜ ⎟⎝ ⎠

Leonard Euler He made various discoveries regarding e in the following years, but it was not until 1748 when Euler published Introductio in Analysis in infinitorum that he gave a full treatment of the ideas surrounding e. He showed that e = 1 + 1/1! + 1/2! + 1/3! + ...

and that e is the limit of (1 + 1/n)^n as n tends to infinity. Euler gave an approximation for e to 18 decimal places, e = 2.718281828459045235

Graph of

Graph is similar to the graphs of

and

Has same characteristics as these graphs

graph of y = e^x

0

5

10

15

20

25

-4 -2 0 2 4

Series1

( ) xf x e=

( ) 2xf x =

( ) 3xf x =

Growth and Decay applicationsThe atmospheric pressure pdecreases with increasing height. The pressure is related to the number of kilometers h above the sea level by the formula:

Find the pressure at sea level ( h =1)

Find the pressure at a height of 7 kilometers.

0.145( ) 760 hP h e−=

Solution: Find the pressure at sea level ( h =1)

Find the pressure at a height of 7 kilometers

0.145(1)(1) 760 657.42P e−= =0.145(7 )(7) 760 275.43P e−= =

Depreciation of a machineA machine is initially worth dollarsbut loses 10% of its value each year. Its value after t years is given by the formula

Find the value after 8 years of a machine whose initial value is $30,000

Solution:

0( ) (0.9 )tV t V=

0V0

( ) (0.9 )tV t V=

8(8) 30000(0.9 ) $12,914V = =

Compound interest

The compound interest formula is

Here, A is the future value of the investment, P is the initial amount (principal), r is the annual interest rate as a decimal, n represents the number of compounding periods per year and t is the number of years

1ntrA P

n⎛ ⎞= +⎜ ⎟⎝ ⎠

Problem: Find the amount to which $1500 will grow if deposited in a bank at 5.75% interest compounded quarterly for 5 years.

Solution: Use the compound interest formula:

Substitute 1500 for P, r = 0.0575, n = 4 and t = 5 to obtain

=$1995.55

1ntrA P

n⎛ ⎞= +⎜ ⎟⎝ ⎠

(4)(5)0.05751500 14

A ⎛ ⎞= +⎜ ⎟⎝ ⎠

Logarithmic Functions

In this section, another type of function will be studied called the logarithmic function. There is a close connection between a logarithmic function and an exponential function. We will see that the logarithmic function and exponential functions are inverse functions. We will study the concept of inverse functions as a prerequisite for our study of logarithmic function.

One to one functions We wish to define an inverse of a function.

Before we do so, it is necessary to discuss the topic of one to one functions.

First of all, only certain functions are one to one.

Definition: A function is said to be one to one if distinct inputs of a function correspond to distinct outputs. That is, if

Graph of one to one functionThis is the graph of a one to one function. Notice that if we choose two different x values, the corresponding values are also different. Here, we see that if x =- 2 , y = 1 and if x = 1, y is about 3.8.

Now, choose any other pair of x values. Do you see that the corresponding yvalues will always be

different?

Horizontal Line Test Recall that for an equation to be a function, its graph must pass the vertical line test. That is, a vertical line that sweeps across the graph of a function from left to right will intersect the graph only once.

There is a similar geometric test to determine if a function is one to one. It is called the horizontal line test. Any horizontal line drawn through the graph of a one to one function will cross the graph only once. If a horizontal line crosses a graph more than once, then the function that is graphed is not one to one.

Which functions are one to one?

-30

-20

-10

0

10

20

30

40

-4 -2 0 2 4

0

2

4

6

8

10

12

-4 -2 0 2 4

Definition of inverse functionGiven a one to one function, the inverse function is found by interchanging the x and y values of the original function. That is to say, if ordered pair (a,b) belongs to the original function then the ordered pair (b,a) belongs to the inverse function. Note: If a function is not one to one (fails the horizontal line test) then the inverse of such a function does not exist.

Logarithmic FunctionsThe logarithmic function with base two is defined to be the inverse of the one to one exponential function

Notice that the exponential function

is one to one and therefore has an inverse.

0

1

2

3

4

5

6

7

8

9

-4 -2 0 2 4

graph of y = 2 (̂x)

approaches the negative x-axis as x gets large

passes through (0,1)

2 xy =

2xy =

Inverse of exponential function

Start with

Now, interchange x and y coordinates:

There are no algebraic techniques that can be used to solve for y, so we simply call this function y the logarithmic function with base 2.

So the definition of this new function isif and only if

(Notice the direction of the arrows to help you remember the formula)

2xy =

2 yx =

2log x y=

2log x y= 2 yx =

Graph, domain, range of logarithmic function

1. The domain of the logarithmic function is the same as the range of the exponential function

(Why?) 2. The range of the logarithmic function is the same as

the domain of the exponential function (Again, why?) 3. Another fact: If one graphs any one to one function and its inverse on the same grid, the two graphs will always be symmetric with respect the line y = x.

2xy =

2xy =

•

•

•

• •

Three graphs: , , y = x Notice the symmetry:

•

•

•

••

•

•

•

2xy = 2log x y=

Logarithmic-exponential conversions

Study the examples below. You should be able to convert a logarithmic into an exponential expression and vice versa. 1.

2.

3.

4.

4log (16) 4 16 2xx x= → = → =

( )3 3 33

31 1log log log 3 327 3

−⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3125 5= ( )5log 125 3→ =

( )12

81181 9 81 9 log 92

= → = → =

Solving equationsUsing the definition of a logarithm, you can solve equations

involving logarithms: See examples below:

3 3 3log (1000) 3 1000 10 10b b b b= → = → = → =

( ) 56log 5 6 7776x x x= → = → =

Properties of logarithms These are the properties of logarithms. M and N are positive real numbers, b not equal to 1, and p and x are real numbers.

log

. log (1) 0log ( ) 1log 1

12.3.4. b

b

bx

bx

bb

b x

====

log log log

log log log

5

log loglog

.

6.

7.8 l g. o

b b b

b b b

pb b

b b

MN M NM M NN

M p MM N iff M N

= +

= −

== =

Solving logarithmic equations1. Solve for x:

2. Product rule

3. Special product

4. Definition of log

5. X can be 10 only6. Why?

( )

4 4

4

24

3 2

2

2

log ( 6) log ( 6) 3log ( 6)( 6) 3

log 36 3

4 3664 36100

1010

x xx x

x

xxxx

x

+ + − = →+ − = →

− = →

= − →

= − →

= →± = →=

Another exampleSolve:

2. Quotient rule

3. Simplify

(divide out common factor of pi)

4. rewrite

5 definition of logarithm

6. Property of exponentials

410

4

log log(10000 )

log10000

1log10000

log 10

10 104

x

x

x

x

x

x

π ππ

π

−

−

− = →

= →

⎛ ⎞ = →⎜ ⎟⎝ ⎠⎡ ⎤ = →⎣ ⎦= →

= −

Common logs and Natural logs

Common log Natural log

10log logx x= ln( ) logex x=

2.7181828e ≈

Solving an equation1. Solve for x. Obtain the

exact solution of this equation in terms of e (2.71828…)

2. Quotient property of logs

3. Definition of (natural log)4. Multiply both sides by x 5. Collect x terms on left side6. Factor out common factor7. Solve for x

Solution:

1

ln( 1) 1 ln( )ln( 1) ln( ) 1

1ln 1

1

1 01

( 1) 11

1

x xx xx

xxe

xex xex xx e

xe

+ = = →+ − =

+⎛ ⎞ = →⎜ ⎟⎝ ⎠

+= →

= + → →− = →− = →

=−

Solving an exponential equation

Solve the equation

1. Take natural logarithm of both sides

2. Exponent property of logarithms

3. Distributive property

4. Isolate x term on left side

5. Solve for x

Solution: 2 15 80x− − =

( )2 1

2 1

5 80

ln 5 ln(80)

( 2 1) ln(5) ln(80)2 ln(5) 1ln(5) ln802 ln(5) ln80 ln 5

ln80 ln 52 ln(5)

x

x

xxx

x

− −

− −

= →

= →

− − = →− − = →− = − →

−=

−

Application How long will it take money to double if compounded monthly at 4 % interest ? 1. compound interest formula2. Replace A by 2P (double the amount) 3. Substitute values for r and m 4. Take ln of both sides 5. Property of logarithms 6. Solve for t and evaluate expression

Solution:

( )

12

12

12

1

0.042 112

2 (1.003333...)

ln 2 ln (1.003333...)

ln 2 12 ln(1.00333...)ln 2 17.36

12ln(1.00333...)

mt

t

t

t

rA Pm

P P

t

t t

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

=

=

=

= → =