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 fx()=− x3 27 x

fx()= x53−++ 5 x 4 x 1 Graph of Odd polynomial

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

5 4 3 2 1 0 -3 -2 -1-1 0 1 2 3 -2 -3 Graph of Odd Polynomial

fx()=− x3 27 x

y=x^3-27x

60

40

20

0 0 102030405060 -20

-40

-60 Graphs of even degree polynomials

† Now, look at the y=x^4-6x^2

shape of the even 30

degree polynomial 25

20 42 fx()=− x 6 x 15 10

5

0 -4 -3 -2 -1 0 1 2 3 4 -5

-10

-15 Graph of even degree polynomial

† Here is another f(x)=3x^2+6x-1 example of an 50 even degree 40 polynomial : 30

20 2 fx()=+− 3 x 6 x 1 10 0 -6 -4 -2 0 2 4 -10 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, px()= x2 − 4 x

† is a second degree polynomial . and p(4)= 42 −= 4(4) 0 † , so r = 4 is a zero of the polynomial as well as being an x-intercept of the graph of p(x). 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 p()xx=−2 4 x Cauchy’s Theorem

† According to this theorem

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

† The numbers within the absolute value symbols are the coefficients of the polynomial p(x). p()xx= 2 − 4 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 px()=− x2 4x 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 Px()=+ 3 x32 12 x ++ 9 x 4 † First step: Coefficient of cubic term must equal one, so divide each term by three to get a new polynomial Q(x)= 4 xxx32+++43 3 † Roots of new polynomial are the same as the roots of P(x). Example, continued

† Step 2: Use the theorem: ⎧ 1 ⎫ r <+1max4,3,⎨ ⎬ ⎩⎭3 r <+14 = 5 r < 5 Example, continued

† Step3: 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)= x + 5 x − 2 is a rational function that is defined for all real values of x with the exception of 2 (Why?) Domain of rational functions

† Domain : {xx≠ 2} † 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 = x + 5 x − 2

† This implies that x + 5 = 0 or x = -5 . 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 y=(x+5)/(x-2) approaches10 zero as x gets small approaches zero as x gets large 0 Series1 -2-101234567891011-10 -20 undefined at -30 x = 2 -40 -50 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 f ()xb= x 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. Riddle

† Here 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)

Day No. pennies 1 1 2 2 2^1 3 4 2^2 4 8 2^3 5 16 6 32 7 64 Generalization

† Now, 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 fx()= 2x

† is used to model this problem. x Graph of fx()= 2

† Use a table to graph the exponential function above. Note: x is a real number and can be replaced with numbers such as 2 as well as other irrational numbers. We will use integer values for x in the table: Table of values

x y

-4 −4 11 2 = 4 = 1216 -3 2−3 = 8 -2 1 2−2 = 4 -1 1 2−1 = 2 0 210 = 1 221 = 2 242 = Graph of y = fx()= 2x Characteristics of the graphs of fx()= bx 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. Graphs of f () xb = x 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. 1 fx()= 2−x = Graph of 2x

† Using a table of values once again, you will obtain the following graph. The graphs of f () xb = x and fx () = b − x will be symmetrical with respect to the y-axis, in general.

12

10

8

6 graph of y = 2^(-x) approaches the positive x-axis as x gets large 4

2 passes through (0,1)

0 -4 -2 0 2 4 Graphing other exponential functions † Now, let’s graph fx()= 3x

† 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. x Preliminary graph of fx()= 3 Complete graph30 25

y = 3^x 20

15 Series1 10

5

0 -4-2024 Other exponential graphs

† This is the graph of fx()= 4−x

† It is symmetric to the graph of fx()= 4x † with respect to the y-axis † Notice that it is always decreasing. † It also passes through (0,1). Exponential function with base e

† The table to the left 1 2 illustrates what happens to the expression 10 2.59374246 x ⎛⎞1 ( ⎜⎟1+ 100 2.7048138291 ⎝⎠x + 1 1000 2.716923932 † as x gets increasingly / x larger. As we can see 10000 2.718145927) from the table, the ^ values approach a 1000000 2.718280469x number whose approximation is 2.718 Leonard Euler

x † Leonard Euler first demonstrated that ⎛⎞1 ⎜⎟1 + ⎝⎠x † will approach a fixed constant we now call “e”. † So much of our mathematical notation is due to Euler that 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 Euler used 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) 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 fx()= ex

graph of y = e^x † Graph is similar to the graphs of fx()= 2x 25 † and 20 x Series1 fx()= 3 15 10 Has same 5 characteristics as 0 -4 -2 0 2 4 these graphs Growth and Decay applications

† The atmospheric pressure p „ Find the pressure at decreases with increasing sea level ( h =1) height. The pressure is related to the number of kilometers h above the sea level by the formula: „ Find the pressure at Ph( )= 760 e−0.145h a height of 7 kilometers. Solution:

„ Find the pressure at „ Find the pressure at sea level ( h =1) a height of 7 kilometers

−0.145(7) Pe(1)== 760−0.145(1) 657.42 Pe(7)== 760 275.43 Depreciation of a machine

† A machine is initially † Solution: worth dollars V0 † but loses 10% of its Vt()= V (0.9)t value each year. Its 0 value after t years is given by the formula V(8)== 30000(0.98 ) $12,914 Vt()= V (0.9)t 0

† Find the value after 8 years of a machine whose initial value is $30,000 Compound interest † The compound interest formula is nt ⎛⎞r AP=+⎜⎟1 ⎝⎠n † 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 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: nt ⎛⎞r AP=+⎜⎟1 † ⎝⎠n Substitute 1500 for P, r = 0.0575, † n = 4 and t = 5 to obtain (4)(5) ⎛⎞0.0575 A =+1500⎜⎟ 1 ƒ =$1995.55 ⎝⎠4 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 function

† This 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 y † values 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?

12 40 10 30 8 20 6 10 4 0 -4 -2 0 2 4 -10 2

-20 0 -4 -2 0 2 4 -30 Definition of inverse function

† Given 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 Functions

† The logarithmic function with base two is defined to be the inverse of the one to one exponential function y = 2 x † Notice that the exponential † function x y = 2 9 8 7 6 graph of y = 2^(x)

† is one to one and therefore has approaches the5 negative x-axis as x gets an inverse. 4 large 3 passes through (0,1) 2 1 0 -4 -2 0 2 4 Inverse of exponential function

† Start with y = 2x

† Now, interchange x and y coordinates: x = 2 y † 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. log2 x = y † So the definition of this new function is † if and only if y log2 xy= x = 2 † (Notice the direction of the arrows to help you remember the formula) Graph, domain, range of logarithmic function

† 1. The domain of the logarithmic function is the same as the range of the exponential function x † (Why?) y = 2 † 2. The range of the logarithmic function is the same as the domain of the exponential function x † (Again, why?) y = 2 † 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. •

x Three graphs: y = 2 , log 2 x = y , y = x Notice the symmetry:

•

• • • • • • • • • Logarithmic-exponential conversions

† Study the examples below. You should be able to convert a logarithmic into an exponential expression and vice versa. † 1. x log4 (16)= xx→=→= 4 16 2

† 2. ⎛⎞ 11 ⎛⎞ −3 log33⎜⎟= log ⎜⎟3 ==log3 ()33 ⎝⎠27 ⎝⎠ 3

† 3. 3 125= 5 →=log5 ( 125) 3 † 4. 1 1 81=→ 9 812 =→ 9 log() 9 = 81 2 Solving equations

Using the definition of a logarithm, you can solve equations involving logarithms: See examples below:

333 logb (1000)=→ 3bbb = 1000 → = 10 →= 10

5 log6 (xxx)= 5→=→ 6 7776 = 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. 5. logMNMN=+ log log 1.logb (1)= 0 bbb M 2. log ()b = 1 6. log=− logM log N b bbbN x p 3. logb b =1 7. logbbMp= log M

logb x 4. bx= 8. logbbM ==lgo N iff M N Solving logarithmic equations

1. Solve for x: log44 (xx+ 6)+−=→ log ( 6) 3

2. Product rule log4 (xx+−=→ 6)( 6) 3 3. Special product 2 log4 ()x −=→ 36 3 4. Definition of log 43632=−→x 64=−→x2 36 2 5. X can be 10 only 100 =→x 6. Why? ±=→10 x x = 10 Another example

† Solve: logπ −=→ log(10000π ) x † 2. Quotient rule π log π=→x † 3. Simplify 10000 † (divide out common factor of pi) ⎛⎞1 log ⎜⎟=→x ⎝⎠10000 † 4. rewrite ⎡⎤−4 log10 ⎣⎦ 10 =→x † 5 definition of logarithm 10x =→ 10−4

† 6. Property of exponentials x =−4 Common logs and Natural logs

† Common log † Natural log

logxx= log10 ln(xx )= loge e ≈ 2.7181828 Solving an equation

1. Solve for x. Obtain the † Solution: exact solution of this ln(xx+ 1)== 1 ln( ) → equation in terms of e (2.71828…) ln(xx+− 1) ln( ) = 1 2. Quotient property of logs ⎛⎞x +1 ln⎜⎟=→ 1 ⎝⎠x 3. Definition of (natural log) x +1 e1 =→ 4. Multiply both sides by x x 5. Collect x terms on left side ex=+→→ x 10 6. Factor out common factor ex−=→ x 1 7. Solve for x xe(1)1−=→ 1 x = e −1 Solving an exponential equation

† Solve the equation † Solution: 580−−21x = 580−−21x =→

1. Take natural logarithm of both −−21x sides ln() 5=→ ln(80) 2. Exponent property of logarithms (2− x −=→ 1)ln(5)ln(80) 3. Distributive property −2x ln(5)−=→ 1ln(5) ln80 4. Isolate x term on left side −=−→2ln(5)ln80ln5x 5. Solve for x ln 80− ln 5 x = −2ln(5) Application

† How long will it take money † Solution: to double if compounded mt monthly at 4 % interest ? ⎛⎞r AP=+⎜⎟1 † 1. compound interest ⎝⎠m formula 12t † 2. Replace A by 2P (double ⎛⎞0.04 21PP=+⎜⎟ the amount) ⎝⎠12 † 3. Substitute values for r 12t and m 2= (1.003333...) † 4. Take ln of both sides ln 2= ln() (1.003333...)12t † 5. Property of logarithms † 6. Solve for t and evaluate ln 2= 12t ln(1.00333...) expression ln 2 =→=tt17.36 12ln(1.00333...)