Dr. Nestler - Math 2 - Ch 3: Polynomial and Rational Functions

Transcription

1 Dr. Nestler - Math 2 - Ch 3: Polynomial and Rational Functions Polynomial Functions We have studied linear functions and quadratic functions Defn. A monomial or power function is a function of the form 0ÐBÑ œ +B where 8 is a positive integer and + is a nonzero real number. The degree of the monomial is the integer 8, and the coefficient is the number +. We also consider constant functions to be monomials of degree zero. The domain of a monomial is Graphs of monomials 8 From the graphs, we see that the range of a monomial function 0ÐBÑ œ +B 8 is Example: Graphing transformations of monomials

2 Next we look at sums of monomials. Defn. A polynomial function of degree 8, where 8 is a positive integer, is a function of the form 8 8 " 0ÐBÑœ+ 8B + 8 " B â + " B +! where + ß ÞÞÞß + are real numbers and + is nonzero. The numbers + ß ÞÞÞß + are called! 8 8! 8 coefficients, the number + 8 is called the leading coefficient, and the monomial +B called the zero. leading term. We also consider constant functions to be polynomials of degree A polynomial is just a sum of one or more monomials. Examples: 8 8 is The domain of a polynomial is Graphing polynomials Calculus Fact: The graph of a polynomial is a smooth, unbroken curve.

3 7 7 " Defn. If ÐB <Ñ is a factor of a polynomial 0ÐBÑ and ÐB <Ñ is not, then < is said to be a Examples: zero of 0 of multiplicity 7. Facts: If < is a zero of 0 of odd multiplicity, then the graph of C œ 0ÐBÑcrosses the B -axis at <. Example: If < is a zero of 0 of even multiplicity, then the graph of C œ 0ÐBÑjust touches the B-axis at <. Calculus Fact: If 0 is a polynomial of degree 8, then 0 has at most 8 " turning points (points where a graph changes direction from increasing to decreasing or vice versa). Turning points are also called local maximum and local minimum points. Examples: Calculus Fact (end behavior): For large values of Example: graph of its leading term. lbl, the graph of a polynomial resembles the

4 Four kinds of end behavior: Finally, we graph CœJÐBÑusing the information we found above: Note: We need calculus to Another complete graphing example: [Can do , 126, 127]

5 Dr. Nestler - Math 2-3.4/3.5 - Rational Functions The domain of a polynomial function is one of four types of end behavior. Ð ß Ñ. Its graph is a smooth, unbroken curve with The sum, difference and product of polynomial functions are again polynomial functions. Definition. A rational function is a function of the form 0 ÐBÑ œ where :ÐBÑ and ;ÐBÑ are polynomials and ;ÐBÑ is not the zero function. The domain of a rational function 0ÐBÑ œ :ÐBÑ ;ÐBÑ At first we will consider only rational functions that are is :ÐBÑ ;ÐBÑ reduced, which means Example: 0ÐBÑ œ " B 1. Vertical Asymptotes Definition. A vertical line Bœ-is a vertical asymptote of the graph of 0if, as the input B approaches - from the left or the right, the output 0ÐBÑ gets arbitrarily large positively or negatively. By definition, the graph of a function cannot cross any of its vertical asymptotes. Theorem. If 0ÐBÑ œ :ÐBÑ ;ÐBÑ is a reduced rational function, then the vertical asymptotes of the graph, if any, are the lines Bœ-where - is a real zero of ;ÐBÑ. Example: 0ÐBÑ œ B " BÐB Ñ

6 Example: 0ÐBÑ œ B * "%B "$B " Example: 0ÐBÑ œ B "' % B " 2. Horizontal Asymptotes Example: Back to 0ÐBÑ œ " B Definition. A horizontal line CœPis a horizontal asymptote of the graph of 0if, as the input B gets arbitrarily large positively or negatively, the output 0ÐBÑ approaches the number P. If one exists, it describes the end behavior of the rational function. The graph of a function may intersect its horizontal asymptote, since it is indicating only end behavior. Example: 0ÐBÑ œ B $B " B B "* $

7 :ÐBÑ ;ÐBÑ Theorem. If 0ÐBÑ œ is a rational function and degree Ð:ÐBÑÑ degree Ð;ÐBÑÑ, then C œ! is the only horizontal asymptote of the graph of 0. In general, consider the behavior of leading term of numerator leading term of denominator as lbl gets large. This gives the only horizontal asymptote, if there is one. Example: Find the domain of this function. Find its intercepts, find equations of any asymptotes and B- and C-coordinates of any holes in the graph, and perform a sign analysis. Determine whether the function intersects any of its asymptotes. Finally, sketch a graph of the function using the above information. 0ÐBÑ œ B " B %

8 Example: 0ÐBÑ œ $ %B B "! (B B :ÐBÑ ;ÐBÑ Theorem. If 0ÐBÑ œ is a rational function and degree Ð:ÐBÑÑ degree Ð;ÐBÑÑ, then the graph has no horizontal asymptote. Example: 0ÐBÑ œ $ B &B ( $B$ $B 'B :ÐBÑ ;ÐBÑ Theorem. If 0 ÐBÑ œ is a rational function and degree Ð:ÐBÑÑ œ degree Ð;ÐBÑÑ, then the graph has a horizontal asymptote given by the ratio of the leading coefficients. :ÐBÑ ;ÐBÑ 3. Oblique Asymptotes If 0ÐBÑ œ is a rational function and degreeð:ðbññ œ degree Ð;ÐBÑÑ ", then the graph will have an oblique (slanted) asymptote: a nonvertical, nonhorizontal line that represents the end behavior of the graph. So a graph may intersect its oblique asymptote. Note that a rational function cannot have both a horizontal asymptote and an oblique asymptote.

9 Example: Find the domain of this function. Find its intercepts, find equations of any asymptotes and B- and C-coordinates of any holes in the graph, and perform a sign analysis. Determine whether the function intersects any of its asymptotes. Finally, sketch a graph of the function using the above information. $ 0ÐBÑ œ B B B *

10 4. Holes Now we consider rational functions that are not reduced. Example: 0ÐBÑ œ B " B " Theorem. Suppose 0ÐBÑ is a rational function that is not reduced, and suppose that obtaining its :ÐBÑ reduced form ;ÐBÑ involves cancelling common factors of the form B -where -is a real number. If ;Ð-Ñ Á!, then the graph has a hole at B œ -. Example: 0ÐBÑ œ B " B B Example: 0ÐBÑ œ ÐB "ÑÐB "!Ñ ÐB BÑÐB "!Ñ Example: 0ÐBÑ œ ÐB "ÑÐB "!Ñ ÐB BÑÐB "!Ñ

11 Dr. Nestler - Math Real Zeros of Polynomials Theorem. If 0ÐBÑ is a polynomial and - is a real number, then the following statements are equivalent, meaning that either they are all true or they are all false: (1) - is a zero of 0, meaning 0Ð-Ñ œ!þ (2) B - is a factor of 0ÐBÑ, meaning the remainder of 0ÐBÑ divided by B - is zeroþ (3) - is an B-intercept of the graph of 0ÐBÑ, meaning the point Ð-ß!Ñis on the graph of 0. Example: 0ÐBÑœB B 'œðb $ÑÐB Ñ. B $ is a factor of 0ÐBÑ Í 0Ð $Ñ œ! Í $ is an B-intercept of the graph of 0ÐBÑ. B is a factor of 0ÐBÑ Í 0ÐÑ œ! Í is an B-intercept of the graph of 0ÐBÑ. Since a polynomial of degree 8 has at most 8 linear factors, it has at most 8 real zeros. This means that the number of real zeros must be equal to or less than its degree. Examples: 0ÐBÑ œ ( ) BÐB "Þ$ÑÐB &Ñ 1ÐBÑ œ % B % & "$ 2ÐBÑ œ B *

12 Example: Find the quotient and remainder of & % $ 0ÐBÑ œ $B &B %B (B $ divided by B. Synthetic division: $ Example: Is the number " a zero of 0ÐBÑœB B B $? 0ÐBÑ B " " is a zero of 0ÐBÑ Í B " is a factor of 0ÐBÑ Í remainder of is zero 8 8 " Rational Zeros Theorem. Suppose 0ÐBÑœ+ 8B + 8 " B â + " B +! is a polynomial of degree 8 " where the coefficients are all integers and the constant term + is not zero. If is a reduced rational zero of 0, then : is a factor of the constant! : ; term +!, and ; is a factor of the leading coefficient + 8. Idea of Proof: Outlined in A complete proof requires what is called mathematical induction, a topic in section 11.4 that we will study at the end of the course.

13 Dr. Nestler - Math Complex Zeros of Polynomials Examples: B "' cannot be factored over the real numbers, since $ B "œðb "ÑÐB B "Ñ Fundamental Theorem of Algebra. Every polynomial function 0ÐBÑ of degree 8 " has at least one complex zero. Complete Factorization Theorem. Every polynomial function 0ÐBÑ of degree 8 " can be factored completely into a product of linear factors: 0ÐBÑ œ +ÐB - ÑâÐB - Ñ " 8 where +ß -" ß á ß -8 are complex (possibly real) numbers. So 0 has exactly 8 complex zeros, counted with multiplicity, and 0 can be factored into a product of 8linear polynomials. Examples: B "' œ ÐB %3ÑÐB %3Ñ $ B "œðb "ÑÐB B "Ñ

14 Conjugate Pairs Theorem. Let 0 be a polynomial function with real coefficients. If D œ +,3 is a complex zero of 0, then D œ +,3is also a zero of " 8 8 " " Proof: Write the polynomial in standard form: 0ÐBÑœ+ B + B â + B +. If D is a complex zero of 0, then 0ÐDÑ œ!. That is, 8 8 "!œ+ 8D + 8 " D â + " D +! Take the complex conjugate of both sides of this equation, and use the facts that the conjugate of a real number is unchanged, the conjugate of a sum is the sum of the conjugates, and the conjugate of a product is the product of the conjugates:!œ+ D8 + D8 " 8 8 " â + " D +! œ+ D8 + D8 " 8 8 " â + " D +! œ+ DâD + DâD â + D " "! œ+ DâD + DâD â + D " "! 8 8 " 8 8 " " œ+ ÐDÑ + ÐDÑ â +D + œ0ðdñ So 0ÐDÑœ!œ!. Thus Dalso is a zero of the polynomial function 0.!!

15 Dr. Nestler - Math 2 - Ch 4: Exponential and Logarithmic Functions One-to-One Functions and their Inverses Previously: A function is a rule of correspondence such that each has exactly one Multiple inputs may have the same output. Definition. A function is one-to-one if each has exactly one Example: 0ÐBÑ œ B $ Example: 0ÐBÑ œ B Horizontal Line Test: A function is one-to-one if and only if each horizontal line intersects its graph at most once. Definition. Suppose 0 is a one-to-one function. A function 1 is the inverse of 0 if 1Ð0ÐBÑÑ œ B for all B in the domain of 0, and 0Ð1ÐBÑÑ œ B for all B in the domain of 1. " " We write 1ÐBÑ œ 0 ÐBÑ and 0ÐBÑ œ 1 ÐBÑ.

16 " " Warning: 0 ÐBÑ Á Ð0ÐBÑÑ œ " Cancellation equations: 0 Ð0ÐBÑÑ œ Bfor all Bin the domain of 0 " " 0Ð0 ÐBÑÑ œ Bfor all B in the domain of 0 " Example: If 0 is one-to-one and 0ÐÑ œ (, then 0 Ð(Ñ œ " " Ð0ÐÑÑ œ 0 Ð0ÐÑÑ œ " Example: If 0 is one-to-one and 0 Ð$Ñ œ ", then 0Ð "Ñ œ " " " Ð0 Ð$ÑÑ œ 0Ð0 Ð$ÑÑ œ An inverse function interchanges inputs and outputs. So the domain of 0 is the and the range of 0 is the " " Example: Let C œ 0ÐBÑ œ B. We will find 0 ÐBÑ. We suspect that 0 ÐBÑ œ 1) Interchange Band C: " 2) Solve for Cœ0 ÐBÑ: B " B " " Example: Let Cœ0ÐBÑœ. Find 0, and find the domain and range of 0and 0. The domain of 0 is the range of 0 " : C " C " Solve Bœ for Cœ0 ÐBÑ: " The domain of 0 is the range of 0:

17 Application: In the Fahrenheit temperature system, water freezes at 32 F and boils at 212 F. We will derive the linear function GÐJÑ that gives degrees Celsius when J represents degrees Fahrenheit, such that GÐ$Ñ œ! and GÐ"Ñ œ "!!. We use the fact that the line of slope 7through ÐBßCÑ " " has equation given by C C" œ7ðb BÑ ". The Fahrenheit equivalent of 25 G is J Ð&Ñ œ Graphing inverses: If the point Ð+ß,Ñ is on the graph of 0, this means 0Ð Ñ œ " " If 0 has an inverse, then 0 Ð Ñ œ, so the point Ð ß Ñis on the graph of 0. " Thus the graph of 0 is the graph of 0 reflected about the line C œ B. [Can do 11-14, 19-28, 33-80, 89-96]

18 Dr. Nestler - Math 2 - Carbon Dating This is adapted from Anton, Bivens and Davis, Calculus, 8th ed., Wiley, 2005 It is a fact of physics that radioactive elements disintegrate spontaneously in a process called radioactive decay. Experimentation has shown that the rate of disintegration is proportional to the amount of the element present. When the nitrogen in the Earth s upper atmosphere is bombarded by cosmic radiation, the radioactive element carbon-14 is produced. This carbon-14 combines with oxygen to form carbon dioxide, which is ingested by plants, which in turn are eaten by animals. In this way all living plants and animals absorb quantities of radioactive carbon-14. In 1947 the American nuclear scientist W.F. Libby proposed the theory that the percentage of carbon-14 in the atmosphere and in living tissues of plants is the same [1]. When a plant or animal dies, the carbon-14 in the tissue begins to decay. Thus, the age of an artifact that contains plant or animal material can be estimated by determining what percentage of its original carbon-14 content remains. Various procedures, called carbon dating or carbon-14 dating, have been developed for measuring this percentage. The half-life of carbon-14 is approximately 5600 years. Exercise: In 1988 the Vatican authorized the British Museum to date a cloth relic known as the Shroud of Turin, possibly the burial shroud of Jesus of Nazareth. This cloth, which first surfaced in 1356, contains the negative image of a human body that was widely believed to be that of Jesus. The report of the British Museum showed that the fibers in the cloth contained between 92% and 93% of their original carbon-14. Use this information to estimate the age of the shroud. [1] W.F. Libby, "Radiocarbon Dating," American Scientist, Vol. 44, 1956, pp

19 Dr. Nestler - Math Definitions of the Trigonometric Functions There are two ways to define the trigonometric functions. 1. Definition of the Trigonometric Functions at a Real Number Suppose > is a real number. Let ÐBß CÑ be the point obtained by moving along the unit circle from the point Ð"ß!Ñ a distance of > units counterclockwise if >! and l>l units clockwise if >!. We call ÐBßCÑ the point on the unit circle corresponding to >. The six Note that each number > determines a unique pair of numbers Band C in this manner. trigonometric functions are defined at > as follows: cosine: cos Ð>Ñ œ B secant: secð>ñ œ B ÐB Á!Ñ " sine: sin Ð>Ñ œ C cosecant: csc Ð>Ñ œ C ÐC Á!Ñ C B tangent: tan Ð>Ñ œ ÐB Á!Ñ cotangent: cotð>ñ œ ÐC Á!Ñ B So the point on the unit circle corresponding to > has coordinates Ð cos >ß sin >Ñ. Example: Evaluate the trigonometric functions at >œ!, >œ 1 and >œ 1Þ " C 2. Definition of the Trigonometric Functions at an Angle Suppose ) is an angle in standard position (with vertex at the origin and initial side along the positive B-axis). Let ÐBßCÑ be the point of intersection of the unit circle with the terminal side of ). The six As above, each angle ) determines a unique pair of numbers Band C in this manner. trigonometric functions are defined at ) as follows:

20 " cosine: cos ÐÑœB ) secant: secðñœ ) B ÐBÁ!Ñ " sine: sin ÐÑœC ) cosecant: csc ÐÑœ ) C ÐCÁ!Ñ C B tangent: tan ÐÑœ ) ÐBÁ!Ñ cotangent: cotðñœ ) ÐCÁ!Ñ B C 3. These two definitions are compatible because: If ) is a central angle of radian measure between 0 and 1, then the length > of the arc of the unit circle cut out by the angle equals the radian measure of ). Since ) œ>, the values of the trigonometric functions at ) equal the values of the trigonometric functions at >. 1 $ 1 Example: Evaluate the trigonometric functions at ) œ and ) œ. Example: Evaluate the trigonometric functions at ) œ 1 %.

21 1 1 ' $ Example: Evaluate the trigonometric functions at ) œ and ) œ Þ We can use the symmetries of the circle to compute trigonometric values at many other inputs. Definition. Suppose that ) is an angle in standard position with terminal side not on the axes. The reference angle corresponding to ) is the acute angle between the B-axis and the terminal side of ). By the symmetries of the circle, the values of the trigonometric functions at an angle ) are the same as the values of the trigonometric functions at its reference angle, or are off by a sign.

22 Example: Compute the trigonometric values at ) œ 1 $. To evaluate a trigonometric function at a number/angle: (1) Evaluate the function at the reference angle. (2) Use ASTC to determine the sign. multiples of We must be able to quickly find the values of the trigonometric functions at all integer 1 1 % and ', where they are defined. [Can do , , 117, 118]

23 Dr. Nestler - Math 2 - Chapter 9 - Conic Sections Analytic geometry: using algebra to describe geometric objects Defns. A parabola is the set of all points in the plane that are the same distance from a fixed point as they are from a fixed line. The fixed point is the focus and the fixed line is the directrix. The line through the focus and perpendicular to the directrix is the of intersection of the parabola with the axis of symmetry is the vertex. axis of symmetry. The point Defns. An ellipse is the set of all points in the plane such that the sum of their distances to two fixed points is a constant. The two fixed points are the the ellipse and the line through the foci are the foci. The two points of intersection of vertices. The line containing the vertices is the major axis. The midpoint of the line segment joining the foci is the center. The line through the center and perpedicular to the major axis is the minor axis. Defns. A hyperbola is the set of all points in the plane such that the positive difference of their distances to two fixed points is a constant. The two fixed points are the intersection of the hyperbola and the line through the foci are the foci. The two points of vertices. The line through the vertices is the transverse axis. The midpoint of line segment joining the foci is the center. The line through the center and perpendicular to the transverse axis is the conjugate axis.

24 Dr. Nestler - Math Hyperbolas Definitions. A hyperbola is the set of all points in the plane such that the positive difference of their distances to two fixed points is a constant. The two fixed points are the foci (or focus points). Suppose the foci are Ð-ß!Ñand Ð -ß!Ñwith -!, so the center is G œð!ß!ñ. Note that -œ.ðgßjñwhere Jstands for one of the foci. By definition, there exists a positive constant + such that.ðtßj" Ñ.ÐTßJ Ñ œ + for all points T on the hyperbola. Suppose T is a point on the hyperbola that is closer to focus J than to focus J ". Then we have.ðtßj" Ñ.ÐTßJ Ñ œ +. So From the triangle in the picture, we see that.ðtß J Ñ.ÐTß J Ñ - ".ÐTß J Ñ.ÐTß J Ñ - " By algebra (see p. 681), the equation is C œ " where, œ - +. B +, Let's look for vertices (points on the hyperbola lying on the line through the foci, in this case the B-axis): if C œ! then B œ +. So the vertices are Ð +ß!Ñ. Note that + œ.ðgßzñ. As a reminder, + -. We have three points on the hyperbola so far: Tand the two vertices.

25 C, Let's look for C-intercepts: If B œ! then œ ", which is impossible. So there are no B C B +, + C-intercepts. In fact, œ ", so ", so B +, so lbl +, and so there are no points on the graph for branches. + B +. Thus the hyperbola has a gap in between its two halves, called Next let's look for end behavior, as B gets large in absolute value. B C, +, + Theorem. The hyperbola œ " has the two oblique asymptotes C œ B., + Idea of Proof: Solve the hyperbola equation for C to get C œ B " (see p. 685). + B As B gets large in either direction, this is close to, + B, since the root approaches 1. Now we can finish sketching the graph of the hyperbola equation. We use the asymptotes to draw the graph in the first quadrant. The graph is symmetric across both the B C +, B-axis and the C-axis, since replacing C with C or Bwith Bin œ " results in an equivalent equation. It can be shown using calculus that each branch of the hyperbola is a smooth, unbroken curve with no extra bumps or wiggles. The line through the vertices is called the joining the foci is the transverse axis. The midpoint of line segment center of the hyperbola. The line through the center and perpendicular to the transverse axis is called the conjugate axis.

26 Now suppose the foci are Ð!ß -Ñ, so the center is still Ð!ß!Ñ but the transverse axis is the C-axis. Then an equation is: œ " where, œ - + C B +, This time there are no B-intercepts or points with + C +. Its two oblique asymptotes are Cœ +, B. Example: Analyze and graph the equation C %B œ %. B Put the equation into standard form: C % " œ " So this is a hyperbola, the center is The conjugate axis is +œ so the vertices are,œ so the asymptotes are Cœ - œ so the foci are Ð!ß!Ñ, and the transverse axis is

27 Example: Analyze and graph the equation B %C B "'C "" œ!. Put this equation into standard form by completing the squares: ÐC Ñ ÐB "Ñ " % œ " This has standard form ÐC 5Ñ ÐB 2Ñ +, œ " The center is Ð2ß 5Ñ œ The transverse axis is The conjugate axis is +œ so the vertices are,œ and the asymptotes are of the form reflective property: An object traveling in a straight line toward one focus of a hyperbola will be reflected toward the other focus. [You can do and Ch 9 Review 1-2!, 39, 40]

28 Dr. Nestler - Math Proof of the Binomial Theorem Binomial Theorem. Suppose + and, are real numbers and 8 is a positive integer. Then Ð+,Ñ œ , â +,, " 8 " 8! " 8 " 8 8 œ 8 8 < < +, Þ <œ! Proof: We use induction on 8. < " First, we see that the statement above is true for 8œ" : Ð+,Ñ œ+,œ " " +,. Next, suppose the statement above is true for some integer 8œ5: < < < <œ! Ð+,Ñ œ 5 +,. 5 " 5 " <œ! <! " 5 " We will show the statement is true for 8œ5 "À Ð+,Ñ œ 5 " < < +,. 5 " 5 (1) Ð+,Ñ œ Ð+,ÑÐ+,Ñ (2) œð+,ñ 5 +, 5 <œ! < 5 < < < < 5 < < < < <œ! <œ! (3) œ + 5 +,, 5 +, 5 5 (4) œ 5 5 " < < +, 5 +, <œ! < < <œ! 5 < < " 5 5 " 5 " 5 " < < 5 < < " 5 "! < < 5 <œ" <œ! (5) œ , 5 +, 5, " 5 " < < 5 Ð< "Ñ Ð< "Ñ " 5 "! < < " 5 <œ" <œ" (6) œ , 5 +, 5, " 5 " < < 5 " < < 5 "! < < " 5 <œ" <œ" (7) œ , 5 +, 5, 5 5 " 5 " < < 5 "! < < " 5 <œ" (8) œ , 5,

29 5 5 " 5 " < < 5 "! < 5 " <œ" (9) œ 5 " + 5 " +, 5 ",. 5 " 5 " <œ! < (10) œ 5 " < < +, as desired. Since 5 was arbitrary, we are done by induction.