Previously in this lecture: Roughly speaking: A function is increasing (decreasing) where its rst derivative is positive

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1 BEE14 { Basic Mathematics for Economists Dieter Balkenborg BEE15 { Introduction to Mathematical Economics Department of Economics Week 6, Lecture, Notes: Sign Diagrams 6/11/1 University of Exeter Objective Previously in this lecture: Roughly speaking: A function is increasing (decreasing) where its rst derivative is positive (negative). A function is strictly convex (strictly concave) where its second derivative is positive (negative). In order to apply this to a polynomial function y = P (x) wehavetosolveinequalities of the type P (x) > and P (x) > for x. In general, solving inequalities is messy and prone to errors. Consider, for instance, P (x) = x +1 x>: (1) We may be tempted to divide by x, as we would do with an algebraic equation, to obtain x +1> () which is true for all x since x is non-negative. We may conclude that P (x) > for all x 6=. (Forx =wehavep (x) =.) However, this reasoning is wrong because if x is a negative number then division by x changes the sign on the left-hand side of (1). So, if x<thenx + 1 must also be negative to have P (x) >, i.e., when we divide by negative x the inequality reverts sign and we obtain x +1< (3) instead of (). Inequality 3 is not satis ed for any x. Putting together our results for positive and negative x we see that P (x) > ifandonlyifx>. There is a much simpler way to get to this result which avoids any algebraic manipulation of inequalities: P (x) is the product of the two factors x +1andx. In order for P (x) to be positive either both factors must be positive or both must be negative. Since x +1isalwayspositive,x must be positive. The method of sign diagrams uses this simple type of consideration. It requires us, however, to nd a \nice" factorization of the relevant polynomial P, which, in turn, requires us to nd its roots. In this lecture we will rst discuss

2 1. How sign diagrams can be used to decide where a polynomial is positive- or negative valued.. How to nd the roots of a polynomial (i.e., the solutions to the equation P (x) =) in special cases. 3. How roots help to factorize a polynomial. Thereafter we demonstrate how these methods can be used to determine the important qualitative features of a polynomial function without drawing tables or evaluating the polynomial at many points. We present this information with the aid of a summary sign diagram. 1 Sign diagrams Consider the polynomial P (x) =(x +5)(x ) ( x +6)= x 4 +4x 3 +38x 136x + 1 Obviously, the roots are x = 5, x =andx = 3. To nd out where P (x) is positive or negative we draw a sign diagram. This is a table with one column for each root and one column for each adjacent interval. There is one row for each factor of the polynomial and a nal row for the polynomial itself. The entries in the table are +, or. For each factor it is easy to decide where it is positive, negative or zero and hence to make the corresponding entry in the table. Once we know the signs of all factors in an interval, we know the sign of P (x) inthisinterval.inourexample x< 5 x = 5 5 <x< x = <x<3 x =3 3 <x x x x x P (x) + + The signs for the factor x+ 6 are obtained as follows: A linear factor changes sign only once, namely at the root which is here x =3(since x +6 = yields 6 = x). For x =4 we have x+6 = <. Therefore x+1 is positive to the right of x = 3 and it must be positive to the left of the root. (Check: For x = we have indeed x +6=>.) For x< 5 andfor3<xthe polynomial P (x) is negative because it has an odd number of negative factors. For 5 <x< and for <x<3 the polynomial is positive because it has an even number of negative factors.

3 Alookatthegraphofy = P (x) con rmsourresults: x x Problem 1 Construct the sign diagram of the polynomial P (x) = 3(x +1) 3 (x 1) (x 4) = 3x 6 +9x 5 +18x 4 18x 3 7x +9x +1 Solution 1 5 (x +1) 3 (x 1) x 4 P (x) x< 1 x = 1 1 <x<1 x =1 1 <x<4 x =4 4 <x Finding roots of a polynomial The hard work is to nd the roots of a polynomial and to factorize it. Except for linear or quadratic polynomials, we restrict ourselves to methods which work only in special cases. Nonetheless, we start with a very deep and general result in algebra..1 The fundamental theorem of algebra Gauss (1777 { 1855): Every non-constant polynomial can be written as a product of linear factors and quadratic factors with no real roots. As a consequence, the roots of a polynomial are precisely the roots of its linear factors. Example 1 x 4 1=(x +1)(x 1) = (x +1)(x +1)(x 1) using twice the always important formula a b =(a + b)(a b) : Here the quadratic factor x +1hasnoreal roots. Example x 8 1=(x 4 +1)(x 4 1) = (x 4 +1)(x +1)(x +1)(x 1) where the polynomial x 4 +1 has no real roots and must hence be the product of two quadratic polynomials 3

4 with no real roots. This factorization is harder to nd, however ³ x + p ³ x +1 x p x +1 = x 4 + p x 3 +x p x 3 x p x x + p x +1 = x 4 +1 so x 8 1= x + p x +1 x p x +1 (x +1)(x +1)(x 1) where the quadratic factors are easily seen to have no real roots.. Roots of linear polynomials The root of a linear polynomial P (x) =ax + b with a 6= isx = b a..3 Roots of quadratic polynomials The roots of a quadratic polynomial f (x) =ax + bx + c with a 6= aregivenby x 1= = b p b 4ac : a When the discriminant b 4c is negative there are no real roots. Suppose x 1 ;x are the roots of a quadratic polynomial. Then one has the formulas of Vieta (154 { 163) and the factorization is since by Vieta's formulas. x 1 + x = b a and x 1x = c a f (x) =a (x x 1 )(x x ) a (x x 1 )(x x )=a x (x 1 + x ) x + x 1 x = ax + bx + c Supplementary useful information on quadratic function: The graph of a quadratic function is called a parabola. If a> the function is strictly convex with a unique minimum at x = b. If a< the function is strictly concave a with a unique maximum at x = b. The parabola is mirror-symmetric to the vertical a line through the maximum/minimum (x ; ), i.e., one has f (z + x )=f( z + x ) for all z. The minimum or maximum is always in the middle between the two roots x 1 ;x when the two exist because x = x 1+x by Vieta's formula. 4

5 x x a convex parabola aconcaveparabola Reading: Ho mann and Bradley (), Appendix A Problem Suppose the government imposes an excise tax t, where t is the percentage of the price charged to consumers a) What is tax revenue when the tax is t =%? b) What is tax revenue when the tax is t =1%? c) Suppose tax revenue is a quadratic function of the excise tax t imposed. What excise tax does then maximize tax revenue? Solution?.4 Polynomials of higher order: historic remarks Greece (ca. 1 BC { 6 AC): The early Greeks were the masters of geometry. They studied the quadratic curves (parabolas, hyperbolas, ellipses) as intersections with cones. hyperbolas parabolas circles or ellipses 5

6 Orient (ca. 6 AC { 15 AC): Algebra was invented in the islamic countries. Muhamed ibn Musa al-khwarizmi (»85): \Hisab al-jabr wal-muqabala" (\The Science of Reduction and Mutual Cancellations") `al-jabr'! `algebra'. He verbally discussed equations like 3x +4=x. Omar Khayyam (ca. 138 {113): He found solutions to cubic equations in special cases. Europe (Renaissance): \mathematical entrepreneurs" Scipio del Ferro (died 156): reported to have known the solutions to all cubic equations, but never told anyone how to do it. Tartaglia (the \stutterer") rediscovered the general method to solve all cubic equations and announced that he could do it (but not how) in a public lecture. He explained his ndings to Hieronimo Cardano in a private conversation. To Tartaglia's great dismay Cardano published the results in a book (1546). The formulae are now named after Cardano. For instance, the equation has a unique root, namely x = 3 s r p 3 x 3 + px = q p;q > 5 + q 4 + q 3 s r p q 4 q These formulae are too bulky for exam purposes! In 1547 Ferrari found the general solution to polynomial equations of order 4 (`biquadratic equations'). Evariste Galois (1811 { 183): Was not considered to be a student who could express himself very well. The night before he died in a duel (aged 1) he wrote down his mathematical ideas and asked for them to be sent to Gauss. This \gibberish" (his own words) is now known as \Galois theory". It could be solved many open problems, for instance that squaring the circle is impossible and that polynomial equations of degree 5 or higher cannot be solved by formulae using roots ( p p + p etc.). Niels Hendrik Abel (18 { 189): Thought to have solved polynomial equations of degree 5 as a student, but later proved that this is not possible. He died from consumption and poverty. Of course, numerical methods approximately to solve such equations exist..5 Integer roots of integer polynomials Integer roots of a polynomial with integer coe±cients divide the constant term. Proof: Suppose that x is an integer root of the polynomial with integer coe±cients. Then P (x) =a n x n + :::+ a 1 x + a P (x )=a n x n + :::+ a 1 x + a = 6

7 which can be rewritten as a n x n ::: a 1x = a or as So x is a factor of a. an x n 1 ::: a 1 {z } integer x {z} integer = a {z} integer Problem 3 Find a root of f (x) =x 3 +1x +31x +3: Solution 3 3 = 3 5. Try x = 1; ; 3; 5; 6; 1; 15; 3. f (1) > f ( 1) = = < f () > f ( ) = = HIT! f (3) > f ( 3) = = HIT! f (5) > f ( 5) = = HIT! The factorization must be (x ) (x 3) (x 5) Problem 4 Find all roots of P (x) =x 3 3x 5x +75. Solution 4?.6 Reducing the degree for even polynomials A polynomial is called even if it has only even powers. For such polynomials the substitution z = x leads to a polynomial of only half the degree, for which roots are easier to nd. Problem 5 Find at least two roots of P (x) =x 5 5x 3 +6x. Solution 5 This is an odd polynomial with factor x and root x =. To nd further roots we divide by x and obtain the polynomial Q (x) =x 4 5x +6x which is even. Using the substitution z = x we have Q p z = z 5z +6: This is a quadratic polynomial. Since 6 = 3and 3= 5 it has the roots z = and z = 3. Hence x = p andx = p 3 are roots of the original polynomial. In fact, ³ x p ³ x + p ³ x p ³ 3 x + p 3 = x x 3 = x 4 x 3x +6=Q (x) 7

8 3 Factorization 3.1 Polynomial division Polynomial division is similar to the division of two integers with remainder (1 7 =14 7 etc.), only simpler. To divide the polynomial by the polynomial we proceed as follows: P (x) =6x 6 +5x 5 +4x 4 +3x 3 +x +1 Q (x) =x 3 + x + x + x +1 6x 3 x x 1 x 3 +x +x +1 6x 6 +5x 5 +4x 4 +3x 3 +x +x +1 R = P 6x 6 +6x 5 +6x 4 +6x 3 x 5 x 4 3x 3 +x +x +1 R 1 x 5 x 4 x 3 x x 4 x 3 +x +x +1 R x 4 x 3 x x x 3 x +x +1 R 3 x 3 x x 1 3x +3x + R 5 We divide the leading term 6x 6 of the polynomial P (x) bytheleadingtermx 3 of Q (x), which gives 6x 3. We write 6x 3 above the term 6x 6 of P (x). Then we multiply each term of Q (x) with6x 3 and write it below P (x). Then we form the di erence between P (x) and the polynomial below. This gives our rst, intermediate remainder R 1 (x). R 1 (x) has onedegreelessthanp because the term 6x 6 cancels in the subtraction. We now proceed with R 1 (x) { and with every successive remainder { in the same fashion as we did before with P (x). Namely, we divide the leading term x 5 of R 1 (x) by the leading term x 3 of Q (x) and write the result x above P (x). Then we multiply each term of Q (x) with x and write the result below R 1 (x). We subtract to obtain the next remainder R (x), which is a polynomial of degree 4. We continue in this fashion to obtain R 3 (x), R 4 (x) andr 5 (x). R 5 (x) isapolynomial of degree. If we would try to divide the leading term 3x of R 5 (x) bytheleadingterm x 3 of Q (x) we would get 3 which is no longer a polynomial. Hence we stop once the x degree of the remainder is less than the degree of the polynomial Q (x) bywhichwewant to divide. Our nal remainder is R (x) =R 5 (x) Denoting the polynomial above P (x) by S (x) =6x 3 x x 1 8

9 we have P (x) Q (x) = S (x)+r (x) Q (x) where in the last fraction the degree of the numerator R 5 (x) issmaller than the degree of the denominator. Notice again the similarity to the division of integers: If we divide 1 by 7 we obtain 14 where in the fraction the numerator is smaller than the denominator. 7 7 Suppose we can divide a polynomial P (x) byapolynomialq (x) without rest, i.e., the remainder R (x) is zero. Then or P (x) Q (x) = S (x) P (x) =S (x) Q (x) : So we have factorized P (x). 3. The factorization theorem Suppose we divide a polynomial P (x) by a linear polynomial x a. The result can be written as P (x) x a = S (x)+ R (x) x a whereby R (x) is a polynomial with a lower degree than x a: This means that the remainder must be a constant, say c. Sowehave or P (x) x a = S (x)+ c x a P (x) =S (x)(x a)+c: Therefore, if we evaluate all expressions at x = a P (a) =S (a)(a a)+c = c: Consequently, the division is without rest (c = ) if and only if a is a root of P (x) (P (a) =). This is the factorization theorem: Theorem 6 If a is a root of the polynomial P (x) then P (x) can be divided by (x a) without rest. The result of the division is a polynomial of one degree lower than P (x). It follows that a polynomial of degree n can have at most n roots. 9

10 4 The summary sign diagram We now aim at describing the main qualitative features of a polynomial function with the aid of a summary sign diagram basedonthe rstandsecondderivative. The example we consider is Here is a rst glimpse at its graph: y = P (x) = x 4 +x 3 x +1= x The roots Checking for integer roots we nd that P (x) has the roots +1 and 1. Hence P (x) has the factors (x +1) and (x 1). It follows that the product (x +1)(x 1) = x 1is also a factor. Division yields Since x x +1=(x 1) we get x +x 1 x 1 x 4 +x 3 x +1 x 4 +x x 3 x x +1 x 3 x x +1 x +1 P (x) = (x 1) 3 (x +1) A sign diagram shows where the polynomial P (x) has positive or negative values: x< <x<1 1 1 <x (x 1) x P (x) + 1

11 4. Critical points A critical point 1 of a di erentiable function is a root of the rst derivative, i.e., a point where the graph of P (x) has a horizontal tangent. The rst derivative of P (x) is P (x) = 4x 3 +6x = Clearly, x =1isarootofP (x). Polynomial division yields 4x +4x x x 1 4x 3 +6x 4x 3 +4x x x x x x i.e., P (x) = (x 1) (x x 1). The roots of x x 1are x 1= = +1 p 1 4()( 1) 4 = 1 p 9 4 = ; i.e., x 1 =1,x = 1. Hence x x 1=(x 1) x + 1 and µ P (x) = 4(x 1) x + 1 : We obtain the sign diagram x< 1 x = 1 1 <x<1 x =1 1 <x 4(x 1) x P (x) + We conclude: The function y = P (x) is increasing for x< 1 and decreasing for 1 <x. The behaviour of the function near a critical point deserves some extra terminology: a) A turning point is a citical point where the function turns from being increasing to being decreasing, i.e., where the rst derivative switches sign. Turning points come in two varieties: a1) A relative (or a local) maximum is a point where the function turns from being increasing to being decreasing or vice versa, i.e., where the rst derivative changes sign from + to. a) A relative (or local) minimum is a point where the function turns from being decreasing to being increasing, i.e., where the rst derivative changes sign from to +. 1 The term stationary point is also often used. 11

12 (I emphasize the term \relative" for reasons I will explain in the lecture on optimization. I will also have to explain why the distinction relative versus absolute is not made in A-level mathematics.) b) A saddle point is a critical point which is not a turning point, i.e., the rst derivative does not switch sign. In our example, there is a relative maximum at x = 1 and a saddle point at x =1. Thereisnorelativeminimum. Reading: Ho mann and Bradley (), Chapter 3, Section 1, in particular Examples 1.1 and In ection points The second derivative is P (x) = 1x +6x =( 1x +6)x = 6(x 1) x for which we obtain the sign diagram x< x = <x<1 x =1 1 <x 6(x 1) x P (x) + Thus the function is strictly concave for x< and for x>1. It is strictly convex in between. Points where the second derivative changes sign are called in ection points. In our example we have in ection points at x =andatx =1. Reading: Ho mann and Bradley (), Chapter 3, Section 4.4 Behaviour for very large positive or negative x. A quick glance at a polynomial reveals its behaviour for very large positive or negative values of x. Consider rstapolynomial(or\monomial") Q (x) =x n with n 1: When x takes on larger and larger positive values then the value of x n also becomes arbitrarily : 1

13 large. This is so because x n >xwhenever x> x We say \As x goes to in nity, x n also goes to in nity" and write (without bothering much about the precise de nition of limits) lim x!+1 xn =+1. For very large negative values of x the behaviour of x n depends on the index n. Ifn is odd then x n is negative for negative x and x n gets more negative when x gets more negative. We write lim x! 1 xn = 1 for n odd. If n is even then x n is positive for negative x (since ( a) n = a n )andx n gets more positive when x gets more negative. We write lim x! 1 xn = 1 for n even. Assume x>1. Since x is positive we obtain a valid inequality when we multiply both sides by x. So x >xfollows, hence x >x>1andx > 1. Multiplying both sides of x > 1withx yields x 3 >x > 1, so x 3 > 1. continuing in this fashion we conclude x n >x. 13

14 x x an odd power an even power From this the behaviour of a monomial a n x n with a coe±cient a n deduced. To say it brie y 6= can be easily lim nx n x!+1 = sign(a n ) (+1) (4) lim nx n x! 1 = ( 1) n lim nx n x!+1 In words: When x goes to plus in nity the limit of a n x n is plus in nity when a n is positive and minus in nity when a n is negative. The limit as x goes to minus in nity is the same when n is even and the opposite when n is odd. For arbitrary polynomials P (x) the behaviour for very large positive or negative x is determined by the leading term a n x n and hence by the degree of the polynomial and the sign of the leading coe±cient. In our example the leading term of our polynomial P (x) is x 4,so To see why this is so notice that lim P (x) = 1 lim x!+1 P (x) = 1 x! 1 P (x) = x 4 +x 3 x +1= x 4 µ 1 x + x 1 x 3 : When x is very large and hence 1 very small all terms ; ; 1 are very small and hence x x x x 3 the term in brackets is very close to 1. Then P (x) is approximately the same as x Summary Finally we can gather all our ndings in a summary sign diagram: x 1 1 <x< <x< <x< <x<1 1 1 <x<+1 +1 P (x) P (x) P (x)

15 Here I made columns for all roots, critical points and in ection points, for 1 at the ends and for all intervals in between. There are rows for P (x) and its rst and second derivative. I basically lled in all the information we found above. In addition I evaluated the function at the critical points and at the in ection points to ease the drawing of a rough sketch. We see that the graph of the function comes from 1 on the left, crosses the horizontal axis at x = 1 and keeps moving upward until it reaches a peak at x = 1. From there onwards it falls, but it has a saddle point at x = 1. For this to be possible the graph has to have a \dent", which is re ected in the fact that the otherwise concave function is convex between x =andx =1. A nal glance at the graph con rms all this x Conclusion Central for this lecture was the method of sign diagrams. A sign diagram of a function tells us where the function is positive-valued, where it is negative-valued and where it has roots. A sign diagram for the rst derivative tells us where the function is increasing or decreasing, and where it has relative maxima, relative minima (the derivative changes sign in these cases) or saddle points (the rst derivative has a root, but does not change sign). A sign diagram of the second derivative tells us where the rst derivative is increasing or decreasing and hence where the original function is strictly convex or concave. It also informs us where the function has in ection points (the second derivative changes sign). A summary sign diagram is a convenient way to present this information in a compact form. We can include information on the behavior of the function for very large positive or negative values of the independent variable. To obtain a sign diagram of a polynomial it is necessary to factorize it. To do so it is necessary to nd the roots since a number a isarootofthepolynomialifandonlyif (x a) is a factor of the polynomial. For quadratic polynomials we can always nd the roots by an explicit formula. Sometimes Vieta's rules give a quicker way to nd the roots (see Appendix A, p. 66, inho mann and Bradley ()). For integer polynomials of higher degree we seek for integer roots. These must divide the constant term. The substitution z = x helps to nd roots of even polynomials. Once we have found a root a we divide by (x a) to obtain a polynomial of one degree less which we then try to factorize by the same methods. 15

16 References Hoffmann, L. D., and G. L. Bradley (): Calculus for Business, Economics and the Social Sciences. McGraw Hill, Boston, 7th, international edn. 16

17 Index critical point, 11 in ection point, 1 maximum relative (or local), 11 minimum relative (or local), 11 monomial, 1 parabola, 4 polynomial even, 7 root of a polynomial, saddle point, 1 sign diagram, turning point, 11 Vieta, 4 17

2 Roots, peaks and troughs, in ection points

2 Roots, peaks and troughs, in ection points BEE1 { Basic Mathematical Economics Week 9, Lecture Wednesday.1. Discussing graphs of functions Dieter Balkenborg Department of Economics University of Exeter 1 Objective We show how the rst and the second