BEE1020 { Basic Mathematical Economics Dieter Balkenborg Week 9, Lecture Wednesday 04.12.02 Department of Economics Discussing graphs of functions University of Exeter 1Objective We show how the ¯rst and the second derivative of a function can be used to describe the qualitative properties of its graph. This will be used in the next lecture to solve optimization problems. 2 Roots, peaks and troughs, in°ection points 2.1 Roots: f (x)=0 A solution to the equation f (x)=0wheref is a function is called a root or x-intercept. At a root the function may or may not switch sign. In the ¯gure on the left the function the function has a root at x = 1and switches sign there. In the ¯gure on the right the function has a root at x = 1,¡ but does not switch sign there. ¡ 8 6 3 4 2 0 -2 -1x 1 2 -2 1 -4 -6 0 -2 -1x 1 2 f (x)=x3 +1 f (x)=x2 4 10 2 8 6 0 -2 -1x 1 2 4 -2 2 -4 0 -2 -1x 1 2 2 f 0 (x)=3x f 0 (x)=2x Lemma 1 Suppose the function y = f (x) is di®erentiable on the interval a x b with · · a<b.Supposex0 with a<x0 <bisarootoftheafunctionandf 0 (x0) =0.Thenthe 6 function switches sign at x0 2.2 Critical points: f 0 (x)=0 A solution to the equation f 0 (x)=0,wheref 0 is the ¯rst derivative of a function f,is called a critical point or a stationary point of the function f. At a stationary point the graph of the function has a horizontal tangent. 0.8 0.6 0.4 0.2 0 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4x 0.6 0.8 1 x = 0 is a critical point Stationary points come in three varieties: Suppose the di®erentiable function is de¯ned on an interval a x b with a<band · · that a<x0 <bis a critical point of the function. We call x0 a peak or a relative maximum or a local maximum if the function turns ² from being increasing on the left to being decreasing on the right at x0. We call x0 a trough or a relative minimum or a local minimum if the function turns ² from being decreasing on the left to being increasing on the right at x0. We call x0 a saddle point if the function is increasing (or decreasing) both to the ² left and to the right of x0. In the following graphs x0 = 0 is a critical point. The ¯rst row shows the graphs of the functions. The second row shows the graphs of the derivative. -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4x 0.6 0.8 1 0 1.8 0.8 -0.2 1.6 1.4 0.6 -0.4 1.2 1 0.4 -0.6 0.8 0.6 0.2 -0.8 0.4 0.2 0 0 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4x 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4x 0.6 0.8 1 Apeak,f (x)= x4 + 1 A trough, f (x)=x4 1 A saddlepoint, f (x)= x3 +1 ¡ ¡ ¡ -1 -0.8 -0.6 -0.4 -0.2 0.2 0.4x 0.6 0.8 1 4 4 0 -0.2 -0.4 2 2 -0.6 -0.8 -1 -1 -0.8 -0.6 -0.4 -0.20 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.20 0.2 0.4 0.6 0.8 1 x x -1.2 -1.4 -2 -2 -1.6 -1.8 -4 -4 3 3 2 f 0 (x)= 4x f 0 (x)=4x f 0 (x)= 3x ¡ ¡ Proposition 1 Suppose the function y = f (x) is continuously di®erentiable on the in- terval a x b with a<band that a<x0 <bis a critical point of the function. Then · · 2 a) x0 is a peak of the function if and only if the ¯rst derivative f 0 (x) changes sign from + to at x0. ¡ b) x0 is a trough of the function if and only if the ¯rst derivative f 0 (x) changes sign from to + at x0. ¡ 2.2.1 A su±cient criterion for peaks and troughs Often the second derivative can be used to determine whether a critical point x0 is a peak or a trough of the function. Question: Suppose f 0 (x)=0.Isx0 a peak, a trough or a saddle point if a) f 00 (x0) > 0, b) f 00 (x0) < 0orc)f 00 (x0)=0? Correct answer: to a) The function has a trough at x0 (because it is upward bowed near x0). to b) The function has a peak at x0 (because it is downward bowed near x0). to b) I don't know. It could be a trough a peak or a saddle point. Ask Dieter. Exercise 1 a) Verify that x = 2 is a critical point of the function y = f (x)=3x4 16x3 +30x2 24x +5: ¡ ¡ Is x = 2 a trough, a a peak or a saddle point? b) Verify that x = 1 is a critical point of the function y = f (x)=3x4 16x3 +30x2 24x +5: ¡ ¡ Is x = 1 a trough, a peak or a saddle point? 2.3 Candidates for in°ection points: f 00 (x)=0 x0 is an in°ection point of the function f (x) if the function changes from being concave to being convex or vice versa at x0. For instance, f (x)=x (x +1)(x 1) has an in°ection ¡ point at x0 =0. 0.4 0.2 0 -1 -0.5 0.5x 1 -0.2 -0.4 f (x)=x (x +1)(x 1) ¡ We call x0 a candidate for an in°ection point if f 00 (x0)=0. Proposition 2 Suppose the function y = f (x) is twice continuously di®erentiable on the interval a x b with a<band that a<x0 <bis a candidate for an in°ection point, · · i.e., f 00 (x0)=0.Thenx0 is indeed an in°ection point if and only if the second derivative changes sign at x0. Proposition 3 Suppose the function y = f (x) is three times di®erentiable on the interval a x b with a<b. Suppose that a<x0 <bsatis¯es f 00 (x0)=0and f 000 (x0) =0. · · 6 Then x0 is an in°ection points. 3 3 An odd polynomial We will study the following function 1 13 y = f (x)= x5 + x3 36x ¡5 3 ¡ Such a polynomial is called odd because it contains no powers with even index like x4, x2 and x0 =1. 4Theroots Clearly, x = 0 is a root and hence x a factor, so 1 65 y = f (x)= x x4 x2 + 180 ¡5 ¡ 3 µ ¶ We claim that the fourth order polynomial 65 g (x)=x4 x2 +180 ¡ 3 has no roots and is everywhere strictly positive. Hence f (x) has only a root at x =0;is positive to the left and negative to the right. To see this, compare g (x) with the quadratic polynomial 65 z2 z +180 ¡ 3 65 2 The latter has no roots because the discriminant is negative 3 4 180 250:56 < 0. If x were a root of g (x)thenz = x2 would be a root of¡ the¡ quadratic£ ¼¡ polynomial, which is impossible. So g (x) has no root. Moreover, g (0) =¡ 180 ¢> 0. If there would exist anumberx1 with g (x1) < 0 then (since g (x) is di®erentiable and hence continuous) there would be a number between x =0andx1 where the graph of g (x) cuts the horizontal axis. Again this is impossible because g (x) has no root. Therefore g (x) > 0forall numbers x. 4.1 The critical points The ¯rst derivative of f (x)is 1 5 13 3 0 y0 = f 0 (x)= x + x 36x ¡5 3 ¡ µ ¶ = x4 +13x2 36 ¡ ¡ Compare this polynomial with the quadratic polynomial z2 +13z 36 ¡ ¡ 4 If we replace in the latter z with x2 we obtain our above derivative. It is easily seen that z0 =4andz1 = 9 are the roots of the quadratic polynomial. Therefore we have the factorization z2 +13z 36 = (z 4) (z 9) . ¡ ¡ ¡ ¡ ¡ Consequently, 4 2 2 2 f 0 (x)= x +13x 36 = x 4 x 9 ¡ ¡ ¡ ¡ ¡ (You should verify this by expanding the product on¡ the right.)¢¡ x2¢ 4isadi®erenceof two squares and therefore x2 4=(x +2)(x 2). Using the same¡ trick on x2 9we obtain the factorization ¡ ¡ ¡ y0 = f 0 (x)= (x +2)(x 2) (x +3)(x 3) ¡ ¡ ¡ We conclude that our polynomial f (x) has four critical points, namely x = 3, x = 2, x =+2andx = +3. To see where our function is increasing / decreasing¡ etc. we draw¡ a sign diagram for the derivative. Our sign diagram contains one row for each factor and one row for f 0 (x)itself.It contains one column for each root and additional columns for the intervals between the roots. It also contains a column for all numbers to the left of all roots and a column for all numbers to the right. x< 3 x = 3 3 <x< 2 x = 2 2 <x<2 x =2 2 <x<3 x =3 3 <x ¡ ¡ ¡ ¡ ¡ ¡ 1 x¡+3 ¡ ¡0 +¡ +¡ +¡ +¡ +¡ +¡ +¡ x +2 ¡ 0 + + + + + x 2 ¡ ¡ ¡ 0 + + + x ¡ 3 ¡ ¡ ¡ ¡ ¡ 0 + ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ f (x) 0 + 0 0 + 0 0 ¡ ¡ ¡ The last line of the diagram tells us precisely where the derivative is negative positive or zero.