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. ¡

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2 0 -2 -1x 1 2 -2 1 -4 -6 0 -2 -1x 1 2 f (x)=x3 +1 f (x)=x2

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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

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.

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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

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

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.

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0 -1 -0.5 0.5x 1

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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

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

4 2 y00 = f 00 (x)= x +13x 36 0 ¡ ¡ = 4x3 +26x ¡ ¡ ¢ 5 2 13 We have y00 = 4x x .So,x = 0 is clearly a candidate for an in°ection point. ¡ ¡ 2 Moreover, x = 13 2:55 are the solutions to x2 13 = 0. These solutions are also § 2¡ ¼§ ¢ ¡ 2 candidates for in°ectionq points. We obtain the factorization

13 13 y = f (x)= 4x x + x 00 00 2 2 ¡ Ã r !Ã ¡ r ! and hence the sign diagram

x< 13 x = 13 13

We see that all candidates for in°ection points are indeed candidates for in°ection points because the second derivative changes sign. Moreover, the function is convex for x< 13 and for 0

x : 3 13 2 0 2 13 3 ¢¢¢ ¡ ¢¢¢ ¡ 2 ¢¢¢ ¡ ¢¢¢ ¢¢¢ ¢¢¢ 2 ¢¢¢ ¢¢¢ f (x) + + + q+ + + + 0 q ¡ ¡ ¡ ¡ ¡ ¡ ¡ f 0 (x) 0 + + + 0 0 + + + 0 ¡ ¡ ¡ ¡ ¡ f 00 (x) + + + 0 0 + + + 0 ¡ ¡ ¡ ¡ ¡ ¡

If you have not seen it before, this was quite some bit of analytic geometry. Without drawing any graphs we determined the main qualitative features of the graph of the function. We used some very basic di®erentiation and some slightly more sophisticated algebra (although, in the end it was just solving quadratic equations). You should take some time out to check how the information in the tables is re°ected in the graphs below. You should also check that what we have said about peaks, in°ection points etc. is visible in the graphs.

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0 -3 -2 -1 1x 2 3 -20

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y = f (x)

-3 -2 -1 1x 2 3 0

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y = f 0 (x)

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0 -3 -2 -1 1x 2 3

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-100

y = f 00 (x) 5 A rational function

This is Example 3.7 from the textbook

x y = f (x)= (x +1)2

1. The numerator has degree 1, the denominator degree 2. Hence the horizontal axis is an asymptote.

2. The denominator is always non-negative. Hence f (x) is positive or negative accord- ing to whether x is positive or negative.

3. The denominator is zero for x = 1. Hence the vertical line at this point is an asymptote. We have f (x)= 1 ¡ x where the ¯rst part is not de¯ned at x = 1 (x+1)2 £ 1 ¡ and the second part is 1whenx = 1. Therefore the function behaves near ¡ 1 ¡ x = 1 similar to the function y = x2 near x =0.Hence,asweapproachx = 1 from¡ the left or the right, the values¡ of the function explode towards . ¡ ¡1

7 4. Summarizing the information so far:

x x< 1 x = 1 1

5. The ¯rst derivative is (using the quotient rule and the general power rule)

2 2 2 (x)0 (x +1) x (x +1) 0 (x +1) x 2(x +1)(x +1)0 y0 (x)= ¡ = ¡ 2 2 4 (x +1)¡ ¢ (¡x +1) ¢ (x +1)2 2x (x +1) (x +1) 2x x +1 = ¡¡ ¢ = ¡ = ¡ (x +1)4 (x +1)3 (x +1)3 which gives the sign diagram

x< 1 x = 1 1

which gives the sign diagram

x< 1 x = 1 1

f 00 (x) 0 + ¡ £ ¡ Instead of giving a summary sign diagram, let me describe verbally the information obtained by the above calculations: Coming from on the left the graph of the function is close to, but below the horizontal axis. As¡1x increases it remains below it and is decreasing and downward- bowed (concave) until it \explodes" towards to the left of 1. To the right of 1 the function shoots up from and is increasing¡1 and downward-bowed.¡ It crosses the¡ horizontal axis at x =0; but¡1 remains increasing until it reaches a peak at x =1.From there onward it is decreasing but it remains positive. There is an in°ection point at x =2 where the graph turns from being downward bowed to being upward bowed. From there onwards the graph approaches more and more the horizontal axis from above.

8 We can deduce all this before turning on the graphic calculator. Of course it is useful to draw the graph in order to see that we are correct:

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-4 -2 2x 4 0

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Fig. 12: y (x)= x (x+1)2

Exercise 2 Read carefully through example 3.8 of the textbook. Verify that the ¯rst and second derivative are correct. Unluckily the in°ection point can only be found numerically in this example. Reading: (Ho®mann and Bradley 2000), Chapter 3, Sections 1 - 3

References

Hoffmann, L. D., and G. L. Bradley (2000): Calculus for Business, Economics and the Social Sciences. McGraw Hill, Boston, 7th, international edn.

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