12.2. Maxima and Minima. Introduction. Prerequisites. Learning Outcomes

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1 Maima and Minima 1. Introduction In this section we analse curves in the local neighbourhood of a stationar point and, from this analsis, deduce necessar conditions satisfied b local maima and local minima. Locating the maima and minima of a function is an important task which arises often in applications of mathematics to problems in engineering and science. It is a task which can often be carried out using onl a knowledge of the derivatives of the function concerned. The problem breaks into two parts finding the stationar points of the given functions distinguishing whether these stationar points are maima, minima or so-called points of inflection. Prerequisites Before starting this Section ou should... Learning Outcomes After completing this Section ou should be able to... 1 be able to take first and second derivatives of simple functions be able to find the roots of simple equations understand the difference between local and global maima and minima appreciate how tangent lines change near a maimum or a minimum locate the position of stationar points use knowledge of the second derivative to distinguish between maima and minima

2 1. Maima and Minima Consider the curve = f() shown in the following figure a b f(a) f(b) a 0 1 b B inspection we see that for this curve there is no value greater than that at = a (i.e. f(a)) and there is no smaller value than that at = b (i.e. f(b)). However, the points on the curve at 0 and 1 merit comment. It is clear that in the near neighbourhood of 0 all the values are greater than the value at 0 and, similarl, in the close neighbourhood of 1 all the values are less than the value at 1. We sa f() has a global maimum at = a and a global minimum at = b but also has a local minimum at = 0 and a local maimum at = 1. Our primar purpose in this section is to see how we might locate the position of the local maima and the local minima for a smooth function f(). A stationar point on a curve is one at which the derivative has a zero value. In the following diagram we have sketched a curve with a maimum and a curve with a minimum. 0 0 B drawing tangent lines to these curves in the close neighbourhood of the local maimum and the local minimum it is obvious that at these points the tangent line is parallel to the ais so that =0 0 Points on the curve = f() atwhich Ke Point =0are called stationar points of the function HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima

3 However, be careful! A stationar point is not necessaril a local maimum or minimum of the function but ma be an eceptional point called a point of inflection: see diagram. 0 Eample Sketch the curve =( ) + and locate the stationar points on the curve. Solution Here f() =( ) +so =( ). At a stationar point =0sowehave ( ) = 0 so =.Weconclude that this function has just one stationar point located at =(where = ). B sketching the curve = f() itisclear that this stationar point is a local minimum. Locate the position of an stationar points of f() = First find = Now locate the stationar points b solving =0 = =3( + 1)( ) =0 so = 1 or = 3 HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima

4 Finall the coordinates of the stationar points can be found b substituting these values into the equation for. The stationar points are (0, ) and ( 1, 13.5). We have, in the following diagram, sketched the curve which confirms our deductions. ( 13.5,1).5 Sketch the curve = cos 0.1 3π 4 and locate the position of the global maimum, global minimum and an local maima and/or minima. 0.1 π/4 π/ 3π/4 global maimum 0.1 π/4 π/ 3π/4 local minimum and global minimum HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima 4

5 . Distinguishing between local maima and minima We might ask if it is possible to predict when a stationar point is a local maimum, a local minimum or a point of inflection without the necessit of drawing the curve. To do this we highlight the general characteristics of curves in the neighbourhood of local maima and minima. For eample: at a local maimum (located at 0 sa) the following diagram correctl describes the situation: f() 0 to the left of the maimum and to the right of the maimum > 0 < 0 If we draw a graph of the derivative against then, near a local maimum, it must take one of two basic shapes described in the following diagram: or 0 α 0 (a) (b) ( ( d d In case (a) tan α<0 whilst in case (b) =0 ) 0 ) 0 We reach the conclusion that at a stationar point which is a maimum the value of the second derivative d f is either negative or zero. Near a local minimum the above graphs are inverted f() 0 whilst the graphs of the derivative are either: to the left of the minimum and to the right of the minimum < 0 > 0 or 0 β 0 (a) (b) 5 HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima

6 > > Here, for case (a) ( d = tan β>0 ) 0 whilst in (b) ( d =0. ) 0 In this case we conclude that at a stationar point which is a minimum the value of the second derivative d f is either positive or zero. For the remaining possibilit for a stationar point, a point of inflection, the graph of f() against and of against take one of two forms: f() f() to the left of 0 > 0 to the left of 0 0 to the right of 0 > 0 to the right of 0 0 ( d For either of these cases =0 ) 0 The sketches and analsis of the shape of a curve = f() inthe near neighbourhood of stationar points allow us to make the following important deduction Ke Point If 0 locates a stationar point of the function f(), so that =0, then the stationar 0 point is a local minimum if > 0 0 and is a local maimum if < 0 0 and is inconclusive if =0 0 HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima 6

7 Eample Find the stationar points of the function f() = 3 6 Are these stationar points local maima or local minima? Solution =3 6 At astationar point =0so3 6=0impling = ±. Thus f() has stationar points at = and =. To see if these are maima or minima we eamine the value of the second derivative of f() atthe stationar points. Now d f =6 so d f =6 > 0. Hence = locates a local minimum. = Similarl d f = 6 < 0. Hence = locates a local maimum. = A simple sketch of the curve confirms this analsis. f() For the function f() =cos, 0.1 6, find the positions of an local minima and/or maima and distinguish between each kind. = stationar points are located at: Hence stationar points are at values of for which sin =0i.e. at = π or =π or =3π (make sure is within the range Now calculate the second derivative = possible stationar points at = π,= π, = 3π = sin. 7 HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima

8 = 4 cos Finall: evaluate the second derivative at each of the stationar points and draw the appropriate conclusions. = = π = =π = = 3π = π = 4 cos π =4> 0 = π locates a local minimum. = 4 cos π = 4 < 0 = π locates a local maimum. =π = 3π = 4 cos 3π =4> 0 = 3π f() locates a local minimum. 0.1π/4 π/ 3π/4 3π/ 6 Determine the local maima and/or minima of the function = First obtain the positions of the stationar points. f() = Thus =0when: = HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima 8

9 =43 = (4 1) =0when =0or when = 1 4 Now obtain the value of the second derivatives at the stationar points =0 = = 1 4 = =1 =0, =0 Hence = 1 4 = 1 4 = 1 1 = 1 > locates a local minimum. However, using this analsis we cannot conclude that the stationar point at = 0isalocal maimum, minimum or a point of inflection. However just to the left of =0the value of (which equals (4 1)) is ve whilst just to the right of =0the value of is ve. Hence the stationar point at =0is a point of inflection. This is confirmed b sketching the curve. f() /4 Eercises Locate the stationar points of the following functions and distinguish between maima and minima. (a) f() = ln. (b) f() = 3 (c) f() = ( 1) (+1)( ) Remember d (ln ) = 1. 1 << 9 HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima

10 Answers (a) =1 1 =0when =1 = 1 =1> 0 =1 = 1, = 1 locates a local minimum f() 1 (b) =3 =0when =0 =6 =0when =0 However, > 0oneither side of =0so (0,0) is a point of inflection. f() (c) = (+1)( ) ( 1)[ 1] (+1)( ) This is zero when ( + 1)( ) ( 1)( 1) =0 i.e. + 3=0 However, this equation has no real roots (b < 4ac) and so f() has no stationar points. The graph of this function confirms this. f() 1 1 HELM (VERSION 1: March 18, 004): Workbook Level 1 1.: Maima and Minima 10