MAXIMA, MINIMA AND POINTS OF INFLEXION

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1 Page of 8 MAXIMA, MINIMA AND POINTS OF INFLEXION Introduction 0 M Q S f( x) P I I R I I T 5 7 a 0 b x The diagram shows the graph of y domain a x b. The points Q and S are called local maxima. The points P, R and T are called local minima. x = f ( x) where f ( x) is a continuous function defined on the The points are designated local maxima or local minima to distinguish from the gobal maximum and global minimum. The latter refer to the greatest and least values attained by f ( x) over the domain. Thus in the diagram M is the global maximum and R, in addition to being a local minimum is also the global minimum. At P, Q, R, S and T the tangent to the curve is horizontal, i.e. the gradient of the curve is zero, so that at any local maxima or local minima the following condition holds : dy dx ( x) = f All points at which f ( x) are called stationary points but as shown in the sketch below there are stationary points which are neither local maxima nor local minima.. P dy At P 0 dx = P is in fact an example of a type of point known as a point of inflexion.

2 Page of 8 Other examples of points of inflexion are the points I, I, I and I in the diagram at the top of page. These are the points at which the curve crosses the tangent to the curve as can be seen in the blown up sketch of the curve in the region of I shown below. I To the left of increasing x so that I the gradient, f ( x), decreases with increasing x and to the right increases with I is a local minimum of f ( x). Similarly and so on for I and I. Hence finding the points of inflexion of f( x) local maxima and minima of f ( x) I is a local maximum of f( x) is equivalent to finding the. For this reason we do not consider points of inflexion further in these notes and concentrate solely on the max/min problem. Local Maxima and L P R At P, f ( x) y = f ( x) At L, f ( x) > 0 At R, f ( x) < 0 As x increases from L to R the gradient of the curve decreases or in other words the rate of change of the gradient with respect to x is non-positive. We cannot say that the rate of change of the gradient is strictly negative since, as we shall see, in some cases it is zero at P. But the rate of change of the gradient with respect to x is d f ( x ) = f ( x ) dx From this we can deduce the following:- At a local maximum f( x) and f( x) 0 y = f ( x) At P, f ( x) At L, f ( x) < 0 L P R At R, f ( x) > 0 In this case the gradient is increasing as x increases from L to R from which we can deduce the following:- At a local minimum

3 Page of 8 f( x) and f( x) 0 Thus to identify the local maxima and minima of a given function we proceed as follows:- Find all the stationary points i.e. solve f ( x). At each stationary point evaluate f x ( ), then (i) if f ( x) < 0; Local Maximum (ii) if f ( x) > 0; Local Minimum (iii) if f ( x) ; examine the behaviour of f ( x) in the neighbourhood of the point as Examples shown in the examples below.. Find and classify the stationary points of the following functions and give also the global maximum and global minimum. (i) f ( x) = x x x+ 5, x The stationary points are given by f ( x) = 6x 6x i.e. 6( x )( x + ) i.e. x = and x = f ( ) = 5 and f ( ) = To classify:- f ( x) = x 6 f ( ) = 8> 0 and f ( ) = 8< 0 On the graph of y = f ( x) the point (,-5) is a local minimum the point (-,) is a local maximum To help sketch the graph we note that at x, y = f(0) = 5 x=, y = f( ) = x=, y = f() = 7 Using all the above information we obtain the graph below.

4 Page of 8 0 ( 7, ) (, ) 0 (, ) 0 0 (, 5) The global minimum is at (,-5) and the global maximum at (,7) (ii) f( x) = ( x ) (x+ ), x To find the stationary points f ( x) = ( x ) ( x+ ) + ( x ) ( x ) ( x+ + x ) 0xx ( ) i.e. x and x = f ( 0) = and f () To classify, f ( x) ( x ) + 60x( x ) f ( 0) < 0 f () Thus (0,) is a local maximum To classify the point (,0) we need to examine the behaviour of f ( x) near x =. = so that if x is slightly less than, f ( x) < 0 We have f ( x) 0 x( x ) and if x is slightly greater than, f ( x) > 0. (,0) is a local minimum In addition we have that f and f() = 9 Hence we obtain the sketch of y f( x) = shown below..

5 Page 5 of 8 ( 9, ) y = (x-) (x+) 5 ( 0, ) ( 0.5, 0) Global maximum at (, 9) Global minimum at, 0 and (, 0 ). (iii) f x x x x x x ( ) = 8 6 +, f x = x x x+ = x x x+ ( ) ( ) We can cancel the so that the stationary points are given by x x x+ = 0 To find the roots notice that putting x = satisfies the equation so that x is a factor of the left hand side. This leads to ( x )( x x ) = ( x )( x+ )( x ) The stationary points are x=, x=, x= f () =, f ( ) = 9, f () = 8 To classify them we need the second derivative f x = x x ( ) ( ) From which we find that (,) is a local maximum, (-,-9) is a local minimum, (,8) is a local minimum. To help sketch the curve y = f( x) notice that f(0), f( ), f() = 5 5

6 Page 6 of 8 (, 0) 0 y = x - 8x -6x + x ( 5, ) 0 (, ) ( 8, ) 0 (, 9) 0 Global maximum at (,5) Global minimum at (-,-9) (iv) f ( x) ( x ) ( x ) 5, x = + + < < ( ) f x = x + x x+ = x x + x+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) = + = + The stationary points are x=, x= f () = 5, f ( ) = ( x ) (x 8) ( x ) ( x ) To classify : f ( x) = {( x ) + ( x )( x+ ) } f () and f (-)=6>0 Hence (-,-) is a local minimum. To classify the point (,5) we need to examine the behaviour of f( x) f ( x) = ( x ) ( x+ ) so in the neighbourhood of x =. We have that f ( x) > 0 whether x is slightly bigger than or slightly less than from which we deduce that (,5) is a point of inflexion. To help sketch the curve y = f( x) notice that f (0) = and also As x ±, f( x) +. Hence we obtain the sketch below : 6

7 Page 7 of 8 y = (x - ) (x + ) (, ) 0 (, ) 0 Global maximum is at ( ±, ) Global minimum is at (-,-). A company is producing closed cylindrical containers of volume of m. In order to minimise costs they need the surface area of a container to be as small as possible. Find the required dimemions (radius and height) correct to four decimal places. If we neglect the thickness of the material from which the containers are constructed we may proceed as follows: Let the radius be r m., the height h m., the surface area A m. and the volume V m. A r rh = + and V = r h= r A= r + = r + r r r da = r for stationary points. dr r This gives r = 0.59 = m. d A = + which is positive for the above value of r dr r Hence h = so that Therefore this value of r gives a local minimum of A.In this type of problem we need to ask is it also the global minimum. In this case we can argue as follows. If a function has only one stationary point and that point is a local minimum it is clearly also the global minimum. r.59m. and h= =.080m. r Therefore the required dimensions are Tutorial ( Solutions on page 9.). Find and classify the stationary points of the following functions and sketch their graphs. Give the global maximum and global minimum. (i) f( x) = x x+, x (ii) f( x) = ( x ) ( x+ ) 0, x (iii) f ( x) x x x 60, x = + + < < (iv) f x x x x ( ) = ( ) +, 7

8 Page 8 of 8. A piece of flexible wire of length 0cm is bent into the shape of an isoscelese triangle.what is the maximum area of the triangle. (Hint: Let the two equal sides be x cm long and the third side y cm so that y+ x. Express the area of the triangle in terms of x and y and substitute for y.). A company produces closed rectangular boxes of height h m with a square base and top of side length x m. (a) If each box is to contain a volume of m find the minimum surface area of material used in its construction. (b) If each box is to be constructed from 0m of material find the dimensions which will maximise the volume contained. (Neglect the thickness of the material.) To go to the tutorial solutions directly click here. (To return to this spot after viewing the solutions just use the Back Button). To return to the main list click here. 8