Absolute Maxima and Minima Definition. A function f is said to have an absolute maximum on an interval I at the point x 0 if it is the largest value of f on that interval; that is if f( x ) f() x for all x in I. 0 A function f is said to have an absolute minimum on an interval I at the point x 0 if it is the smallest value of f on that interval; that is if f( x ) f() x for all x in I. 0 If f has either an absolute maximum or an absolute minimum on an interval I at the point x 0, we say that it has an absolute extremum at that point.
Absolute maximum on the interval [-8, 8] Absolute minimum on the interval [-8, 8]
Two facts appear in the previous example. 1. The function f was continuous on a closed interval (a finite interval and its end points), and it had an absolute maximum and an absolute minimum. 2. The absolute maximum and minimum occurred either at an end point or at a local maximum or minimum point. The two properties of f in part 1. above turn out to be critical. In general the existence of absolute maxima and minima will depend on the nature of the function f and the type of interval. The previous example demonstrates the following theorem.
Theorem. If f(x) is a continuous function on a closed interval [a, b], then f always has an absolute maximum and an absolute minimum value on [a, b]. This is a difficult theorem to prove, but it is intuitively clear. It says intuitively that if you put a pencil at f(a), and then draw a curve ending at f(b), without lifting the pencil from the paper, then that curve will have a highest and a lowest height. The previous example also illustrates the proper procedure to find the absolute maximum and minimum values in this case.
To Find the Absolute maximum and minimum value of a continuous function f on a closed interval [a, b]. 1. Locate all critical points in the interval [a, b] 2. Evaluate f at all of the critical points and at the points a and b. 3. The largest value found in step 2 is the absolute maximum of f on [a, b] and the corresponding point is the point where that maximum is achieved. A similar statement is true for the absolute minimum.
Example. Find the absolute extrema of In the interval [ 1, 1]. 4 1 3 3 f() x = 3x 9x Solution. 1 2 2 3 3 3 x 2 3 f () x = 4x 3 x = x (4x 3) = (4 3) x The derivative of f does not exist at x = 0, and is 0 at the point x = 3/4. Thus we must consider the four points 1, 0, ¾, 1. The values of the function at these four points are respectively: f ( 1) = 3+ 9= 12 3 f (0) = 0 f 6.133 4 = f (1) = 3 9= 6 Thus the absolute maximum is 12 and it occurs at the left hand endpoint, and the absolute minimum is 6.133, occurring at x = ¾.
Example. Find the absolute extrema of in the interval [ 1, 1]. f() x = sin( x) cos( x) Solution. f () x = cos( x) + sin( x) This function is 0 at the point x = π and at every point that 4 differs from this point by a multiple of π. Thus x = π 4 is the only such point within the desired interval. We must therefore consider the points 1, π/4, and 1. f ( 1) = 1.38 π f = 1.414 4 f (1) = 0.301 Thus the absolute maximum is 1.414 and it occurs at π/4, and the absolute minimum is 0.301, occurring at x = 1.
This is the graph of that function.
Example. Find the absolute extrema of In the interval [0, 6]. f() x = 8x x2 Solution. f () x = 8 2x This function is 0 at the point x = 4. Thus we must check the three points 0, 4, 6. At these points the value of f is resp. f (0) = 0 f (4) = 32 16= 16 f (6) = 48 36= 12 Thus the absolute maximum is 16 and occurs at x = 4, and the absolute minimum is 0, occurring at x = 0.
Here is the graph.
Example. Find the absolute extrema of In the interval [ 2, 1]. f() x = 2x3 3x2 12x Solution. f () x = 6x2 6x 12= 6( x2 x 2) = 6( x 2)( x+ 1) The derivative of f is 0 at the points x = 2 and x = 1. However, only one of these points is in the desired interval, namely x = 1. Thus we must check the three points 2, 1, 1. At these points the value of f is resp. f ( 2) = 2( 8) 3(4) 12( 2) = 4 f ( 1) = 2 3+ 12= 7 f (1) = 2 312 = 13 Thus the absolute maximum is 7 and occurs at x = 1, and the absolute minimum is 13, occurring at x = 1.
Function in the interval [ 2, 3] Function in the interval [ 2, 1]
Example. Find the absolute extrema of In the interval [ 3, 3]. f() x = 6 4x Solution. The derivative of f is never 0, since it is equal to 4 if x < 3/2, and to 4 when x > 3/2. The derivative fails to exist at 3/2, so this is a critical point in the desired interval. We must therefore test at the three points 3, 3/2, and 3. 3 f ( 3) = 6+ 12= 18 f 2 = 6 6= 0 f (3) = 6 12= 6 Thus the absolute maximum is 18 and it occurs at x = 3, and the absolute minimum is 0, occurring at x = 3/2.
Here is the graph.
Absolute maxima and minima on infinite or non closed intervals If an interval is not open or not finite, then there may be no absolute maximum or minimum. The following diagrams illustrate this problem. f(x) = 1/x has no absolute maximum or minimum in (0, 1). It has an absolute minimum of 1 in (0, 1], but no absolute maximum.
x = π/2 x = π/2 f(x) = tan(x) has no absolute maximum or minimum in the open interval ( π/2, π/2).
f() x 1 = 1+ x 2 has no absolute minimum in the open interval (, ), but has an absolute maximum at x = 0.
f() x = x3 has no absolute maximum or minimum in the interval (, ).
f() x = x2 has no absolute maximum in the interval (, ), but has an absolute minimum at x = 0. It also has no absolute maximum in the interval ( 1, 1).
There are, however, conditions where something can be said. Theorem. Suppose that a function f is continuous on an interval I (of any kind) and has exactly one relative extremum in that interval at a point x 0. If the point x 0 is a relative maximum, then it is an absolute maximum. If it is a relative minimum, then it is an absolute minimum. We will not try to prove this, but instead will illustrate it with a diagram. Relative maximum If it is eventually higher, it must pass through a relative minimum.
Example. Let f() x = x 3 3x 2 + 5 Find all absolute extrema (if any) for f on the interval (0, ). Solution. f () x = 3x2 6x= 3( xx 2) This function is zero at 0 and 2, and only x = 2 is in the given interval. This is the only critical point of the function in (0, ). Since f () x = 6x 6= 6( x 1) and this is greater than 0 at 2, we know that there is a relative minimum at x = 2, and by the theorem, this must be an absolute minimum. There is no absolute maximum since lim f() x =+ x
This is confirmed by the graph.
Example. Let x 2 f() x = (1 + x 2 ) Find all absolute extrema (if any) for f on the interval (, ). Solution. (1 2 )(2 ) 2 f () x = + x x x (2 x) = 2x (1 + x 22 ) (1 + x 22 ) Which is zero only at x = 0. The derivative is positive for x > 0 and negative for x < 0, so by the first derivative test, 0 is a local minimum. Since this is the only critical point in the interval, it is an absolute minimum. We also see that the function is always between 0 and 1 and lim f() x = 1. The function never reaches 1, so there is no x ± absolute maximum.
The graph confirms this.
For curves defined by parametric equations, where the parametric interval is a closed interval [a, b] also have absolute maxima and minima, in fact their graph is contained in a finite rectangle. Example. Let x(t) = t 3sin(t) and y(t) =4 3cos(t), for t in the interval [0, 10]. What are the highest and lowest points on this curve?
We know that curve occur where dy 0. dt = dy dy = dt and so the stationary points of the dx dx dt Thus absolute maxima or minima must occur either at the end points of the parametric interval or at these stationary points. In the previous example, dy 3sin() t 0 dt = = at π, 2 π, and 3 π. If we check these points and 0 and 10, we have values 1, 7, 1, 7, 6.517, resp. Thus the absolute maximum of 7 is reached at points π, and 3 π. The absolute minimum of 1 is reached at 0 and 2 π.