

Maxima and Minima:
Maxima and Minima:
Let us learn about the concepts of Maxima and Minima.Below are also some practice problems related to maxima and minima.
A function f(x) is said to have a local maximum at x = a, if $ is a neighbourhood I of 'a', such that
f(a) f(x) for all x I
The number f(a) is called the local maximum of f(x). The point a is called the point of maxima.
A function f(x) is said to have a local maximum at x = a, if $ is a neighbourhood I of 'a', such that
f(a) f(x) for all x I
The number f(a) is called the local maximum of f(x). The point a is called the point of maxima.
Note that when 'a' is the point of local maxima, f(x) is increasing for all values of x <> a in the given interval.
At x = a, the function ceases to increase.
A function f(x) is said to have a local minimum at x = a, if $ is a neighbourhood I of 'a', such that
f(a) f(x) for all x I
Here, f(a) is called the local minimum of f(x). The point a is called the point of minima.
Note that, when a is a point of local minimum f (x) is decreasing for all x <> a in the given interval. At x = a, the function ceases to decrease.
If f(a) is either a maximum value or a minimum value of f in an interval I, then f is said to have an extreme value in I and the point a is called the extreme point.
Monotonic Function maxima and minima:
A function is said to be monotonic if it is either increasing or decreasing but not both in a given interval.
Consider the function
The given function is increasing function on R. Therefore it is a monotonic function in [0,1]. It has its minimum value at x = 0 which is equal to f (0) =1, has a maximum value at x = 1, which is equal to f (1) = 4.
Here we state a more general result that, 'Every monotonic function assumes its maximum or minimum values at the end points of its domain of definition.'
Note that 'every continuous function on a closed interval has a maximum and a minimum value.'
Example on Local Maxima and Minima:
Find the local maxima and local minima of the function f (x) = 2x3 - 21x2 +36x - 20. Find also the local maximum and local minimum values.
Solution:
f '(x) = 6x2 - 42x + 36
f '(x) = 0
x = 1 and x = 6 are the critical values
f ''(x) =12x - 42
If x =1, f ''(1) =12 - 42 = - 30 < x ="1" value =" 2(1)3" 20 =" -3" x =" 6," 42 =" 30"> 0
x = 6 is a point of local minima of f (x)
Minimum value = 2(6)3 - 21 (6)2 + 36 (6)- 20
= -128
Hope you like the above example of Maxima and Minima.Please leave your comments, if you have any doubts.
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