Global maxima or minima in [a, b]:

Global maxima or minima of f(x) in [a, b] is greatest or least value of f(x) in [a, b].

Global maxima or minima in [a, b] will occur either at critical points of f(x) within [a, b] or at end points of the interval.

Step to find out global maxima or minima in [a, b]

Step 1:           Find out all critical points of f(x) in (a, b). Let C₁, C₂,....C_n be different critical points.

Step 2:           Find out the value of function at these critical points and also at end points of domain. Let the values of function at the critical points be f(C₁), f(C₂)...........f(C_n).

Step 3:           Find out M₁ =max{ f(a), f(C₁), f(C₂)...........f(C_n), f(b)} and M₂= min{ f(a), f(C₁), f(C₂)...........f(C_n), f(b)}. Now M₁is maximum value of f(x) in [a, b] and M₂ is minimum value of f(x) in [a, b].

Global maxima or minima in (a, b):

To find global maxima and minima in (a, b) step 1 and 2 is same but we have to be a bit careful after that. After step 1 and 2 find M₁ =max{ f(C₁), f(C₂)...........f(C_n)} and M₂= min{f(C₁), f(C₂)...........f(C_n)}.

Now if ,

 f(x) would not have global  maximum in (a, b) but if 

then M1 and M2 would respectively be global maximum and global minimum of f(x) in (a,b)

Illustration: Let f (x)= 2x³ - 9x² + 12x + 6. Discuss global maxima and minima of f (x) in   [0, 2] and (1, 3).

Solution:             f (x) = 2x³ - 9x² + 12 x + 6

                              => f'(x) = 6x² - 18x + 12  = 6 (x² - 3x + 2) = 6 (x-1) (x-2)

                              Firstly let us discuss [0, 2].

                              Clearly the critical point of f (x) in [0, 2] is x = 1.

                              f (0) = 6,  f (1) = 11, f (2) = 10

                              Therefore x = 0 is the point of global minimum of f(x) in [0, 2] and x = 1 is point of global maximum.

                              Now let us consider (1, 3).

                              Clearly x = 2 is the critical point in (1, 3).

                              f (2) = 10. 

            Therefore x = 2 is the point of global minimum in (1, 3) and global maximum in (1, 3) does not exist.   

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