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Concept of Global Maxima or Minima


Global Maxima or Minima is an important topic as it fetches some questions in the JEE. It is important to master this topic in order to remain competitive in the IIT JEE. It is crucial to note that the concept is slightly different for functions in open intervals than those in closed intervals.   

Case 1: Global Maxima or Minima in [a, b]

Global maxima or minima of f(x) in [a, b] is basically the greatest or least value of f(x) in [a, b]. Mathematically, it is written as:

The function f(x) has a global maximum at the point ‘a’ in the interval I if f (a) ≥f(x), for all x∈I.

Similarly, f(x) has a global minimum at the point ‘a’ if f (a) ≤f (x), for all x∈I.

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

The graph of the function f(x) depicting the global maximum and minimum.

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

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

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

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

Case 2: Global maxima or minima in (a, b):

To find the global maxima and minima in (a, b) step 1 and 2 is same but after that we have to be a bit careful.   After step 1 and 2 find

M=max{ f(C1), f(C2)………..f(Cn)} and M2= min{f(C1), f(C2)………..f(Cn)}.

Now if x approaches a- or if x approaches b- , the limit of f(x) > M1 or its limit f(x) < M1 would not have global maximum (or global minimum) in (a, b) but if as x approaches a- and x approaches b- , lim f(x) < M1 and lim f(x) > M2, then M1 and M2 will respectively be the global maximum and global minimum of f(x) in (a,b).

Result: If f(x) is a continuous function on a closed bounded interval [a,b], then f(x) will have a global maximum and a global minimum on [a,b]. On the other hand, if the interval is not bounded or closed, then there is no guarantee that a continuous function f(x) will have global extrema.

Example: f(x) =x2 does not have a global maximum on the interval [0, ∞), the function f(x) =-1/x does not have a global minimum on the interval (0, 1).

Result: If f(x) is differentiable on the interval I, then every global extremum is a local extremum or an endpoint extremum.

For more on the topic, you may view the video:


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


f (x) = 2x3 – 9x2 + 12 x + 6

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

First of all 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

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

 Now let us consider (1, 3).

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

 f (2) = 10. Limx–>1+0 f (x) = 11 and Limx–>3–0 f (x) = 15.

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


Find the absolute maximum and minimum value of the function

f(x) = x3-3x2+1 for -1/2≤ x ≤ 4


Since f is continuous on [-1/2, 4], we can use the Closed Interval Method.

Step 1: Evaluate the values of f at the critical points of f in (-1/2, 4)
f(x) = x3 – 3x2 + 1
f ‘(x) = 3x2 – 6x = 3x(x – 2)
When f ‘(x) = 3x(x – 2) we get x = 0 or x = 2

So, the critical numbers are x = 0 and x = 2

The values of f at these critical numbers are
f(0) = 1 and f(2) = –3

Step 2: Find the values of f at the endpoints of the interval.

The values of f at the endpoints of the interval are

f (-1/2) = 1/8 and f(4) = 17.

Step 3: The largest of the values from Steps 1 and 2 is the absolute maximum value and the smallest of these values is the absolute minimum value.
Comparing the four numbers, we see that the absolute maximum value is f(4) = 17 and the absolute minimum is f(2) = –3.

You can also consult the Papers of Previous Years to get an idea about the types of questions asked.


Consider the function f(x) = (x-1)2, for x∈ [0,3].

We first draw the graph of the function to get a clear picture. It is visible from the function itself that the only critical point is x =1. And solving we can find out that it is a point of local minimum. From the graph it follows that the end point i.e. x=3 is the point of global maximum.

Graphical representation of the function f(x) = (x-1)2

To read more, Buy study materials of Applications of Derivatives comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.

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