# Global Maximum , Global Minimum

Let y = f (x) be a given function with domain D. Let [a, b] ⊆ D. Global maximum/minimum of f (x) in [a, b] is basically the greatest/ least value of f (x) in [a, b].
Global maximum and minimum in [a, b] would always occur at critical points of f(x) within [a, b] or at the end points of the interval.

### Global Maximum / Minimum in [a , b] :

In order to find the global maximum and minimum of f(x) in [a, b], find out all the critical points of f(x) in (a, b). Let c1, c2, …., cn be the different critical points. Find the value of the function at these critical points.

Let f(c1) , f(c2) , ….., f(cn) be the values of the function at critical points.

Say, M1 = max {f(a), f(c1), f(c2), …….. , f(cn) , f(b)} and M2 = min {f(a), f(c1), f(c2), …….. , f(cn) , f(b)}.

Then M1 is the greatest value of f (x) in [a , b] and M2 is the least value of f (x) in [a , b].

### Global Maximum / Minimum in (a , b):

Method for obtaining the greatest and least values of f(x) in (a, b) is almost same as the method used for obtaining the greatest and least values in [a, b], however with a caution.

Let y = f (x) be a function and c1 , c2 , ……cn be the different critical points of the function in (a , b).

Let M1 = max {f(c1), f(c2), f(c3) ……, f(cn)}

and M2 = min {f(c1), f(c2), f(c3) ……, f(cn)}.

Now if $\large Lim_{x \rightarrow a+0 \; (or , x \rightarrow b-0)}$   f(x) > M1 or < M2, f(x) would not have global maximum (or global minimum) in (a, b).

This means that if the limiting values at the end points are greater than M1 or less than M2, then f (x) would not have global maximum/minimum in (a, b). On the other hand if  $M_1 > Lim_{x \rightarrow a+0 (and \; x \rightarrow b-0)} f(x)$ and $M_2 < Lim_{x \rightarrow a+0 (and \; x \rightarrow b-0)} f(x)$ , then M1 and M2 would respectively be the global maximum and global minimum of f (x) in (a, b).

Illustration : Let f (x) = 2x3 – 9x2 + 12 x + 6 Discuss the 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).

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.

Illustration : Use f (x) = x1/x to determine the bigger of the two number eπ and πe

Solution: $\large f ‘ (x) = x^{1/x} (\frac{1}{x^2} – \frac{lnx}{x^2})$

$\large = \frac{x^{1/x}}{x^2} (1-ln x)$

Now, f’ (x) > 0 if x < e

and f’ (x) < 0 if x > e

so f (x) has maxima at x = e

so f (e) > f (π)

e1/e > π1/π

(e1/e)πe > (π1/π)πe

eπ > πe

### Exercise :

(i) Discuss the global maxima/minima of following function in the given interval

(a) f (x) = |x2 – 4x + 3| in [0, 5].

(b) f(x) = 3x4 – 2x3 – 6x2 + 6x + 1 in [0, 2]

(c) $\large f(x) = \sqrt{(1-x^2)(1+2x^2)}$ in [-1, 1]

(d) f(x) = sinx sin2x in (-∞, ∞)

(ii) Show that maximum or minimum of f(x) is a minimum or maximum of 1/f(x), where f(x) and 1/f(x) are well defined.

(iii) Let f(x) = sin3x + λ sin2x, x ∈ (-π/2, π/2). Find the values of parameter λ so that f(x) has exactly one maximum and exactly one minimum.

(iv) Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6r√3

(v) Find the least perimeter of a cyclic quadrilateral in which a circle of radius r can be inscribed.

(vi) Find a point on the curve 4x2 + a2y2 = 4a2 , 4 < a2 < 8 that is farthest from the point (0, – 2).

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