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 c_{1}, c_{2}, …., c_{n} be the different critical points. Find the value of the function at these critical points.

Let f(c_{1}) , f(c_{2}) , ….., f(c_{n}) be the values of the function at critical points.

Say, M_{1} = max {f(a), f(c_{1}), f(c_{2}), …….. , f(c_{n}) , f(b)} and M_{2} = min {f(a), f(c_{1}), f(c_{2}), …….. , f(c_{n}) , f(b)}.

Then M_{1} is the greatest value of f (x) in [a , b] and M_{2} 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 c_{1 }, c_{2 }, ……c_{n} be the different critical points of the function in (a , b).

Let M_{1} = max {f(c_{1}), f(c_{2}), f(c_{3}) ……, f(c_{n})}

and M_{2} = min {f(c_{1}), f(c_{2}), f(c_{3}) ……, f(c_{n})}.

Now if $\large Lim_{x \rightarrow a+0 \; (or , x \rightarrow b-0)}$ f(x) > M_{1} or < M_{2}, 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 M_{1} or less than M_{2}, 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 M_{1} and M_{2} would respectively be the global maximum and global minimum of f (x) in (a, b).

Illustration : Let f (x) = 2x^{3} – 9x^{2} + 12 x + 6 Discuss the global maxima and minima of f (x) in [0, 2] and (1, 3)

Solution: f (x) = 2x^{3} – 9x^{2} + 12 x + 6

f ‘(x) = 6x^{2} – 18x + 12

= 6 (x^{2} – 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.

Lim_{x→1 + 0} f (x) = 11

and , Lim_{x→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) = x^{1/x} to determine the bigger of the two number e^{π }and π^{e }

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)

e^{1/e} > π^{1/π}

(e^{1/e})^{πe} > (π^{1/π})^{πe }

e^{π }> π^{e }

__Exercise :__

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

(a) f (x) = |x^{2} – 4x + 3| in [0, 5].

(b) f(x) = 3x^{4} – 2x^{3} – 6x^{2} + 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) = sin^{3}x + λ sin^{2}x, 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 4x^{2} + a^{2}y^{2} = 4a^{2} , 4 < a^{2} < 8 that is farthest from the point (0, – 2).