Local and Global(Absolute) Maxima and Minima

Local and Global(Absolute) Maxima and Minima

Definition 6.1

Local Maximum and local Minimum

A function f has a local maximum (or relative maximum) at c if there is an open interval (a,b) containing c such that f(c)≥f(x) for every x ∈ (a,b)

Similarly, f has a local minimum at c if there is an open interval (a,b) containing c such that f(c) ≤ f(x) for every x ! (a,b).

Definition 6.2

Absolute maximum and absolute minimum

A function f has an absolute maximum at c if f(c)≥f(x) for all x in domain of f. The number f(c) is called maximum value of f in the domain. Similarly f has an absolute minimum at c if f(c)≤f(x) for all x in domain of f and the number f(c) is called the minimum value of f on the domain. The maximum and minimum value of f are called extreme values of f.

NOTE

Absolute maximum and absolute minimum values of a function f on an interval (a,b) are also called the global maximum and global minimum of f in (a,b).

Criteria for local maxima and local minima

Let f be a differentiable function on an open interval (a,b) containing c and suppose that f ‘’ (c) exists.

(i) If f ‘ (c) = 0 and f ‘’ (c) > 0, then f has a local minimum at c.

(ii) If f ‘ (c) = 0 and f ‘’ (c) < 0,then f has a local maximum at c.

NOTE

In Economics, if y = f(x) represent cost function or revenue function, then the point at which dy/dx = 0, the cost or revenue is maximum or minimum.

Example 6.25

Find the extremum values of the function f(x)=2x³+3x²â€“12x.

Solution :

Given

f(x) = 2x³+3x²â€“12x … (1)

f ‘ (x) = 6x²+6x–12

f''(x) = 12x + 6

f ‘ (x) = 0 ⟹ 6x²+6x–12 =0

⟹ 6(x²+x–2)= 0

⟹ 6(x+2)(x–1)= 0

⟹ x = –2 ; x = 1

When

x= –2

f ‘’ (–2) = 12(–2) + 6

= –18 < 0

 f(x) attains local maximum at x = – 2 and local maximum value is obtained from (1) by substituting the value x = – 2

f(–2) = 2 (–2)³+3(–2)²â€“12(–2)

 = –16+12+24

 = 20.

When

x = 1

f ‘’ (1) = 12(1) + 6

 = 18.

f(x) attains local minimum at x = 1 and the local minimum value is obtained by substituting x = 1 in (1).

f(1) = 2(1) + 3(1) – 12 (1)

= –7

Extremum values are – 7 and 20.

Example 6.26

Find the absolute (global) maximum and absolute minimum of the function f(x)=3x⁵–25x³+60x+1 in the interval [–2,2]

Solution :

f(x) = 3x⁵-25x³+60x+1 … (1)

f’ (x) = 15x⁴–75x²+60

= 15(x⁴–5x²+4)

f ‘ (x) = 0 ⟹ 15(x⁴–5x²+4)= 0

⟹ (x²â€“4)(x²â€“1)= 0

x = ±2 (or)  x = ±1

of these four points − 2, ±1 ∈ [−2, 1]  and 2 ∉  [−2,1]

From (1)

f(–2) =  3 ( - 2 )⁵ – 25(- 2 )³ + 60 ( - 2 ) + 1

= –15

When x= 1

f(1) =  3 ( 1 )⁵ – 25(1 )³ + 60 ( 1 ) + 1

= 39

When x = – 1

f(-1) =  3 ( -1 )⁵ – 25(-1 )³ + 60 ( -1 ) + 1

= –37.

Absolute maximum is 39

Absolute minimum is -37