Local Maximum and local Minimum, Absolute maximum and absolute minimum

**Local and Global(Absolute) Maxima and
Minima**

**Definition 6.1**

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*)â‰¥*f*(*x*) for every *x* âˆˆ (*a*,*b*)

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

**Definition 6.2**

A function *f* has an **absolute maximum** at *c* if *f*(*c*)â‰¥*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*)â‰¤*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*).

Let *f* be a differentiable function on an open interval (*a*,*b*)
containing *c* and suppose that *f* â€˜â€™ (*c*) exists.

(i) If *f* â€˜ (*c*) = 0 and *f* â€˜â€™ (*c*) > 0, then *f* has a local minimum at *c*.

(ii) If *f* â€˜ (*c*) = 0 and *f* â€˜â€™ (*c*) < 0,then *f* has a local maximum at *c*.

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.

Find the extremum values of the
function *f*(*x*)=2*x*^{3}+3*x*^{2}â€“12*x*.

*Solution :*

Given

*f*(*x*) = 2*x*^{3}+3*x*^{2}â€“12*x *â€¦ (1)

*f *â€˜* *(*x*) = 6*x*^{2}+6*x*â€“12

*f''*(*x*) = 12*x* + 6

*f *â€˜* *(*x*) = 0 âŸ¹ 6*x*^{2}+6*x*â€“12 =0

âŸ¹ 6(*x*^{2}+*x*â€“2)= 0

âŸ¹ 6(*x*+2)(*x*â€“1)=
0

âŸ¹ *x *= â€“2 ;* x *= 1

When

*x*= â€“2

*f *â€˜â€™* *(â€“2) =
12(â€“2) + 6

= â€“18 < 0

* f*(*x*) attains local maximum at* x *= â€“ 2 and local maximum value is
obtained from (1) by substituting the value *x*
= â€“ 2

*f*(â€“2) = 2
(â€“2)^{3}+3(â€“2)^{2}â€“12(â€“2)

= â€“16+12+24

= 20.

When

x = 1

*f *â€˜â€™* *(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) â€“ 12 (1)

= â€“7

Extremum values are â€“ 7 and 20.

Find the absolute (global)
maximum and absolute minimum of the function *f*(*x*)=3*x*^{5}â€“25*x*^{3}+60*x*+1 in
the interval [â€“2,2]

*f*(*x*) = 3*x*^{5}-25*x*^{3}+60*x*+1 â€¦ (1)

*fâ€™ *(*x*) = 15*x*^{4}â€“75*x*^{2}+60

= 15(*x*^{4}â€“5*x*^{2}+4)

*f *â€˜* *(*x*) = 0 âŸ¹ 15(*x*^{4}â€“5*x*^{2}+4)= 0

âŸ¹ (*x*^{2}â€“4)(*x*^{2}â€“1)= 0

*x** *= Â±2 (or) *x* =
Â±1

of these four points âˆ’ 2, Â±1 âˆˆ [âˆ’2, 1] and 2 âˆ‰ [âˆ’2,1]

From (1)

*f*(â€“2)
= 3 ( - 2 )^{5} â€“ 25(- 2 )^{3} + 60 ( - 2 ) + 1

= â€“15

When *x*= 1

*f*(1)
= 3 ( 1 )^{5} â€“ 25(1 )^{3} + 60 ( 1 ) + 1

= 39

When *x *= â€“ 1

*f*(-1)
= 3 ( -1 )^{5} â€“ 25(-1 )^{3} + 60 ( -1 ) + 1

= â€“37.

Absolute maximum is 39

Absolute minimum is -37

Tags : Maxima and minima - Applications of Differentiation , 11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation

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11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation : Local and Global(Absolute) Maxima and Minima | Maxima and minima - Applications of Differentiation

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