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Que:- Consider the function f(x) = $2x^3 - 3x^2$ in the domain [-1, 2], the "global minimum" value of f(x) is _______?
(A) -5  (B)-1  (C)4   (D) 0

f(x) = $2x^3 - 3x^2$
f'(x) = $6x^2 -6x$ =0, x=0,1 ( critical points)

f"(x) = 12x -6
f"(0) = 12*0 - 6 < 0  [f(x) attains local maximum at x=0]
f"(1) = 12*1 - 6 = 6 >0 [f(x) attains local minimum at x=1]

Consider extreme points also as closed intervals are given:

f(1) = -1 ( local minimum)
f(-1) = -5
f(2) = 4

Min(-1, -5, 4) = -5( global minimum)

In this question "global minium" is asked hence answer is -5.
My question is, if "local minimum" is asked instead of "global minimum" then what will be the answer -5 or -1, as closed intervals are given, so in case of local minimum also we should consider the extreme points, right?

asked in Calculus by Boss (42.4k points) | 233 views
There can be more than one local minimum(as it is considered in the small neighborhood only) but only one global minimum(as it is considered in an entire range).

So (-1, -5, 4) all are local minimum but -5 is the global minimum.
@shivam is not the value of local minimum unique? if local minimum is asked what will be your answer -1 or -5?
@shivam how did you know that 4 is also local minimum?
Local minima is not unique. If a value is smaller than its neighborhood values then it is local minima.
ok, and how do you know it's local minima at 2?


Look at this video. Then look the internet for further explanation.

There is no local minima at 2.. at 2, the graph has a maxima

I think local minimum will be at x=1 and if we talk about global minimum then we have to check all the values, in the sense, all critical points along with closed interval values on which function is defined

Local minimum will be x=1

the global minimum will be x=-1
@akash yes, seems correct to me!

At x=-1, a function has global minimum value -5..

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