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In this maxima - minima question, teacher says that critical point -2 doesn't belong to the interval [-3, 3], isn't this wrong or i am missing something?

asked in Calculus | 336 views
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Put the value of -2, -3, and 3 into the f(x), and see which one is minimum.

Here -2 $\epsilon$ [-3, 3] but at -2 we have local maxima, not minima. That's why we do not get minimum value at -2 in the interval.
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yes Hemant, i only wanted to confirm that -2 belongs to given interval or not.

First of all ,

a) Critical point (or) stationary point : The point where the derivative of a function is zero provided the function is differentiable over its entire domain..

b) Local maxima : Critical point and in addition to that f''(x) < 0 and for local minima , f''(x) > 0 ..These points are also known as points of local extremum..

So given   f(x)  =  x3  -  3x2   -  24x + 100

f'(x)   =  3x2  -  6x  - 24   =   0

==>   x2  -  2x  - 8   =   0

==>   x  =  -2 , 4

Thus in [-3,3] , only -2 is the critical point..

As in the question , we are asked about minimum value simply , so this refers to the global mimimum here ..In such question , besides critical point in the interval , the extreme points are also to be considered..As the given interval is a closed interval , hence both the points x = -3 and x = 3 need to be considered as well in addition to x = -2..

So on substituting x = -3 , 3 and  -2 , we get 118 , 28 and 128 as the value of function respectively

Thus minimum value = 28 at x = 3

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