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) = x^{3} - 3x^{2} - 24x + 100
f'(x) = 3x^{2} - 6x - 24 = 0
==> x^{2} - 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