Answer :

f(x) = (x-2)(x-3)2


f(x) = (x-2)(x2-6x+9)


f(x) = x3-8x2+21x-18.


f’(x) = 3x2-16x+21


f’’(x) = 6x-16


For maximum or minimum value f’(x) = 0.


3x2-9x-7x+21 = 0


3x(x-3)-7(x-3)=0


x = 3 or x =7/3.


f’’(x) at x = 3.


f’’(x) = 2


f’’(x)>0 it is decreasing and has minimum value at x = 3


At x = 7/3


F’’(x) = -2


F’’(x)<0 it is increasing and has maximum value at x = 7/3.


Substituting x = 7/3 in f(x) we get




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