Q. 88

Mark (√) against

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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