Q. 76

# Mark (√) against

Given:

f(x) = (x+1)3.(x-3)3

f’(x) = 3(x+1)2(x-3)3 + 3(x-3)3 (x+1)3

Put f’(x) = 0

3(x+1)2(x-3)3 = -3(x-3)2(x+1)3

x-3 = -(x+1)

2x = 2

x =1

When x>1 the function is increasing.

x<1 function is decreasing.

So, f(x) is increasing in (1, ∞).

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