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find the intervals in which the function is strictly increasing or decreasing f(x)= (x+1)³(x-1)³ |
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Answer» f(x) = (x+1)3 (x-1)3 f'(x) = 3(x+1)2 (x-1)3 + 3(x+3)3(x-1)2 = 3(x+1)2(x-1)2(x-1+x+1) = 6x(x+1)2(x-1)2 ∵ f'(x) > 0 when x > 0 & x ≠ 1 & f'(x) < 0 when x < 0 & x ≠ -1 Hence, f(x) is strictly increasing in (0,∞) - {1} & stictly decreasing in (-∞,0) - {-1} |
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