Q. 6 E3.9( 45 Votes )
Find the interval
Answer :
It is given that function f(x) = –(x + 1)3 (x – 3)3
⇒ f’(x) = 3(x + 1)2 (x – 3)3 +3(x + 1)3 (x – 3)2
⇒ f’(x) = 3(x + 1)2 (x – 3)2[x – 3 + x + 1]
⇒ f’(x) = 6(x + 1)2 (x – 3)2(x-1)
If f’(x) = 0, then we get,
⇒ x = -1,3 and 1
So, the points x = -1, x =1 and x = 3 divides the real line into four disjoint intervals,
So, in interval
f’(x) = 6(x + 1)2 (x – 3)2(x-1) < 0
Therefore, the given function (f) is strictly decreasing in intervals .
So, in interval
f’(x) = 6(x + 1)2 (x – 3)2(x-1) > 0
Therefore, the given function (f) is strictly increasing in intervals.
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