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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