Question

Find the intervals in which f(x)=(x+1)³(x-3)³ is (1) strictly increasing (2) strictly decreasing.

Answer 1

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 possible intervals are : (-∞ ,-1) , (-1,1) , (1, 3) , (3, ∞)

case 1 :- f'(x) < 0 in (-∞, -1) , (-1, 1)

so, function is strictly decreasing in (-∞ , -1) U (-1, 1)

case 2 :- f'(x) > 0 in (1, 3) and (3, ∞)

so, function is strictly increasing in (1, 3) U (3, ∞)

Answer 2

Dear Student:

F(x)=(x+1)³(x-3)³

For,maximum and minimum  

We will find df/dx and then equate to zero.

If df/dx>o then f is strictly increasing.

If df/dx<0 then strictly decreasing.

And then again proceed for second derivative and see it is negative or positive.

If it is positive then f will be minimum

And if negative then f maximum

See the attachment: