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How to find such things? Like, it is negative or positive and all? :/ This pic is just a sample.. am unable to understand how we get (-)(-)(-) or (-)(+)(-) or (+)(+)(-) or (+)(+)(+) and all? :/ (P.S : Don't ask me to check study material coz am NEET member, so am unable to get access to maths)
 
Interval Sign of f'(x)=12x(x+1)(x-2) Nature of function

- ∞ , - 1 		(-)(-)(-)=(-) or <0 Strictly decreasing
(-1,0)
(-)(+)(-)=(+) or >0
Strictly increasing
(0,2) (+)(+)(-)=(-) or <0 Strictly decreasing
(2, ∞ )
(-)(+)(+)=(+) or >0 		Strictly increasing

(a) The given function is strictly increasing in the intervals  - 1 , 0 ∪ 2 , ∞
(b) The given function is strictly decreasing in the intervals  - ∞ , - 1 ∪ 0 , 2

Dear Student,

We have f'x = 12xx + 1x - 2Now, to check the sign of f'x in the interval -∞, -1,we evaluate f'x for values of x in -∞, -1.Let's take x = -2.Then, f'-2 = 12-2-2 + 1-2 - 2 = -24-1-4.Now, multiplying negative numbers thrice gives a negative number.So, we conclude, without actually calculating the actual value of f'x, that it is less than 0 in the interval -∞, -1.The above will be true for any value of x in -∞, -1.So, we can write the sign of f'x as --- = -ve, for all values of x in -∞, -1.Similarly, for x ∈ -1, 0,We can say that f'x = -+- = +ve You can test this by taking different values of x in -1, 0 such as -0.5, -0.75, etc. - they will all yield the same resultYou can check this in another way.Let's take x ∈ 0, 2We have f'x = 12xx + 1x - 2Now, since x is positive, 12x is also positive. Since x is positive, x + 1 is also positive. However, 0 < x < 2 ⇒ 0 - 2 < x - 2 < 2 - 2 ⇒ -2 < x - 2 < 0, or x - 2 is negative.So, for all x ∈ 0, 2, f'x = ++- = -ve.

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