Answer :

Given:- Function f(x) = x3 – 6x2 + 12x – 18


Theorem:- Let f be a differentiable real function defined on an open interval (a,b).


(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)


(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)


Algorithm:-


(i) Obtain the function and put it equal to f(x)


(ii) Find f’(x)


(iii) Put f’(x) > 0 and solve this inequation.


For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain, it is decreasing.


Here we have,


f(x) = x3 – 6x2 + 12x – 18



f’(x) = 3x2 – 12x + 12


f’(x) = 3(x2 – 4x + 4)


f’(x) = 3(x – 2)2


as given


x ϵ R


(x – 2)2> 0


3(x – 2)2 > 0


f’(x) > 0


Hence, condition for f(x) to be increasing


Thus f(x) is increasing on interval x R


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