Answer :

Given:- Function f(x) = 2x3 – 12x2 + 18x + 15


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) = 2x3 – 12x2 + 18x + 15



f’(x) = 6x2 – 24x + 18


For f(x) lets find critical point, we must have


f’(x) = 0


6x2 – 24x + 18 = 0


6(x2 – 4x + 3) = 0


6(x2 – 3x – x + 3) = 0


6(x – 3)(x – 1) = 0


(x – 3)(x – 1) = 0


x = 3 , 1





clearly, f’(x) > 0 if x < 1 and x > 3


and f’(x) < 0 if 1< x < 3


Thus, f(x) increases on (–,1) (3, ∞)


and f(x) is decreasing on interval x (1,3)



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