Answer :

Given:- Function f(x) = 5 + 36x + 3x2 – 2x3


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) = 5 + 36x + 3x2 – 2x3



f’(x) = 36 + 6x – 6x2


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


f’(x) = 0


36 + 6x – 6x2 = 0


6(–x2 + x + 6) = 0


6(–x2 + 3x – 2x + 6) = 0


–x2 + 3x – 2x + 6 = 0


x2 – 3x + 2x – 6 = 0


(x – 3)(x + 2) = 0


x = 3 , – 2





clearly, f’(x) > 0 if –2< x < 3


and f’(x) < 0 if x < –2 and x > 3


Thus, f(x) increases on x (–2,3)


and f(x) is decreasing on interval (–,–2) (3, ∞)



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