Answer :

Given:- Function


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) = 6x3 – 12x2 – 90x


f’(x) = 6x(x2 – 2x – 15)


f’(x) = 6x(x2 – 5x + 3x – 15)


f’(x) = 6x(x – 5)(x + 3)


For f(x) to be increasing, we must have


f’(x) > 0


6x(x – 5)(x + 3)> 0


x(x – 5)(x + 3) > 0


–3 < x < 0 or 5 < x < ∞


x (–3,0)(5, )


Thus f(x) is increasing on interval (–3,0)(5, )


Again, For f(x) to be decreasing, we must have


f’(x) < 0


6x(x – 5)(x + 3)> 0


x(x – 5)(x + 3) > 0


–∞ < x < –3 or 0 < x < 5


x (–∞, –3)(0, 5)


Thus f(x) is decreasing on interval (–∞, –3)(0, 5)


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