Answer :

Given:- Function f(x) = 3x4 – 4x3 – 12x2 + 5


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) = 3x4 – 4x3 – 12x2 + 5



f’(x) = 12x3 – 12x2 – 24x


f’(x) = 12x(x2 – x – 2)


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


f’(x) > 0


12x(x2 – x – 2)> 0


x(x2 – 2x + x – 2) > 0


x(x – 2)(x + 1) > 0


–1 < x < 0 and x > 2


x (–1,0)(2, )


Thus f(x) is increasing on interval (–1,0)(2, )


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


f’(x) < 0


12x(x2 – x – 2)< 0


x(x2 – 2x + x – 2) < 0


x(x – 2)(x + 1) < 0


–∞ < x < –1 and 0 < x < 2


x (–∞, –1) (0, 2)


Thus f(x) is decreasing on interval (–∞, –1) (0, 2)


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