Answer :

Given:- Function f(x) = 2x3 – 24x + 7


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 – 24x + 7



f’(x) = 6x2 – 24


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


f’(x) > 0


6x2 – 24 > 0



x2 < 4


x < –2, +2


x (–,–2) and x (2,∞)


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


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


f’(x) < 0


6x2 – 24< 0



x2 < 4


x> –1


x (–1,∞)


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


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