Find the intervals in which the following functions are increasing or decreasing.

f(x) = 2x3 – 24x + 107


Given:- Function f(x) = 2x3 – 24x + 107


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 + 107



f’(x) = 6x2 – 24


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


f’(x) = 0


6x2 – 24 = 0


6(x2 – 4) = 0


(x – 2)(x + 2) = 0


x = –2, 2


clearly, f’(x) > 0 if x < –2 and x > 2


and f’(x) < 0 if –2 < x < 2


Thus, f(x) increases on (–, –2) (2, ∞)


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


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