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,





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


f’(x) = 0




x =1, –2, 3





Now, lets check values of f(x) between different ranges


Here points x = 1, –2 , 3 divide the number line into disjoint intervals namely, (–, –2),(–2, 1), (1, 3) and (3, ∞)


Lets consider interval (–, –2)


In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0


Therefore, f’(x) < 0 when –∞ < x < –2


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


consider interval (–2, 1)


In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0


Therefore, f’(x) > 0 when –2 < x < 1


Thus, f(x) is strictly increases on interval x (–2, 1)


Now, consider interval (1, 3)


In this case, we have x – 1 > 0, x + 2 > 0 and x – 3 < 0


Therefore, f’(x) < 0 when 1 < x < 3


Thus, f(x) is strictly decreases on interval x (1, 3)


finally, consider interval (3, ∞)


In this case, we have x – 1 > 0, x + 2 > 0 and x – 3 > 0


Therefore, f’(x) > 0 when x > 3


Thus, f(x) is strictly increases on interval x (3, ∞)



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