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