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

f(x) = {x (x – 2)}^{2}

Given:- Function f(x)= {x (x – 2)}^{2}

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)= {x (x – 2)}^{2}

⇒ f(x)= {[x^{2}–2x]}^{2}

⇒

⇒ f’(x)= 2(x^{2}–2x)(2x–2)

⇒ f’(x)= 4x(x–2)(x–1)

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

⇒ f’(x) = 0

⇒ 4x(x–2)(x–1)= 0

⇒ x(x–2)(x–1)= 0

⇒ x =0, 1, 2

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

Here points x = 0, 1, 2 divide the number line into disjoint intervals namely, (–∞, 0),(0, 1), (1, 2) and (2, ∞)

Lets consider interval (–∞, 0) and (1, 2)

In this case, we have x(x–2)(x–1)< 0

Therefore, f’(x) < 0 when x < 0 and 1< x < 2

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

Now, consider interval (0, 1) and (2, ∞)

In this case, we have x(x–2)(x–1)> 0

Therefore, f’(x) > 0 when 0< x < 1 and x < 2

Thus, f(x) is strictly increases on interval (0, 1)∪(2, ∞)

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