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

Solution:

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 $f^{\prime}(x)>0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$

Algorithm:-

(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ 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,

$f(x)=\{x(x-2)\}^{2}$

$\Rightarrow f(x)=\left\{\left[x^{2}-2 x\right]\right\}^{2}$

$\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(\left[x^{2}-2 x\right]^{2}\right)$

$\Rightarrow f^{\prime}(x)=2\left(x^{2}-2 x\right)(2 x-2)$

$\Rightarrow f^{\prime}(x)=4 x(x-2)(x-1)$

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

$\Rightarrow f^{\prime}(x)=0$

$\Rightarrow 4 x(x-2)(x-1)=0$

$\Rightarrow x(x-2)(x-1)=0$

$\Rightarrow x=0,1,2$

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

Here points $x=0,1,2$ divide the number line into disjoint intervals namely, $(-\infty, 0),(0,1),(1,2)$ and $(2, \infty)$

Lets consider interval $(-\infty, 0)$ and $(1,2)$

In this case, we have $x(x-2)(x-1)<0$

Therefore, $f^{\prime}(x)<0$ when $x<0$ and $1

Thus, $f(x)$ is strictly decreasing on interval $(-\infty, 0) \cup(1,2)$

Now, consider interval $(0,1)$ and $(2, \infty)$

In this case, we have $x(x-2)(x-1)>0$

Therefore, $f^{\prime}(x)>0$ when $0

Thus, $f(x)$ is strictly increases on interval $(0,1) \cup(2, \infty)$