#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xxv maths textbook solution

$\text { Increasing interval }(0,1) \cup(2, \infty) \\ \text { Decreasing interval }(-\infty ,0) \cup(1,2)$

Given:

Here given that

$f(x)=[x(x-2)]^{2}$

To find:

We have to find the increasing or decreasing interval for f(x).

Hint:

Put f ‘(x) = 0 to find critical points and then use increasing and decreasing property.

Solution:

We have,

$f(x)=[x(x-2)]^{2}$

Differentiating w.r.t. x, we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left[x^{2}-2 x\right]^{2} \\ &=2\left(x^{2}-2 x\right)(2 x-2) \\ &=4 x(x-2)(x-1) \end{aligned}

For f(x), we have to find critical points.

we must have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 4 x(x-2)(x-1)=0 \\ &\Rightarrow x(x-2)(x-1)=0 \\ &\Rightarrow x=0,1,2 \end{aligned}

$\text { The possible intervals are }(-\infty, 0),(0,1),(1,2) \text { and }(2, \infty)$

\begin{aligned} &\text { In the intervals }(-\infty, 0) \text { and }(1,2) f^{\prime}(x)<0 \text { . }\\ &\text { Clearly, } f^{\prime}(x)<0 \text { if }-\infty

$\text { However, } \\ \text { In intervals }(0,1) \text { and }(2, \infty), f^{\prime}(x)>0 \\ \text { Clearly, } f^{\prime}(x)>0 \text { if } 0

$\text { Thus, the function } f(x) \text { is increasing on }(0,1) \cup(2, \infty) \text { and decreasing on }(-\infty, 0) \cup(1,2) \text { . }$