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Need solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion xv

Answers (1)

Answer:

\text { Increasing interval } \left(-\infty, \frac{4}{3}\right) \cup(2, \infty) \\ \text { Decreasing interval } \left(\frac{4}{3}, 2 \right)

Given:

Here given that

f(x)=(x-1)(x-2)^{2}

To find:

We have to find the intervals in which f(x) is increasing or decreasing.

Hint:

Use product rule of differentiation

\text { i.e } \frac{d}{d x}(u \cdot v)=u \cdot \frac{d v}{d x}+v \frac{d u}{d x}

Solution:

We have,

f(x)=(x-1)(x-2)^{2}

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

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left[(x-1)(x-2)^{2}\right] \\ &\Rightarrow f^{\prime}(x)=(x-1) \frac{d}{d x}\left[(x-2)^{2}\right]+(x-2)^{2} \frac{d}{d x}(x-1) \\ &=2(x-1)(x-2)+(x-2)^{2} \\ &=(x-2)[2(x-1)+(x-2)] \\ &=(x-2)[2 x-2+x-2] \\ &=(x-2)(3 x-4) \end{aligned}

For f(x) we have to find critical points, for this we must have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow(x-2)(3 x-4)=0 \\ &\Rightarrow x-2=0 \text { and } 3 x-4=0 \end{aligned}

\begin{aligned} &\Rightarrow x=2 \text { and } 3 x=4\\ &\text { or } x=\frac{4}{3}\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<\frac{4}{3} \text { and } x>2 \text { and } f^{\prime}(x)<0 \text { if } \frac{4}{3}<x<2 \end{aligned}

\text { So, } f(x) \text { is increasing on the interval }\left(-\infty, \frac{4}{3}\right) \cup(2, \infty) \text { and decreasing on the interval }\left(\frac{4}{3}, 2\right) \text { . }

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