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Explain solution RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion iv maths

Answers (1)

Answer:

\text {Increasing interval}(-\infty, 1) \cup(3, \infty) \\ \text {Decreasing interval}(1,3)

Given:

Here given that

f(x)=2x^{3}-12x^{2}+18x+15

To find:

We have to find out the intervals in which function is increasing and decreasing.

Hint:

Firstly we will find critical points and then use increasing and decreasing property.

Solution:

Given that

f(x)=2x^{3}-12x^{2}+18x+15

On differentiating we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}-12 x^{2}+18 x+15\right) \\ &\Rightarrow f^{\prime}(x)=6 x^{2}-24 x+18 \end{aligned}

Firstly we will find critical points for f(x).

For this we have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}-24 x+18=0 \\ &\Rightarrow 6\left(x^{2}-4 x+3\right)=0 \\ &\Rightarrow x^{2}-3 x-x+3=0\{\therefore 6>0\} \\ &\Rightarrow(x-3)(x-1)=0 \\ &\Rightarrow x-3=0 \text { and } x-1=0 \\ &\Rightarrow x=3 \text { and } x=1 \end{aligned}

\begin{aligned} &\text { Clearly, } f^{\prime}(x)>0, f(x)<1 \text { and } x>3 \text { or } x \in(-\infty, 1) \text { and } x \in(3, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &1<x<3 \text { or } x \in(1,3) \end{aligned}

\text { So, } f(x) \text { is increasing the interval }(-\infty, 1) \cup(3, \infty) \text { and } f(x) \text { is decreasing on interval }(1,3) \text { . }

Posted by

Gurleen Kaur

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