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

Answers (1)

Answer:

\text {f(x) is decreasing on the interval (-2,-1) and} \\ \text { increasing on the interval} (-\infty,-2) \cup (-1, \infty)

Given:

Here given that

f(x)=2x^{3}+9x^{2}+12x+20

To find:

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

Hint:

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

Solution:

We have,

f(x)=2x^{3}+9x^{2}+12x+20

Differentiating w.r.t. x we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}+9 x^{2}+12 x+20\right) \\ &\Rightarrow f^{\prime}(x)=6 x^{2}+18 x+12 \end{aligned}

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

For this we must have,

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}+18 x+12=0 \\ &\Rightarrow 6\left(x^{2}+3 x+2\right)=0 \\ &\Rightarrow x^{2}+3 x+2=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}+2 x+x+2=0 \\ &\Rightarrow(x+2)(x+1)=0 \\ &\Rightarrow x+2=0 \text { and } x+1=0 \\ &\Rightarrow x=-2 \text { and } x=-1 \\ &\text { Clearly, } f'(x)<0, \text { if }-2<x<-1 \text { or } x \in(-2,-1) \end{aligned}

\text { and } f^{\prime}(x)>0 \text { if } x<-1 \text { and } x>-2 \text { or } x \in(-\infty,-2) \text { and } x \in(-1, \infty)

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

Posted by

Gurleen Kaur

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