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

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\text {Increasing interval} (-\infty,-2) \cup(6, \infty) \\ \text {Decreasing interval} (-2,6)


Here given that.


To find:

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


Put f(x)=0 and solve this equation to find critical points of f(x) and use increasing and decreasing property.


We have,


Differentiating with respect to x we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-6 x^{2}-36 x+2\right) \\ &f^{\prime}(x)=3 x^{2}-12 x-36 \end{aligned}

Now we have to find critical points for f(x).

We must have,

\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-12 x-36=0\\ &\Rightarrow 3\left(x^{2}-4 x-12\right)=0\\ &\Rightarrow x^{2}-4 x-12\{\therefore 3>0\}\\ &\Rightarrow x^{2}-6 x+2 x-12=0\\ &\Rightarrow(x-6)(x+2)=0\\ &\Rightarrow x-6=0 \text { and } x+2=0\\ &\Rightarrow x=6 \text { and } x=-2\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<-2 \text { and } x>6 \text { or } x \in(-\infty,-2) \text { and } x \in(6, \infty) \text { and } f^{\prime}(x)<0 \text { if } \end{aligned}

\begin{aligned} &-2<x<6 \text { or } x \in(-2,6)\\ &\text { Thus, } f(x) \text { is increasing on the interval }(-\infty,-2) \cup(6, \infty) \text { and } \\ &\text {f(x) is decreasing on interval (-2,6).} \end{aligned}

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Gurleen Kaur

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