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

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

Answers (1)

Answer:

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

Given:

Here given that

f(x)=3x^{4}-4x^{3}-12x^{2}+5

To find:

We have to find the increasing and decreasing interval.

Hint:

\text { If } f^{\prime}(x)>0 \text { for all } x \in(a ., b), \text { then } f(x) \text { is increasing on }(a, b) \text { and If } f^{\prime}(x)<0 \text { for all } x \in \text { (a., } b) \text { , then } f(x) \text { is decreasing on }(a, b) \text { . }

Solution:

We have,

f(x)=3x^{4}-4x^{3}-12x^{2}+5

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

\begin{aligned} &f^{\prime}(x)=12 x^{3}-12 x^{2}-24 x \\ &=12 x\left(x^{2}-x-2\right) \end{aligned}

Now,

For critical points, we must have,

\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 12 x\left(x^{2}-x-2\right)=0\\ &\Rightarrow x\left(x^{2}-x-2\right)=0,\{\therefore 12=0\}\\ &\Rightarrow x\left(x^{2}-2 x+x-2\right)=0\\ &\Rightarrow x(x-2)(x+1)=0\\ &\Rightarrow x=0 ; x=-1 ; x=2\\ &\text { The points } x=0,-1,2 \text { divide the real line into four disjoint intervals, } \end{aligned}

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

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

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

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

Posted by

Gurleen Kaur

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