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

Answers (1)

Answer:

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

Given:

Here given that

f(x)=x^{3}-6x^{2}+9x+15

To find:

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

Hint:

Put f ‘(x) = 0 to find critical points of f(x) and then use increasing and decreasing property.

Solution:

We have,

f(x)=x^{3}-6x^{2}+9x+15

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

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(x^{3}-6 x^{2}+9 x+15\right) \\ &\Rightarrow f^{\prime}(x)=3 x^{2}-12 x+9 \end{aligned}

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

We must have,

\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow 3 x^{2}-12 x+9=0\\ &\Rightarrow 3\left(x^{2}-4 x+3\right)=0\\ &\Rightarrow x^{2}-4 x+3=0\\ &\Rightarrow x^{2}-3 x-x+3=0\\ &\Rightarrow(x-3)(x-1)=0\\ &\Rightarrow x=3,1\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } 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 on 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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