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

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

Answers (1)

Answer:

\text { Increasing interval }\left(-\infty, \frac{-9}{2}\right) \\ \text { Decreasing interval }\left(\frac{-9}{2}, \infty\right)

Given:

Here given that

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

To find:

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

Hint:

Use increasing and decreasing property.

Solution:

Here given that

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

On differentiating we get,

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

For f(x) to be increasing, we must have

\begin{aligned} &f^{\prime}(x)>0 \\ &\Rightarrow-9-2 x> \\ &\Rightarrow-2 x>9 \\ &\Rightarrow x<\frac{-9}{2} \end{aligned}

By applying (-) ve sign change comparison sign.

\begin{aligned} &x \in\left(-\infty, \frac{-9}{2}\right)\\ &\text { So, } f(x) \text { is increasing on the interval }\left(-\infty, \frac{-9}{2}\right) \end{aligned}

For f(x) to be decreasing, we must have

\begin{aligned} &f^{\prime}(x)<0 \\ &\Rightarrow-9-2 x<0 \\ &\Rightarrow-2 x<9 \\ &\Rightarrow x>\frac{-9}{2} \end{aligned}

By applying (-) ve sign change comparison sign.

\begin{aligned} &\Rightarrow x \in\left(\frac{-9}{2}, \infty\right)\\ &\text { So, } f(x) \text { is decreasing on the interval }\left(\frac{-9}{2}, \infty\right) \end{aligned}

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Gurleen Kaur

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