#### Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 5

$x=0$ is the point of  local maxima and the value of local maxima is $\frac{1}{2}$

Hint:

Use first derivative test to find the value and point of local maxima and local minima.

Given:

$f\left ( x \right )=\frac{1}{x^{2}+2}$

Solution:

$f\left ( x \right )=\frac{1}{x^{2}+2}$

Differentiating  $f\left ( x \right )$ with respect to ‘x’ then,

$\frac{d}{d x}\{f(x)\}=\frac{d\left(\frac{1}{x^{2}+2}\right)}{d x}=\frac{d\left(x^{2}+2\right)-1}{d x} \quad\left[\because \frac{d}{d x}\left(x^{n}+a\right)^{n}=n\left(x^{n}+a\right)^{n-1} \frac{d}{d x}\left(x^{n}+a\right)\right]$

$=-1\left(x^{2}+2\right)-1-1 \cdot \frac{d}{d x}\left(x^{2}+2\right)$

$=-1\left(x^{2}+2\right)^{-2}\left\{\frac{d\left(x^{2}\right)}{d x}+\frac{d(2)}{d x}\right\} \ \ \ \ \ \ \ \ \left[\because \frac{d}{d x}\left(x^{n}+a\right)=\frac{d\left(x^{n}\right)}{d x}+\frac{d(a)}{d x}\right]$

\begin{aligned} &=\frac{-1}{\left(x^{2}+2\right)^{2}}\left[2 x^{2-1}+0\right] \quad\left[\because \frac{d x^{n}}{d x}=n x^{n-1}, \frac{d}{d x} \text { constant }=0\right. \\ &\therefore f^{i}(x)=\frac{-1}{\left(x^{2}+2\right)^{2}} 2 x=\frac{-2 x}{\left(x^{2}+2\right)^{2}} \end{aligned}

By first derivative test, for local maxima or local minima, we have

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{-2 x}{\left(x^{2}+2\right)^{2}}=0 \quad \Rightarrow \quad \frac{x}{\left(x^{2}+2\right)^{2}}=0[\because-2 \neq 0] \\ &\Rightarrow x=0 \end{aligned}

+                                     -

-∞                                     0                                ∞

since ${f}'\left ( x \right )$ changes from +ve to -ve when  through $x=0$
So, $x=0$ is the point of local maxima.
The value of local maxima of ${f}'\left ( x \right )$ at is$-x=0$
$f(0)=\frac{1}{0^{2}+2}=\frac{1}{2}$

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$x=0$ is the point of  local maxima and the value of local maxima is $\frac{1}{2}$

Hint:

Use first derivative test to find the value and point of local maxima and local minima.

Given:

$f\left ( x \right )=\frac{1}{x^{2}+2}$

Solution:

$f\left ( x \right )=\frac{1}{x^{2}+2}$

Differentiating  $f\left ( x \right )$ with respect to ‘x’ then,

$\frac{d}{d x}\{f(x)\}=\frac{d\left(\frac{1}{x^{2}+2}\right)}{d x}=\frac{d\left(x^{2}+2\right)-1}{d x} \quad\left[\because \frac{d}{d x}\left(x^{n}+a\right)^{n}=n\left(x^{n}+a\right)^{n-1} \frac{d}{d x}\left(x^{n}+a\right)\right]$

$=-1\left(x^{2}+2\right)-1-1 \cdot \frac{d}{d x}\left(x^{2}+2\right)$

$=-1\left(x^{2}+2\right)^{-2}\left\{\frac{d\left(x^{2}\right)}{d x}+\frac{d(2)}{d x}\right\}\left[\because \frac{d}{d x}\left(x^{n}+a\right)=\frac{d\left(x^{n}\right)}{d x}+\frac{d(a)}{d x}\right]$

\begin{aligned} &=\frac{-1}{\left(x^{2}+2\right)^{2}}\left[2 x^{2-1}+0\right] \quad\left[\because \frac{d x^{n}}{d x}=n x^{n-1}, \frac{d}{d x} \text { constant }=0\right. \\ &\therefore f^{i}(x)=\frac{-1}{\left(x^{2}+2\right)^{2}} 2 x=\frac{-2 x}{\left(x^{2}+2\right)^{2}} \end{aligned}

By first derivative test, for local maxima or local minima, we have

\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow \frac{-2 x}{\left(x^{2}+2\right)^{2}}=0 \quad \Rightarrow \quad \frac{x}{\left(x^{2}+2\right)^{2}}=0[\because-2 \neq 0] \\ &\Rightarrow x=0 \end{aligned}

+                                     -

-∞                                     0                                ∞

since ${f}'\left ( x \right )$ changes from +ve to -ve when  through $x=0$
So, $x=0$ is the point of local maxima.
The value of local maxima of ${f}'\left ( x \right )$ at is$-x=0$
$f(0)=\frac{1}{0^{2}+2}=\frac{1}{2}$