#### Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 14

$x=2$ is the point of local minima and the value of local minima is 2

Hint:

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

Given:

$f(x)=\frac{x}{2}+\frac{2}{x}, x>0$

Solution:-

$f(x)=\frac{x}{2}+\frac{2}{x}$

Differentiating $f(x) w \cdot r . t^{\prime} x^{\prime}then$

\begin{aligned} \frac{d}{d x}\{f(x)\} &=\frac{d}{d x}\left\{\frac{2}{x}+\frac{x}{2}\right\} \\ &=\frac{d}{d x}\left\{\frac{2}{x}\right\}+\frac{d}{d x}\left\{\frac{x}{2}\right\} \quad\left[\frac{d}{d x}\{f(x)+h(x)\}=\frac{d}{d x}\{f(x)\}+\frac{d}{d x}\{h(x)\}\right] \\ &=2 \frac{d}{d x}\left(x^{-1}\right)+\frac{1}{2} \frac{d}{d x}(x) \quad\left[\frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ &=2(-1) x^{-1-1}+\frac{1}{2} x^{1-1} \quad\left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \\ \end{aligned}

$=-2 x^{-2}+\frac{1}{2} x^{0} \\$

$=\frac{-2}{x^{2}}+\frac{1}{2} \\ f^{\prime}(x) =\frac{1}{2}-\frac{2}{x^{2}}$

By first derivative test, for local maxima and local minima, we ha

$f^{\prime}(x)=0$

$\Rightarrow \frac{1}{2}-\frac{2}{x^{2}}=0 \Rightarrow \frac{1}{2}=\frac{2}{x^{2}}$

$\Rightarrow \quad x^{2}=4 \quad \Rightarrow x=\pm 2$

$\Rightarrow x=2$             $[x=-2 is not possible since x>0]$

-                      +

-∞              2               +∞

Since ${f}'\left ( x \right )$ changes from $-v e to +v e$ when $x$ increases through 2. So $x=2$ is the point of local minima

The value of local minima of $f\left ( x \right )$ at $x=2$  is

\begin{aligned} f(2) &=\frac{2}{2}+\frac{2}{2}=1+1 \\ &=2 \end{aligned}