#### Need solution for R.D.Sharma maths class 12 chapter  Maxima and Minima exercise 17.3 Question 1 sub Question 6.

Point of local minima value is 2 and it’s local minimum value is 2.

Hint:

First find critical values of f(x) by solving f'(x) =0 then find f''(x).

If $f^{\prime \prime}\left(c_{1}\right)>0$ then $\mathrm{c}_{1}$ is point of local minima.

If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .

where $c_1$ &$c_2$ are critical points.

Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.

Given:

$\frac{x}{2}+\frac{2}{x}, \underline{x}>0$

Explanation:

We have,

\begin{aligned} &f(x)=\frac{x}{2}+\frac{2}{x} \quad, x>0 \\ &f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}} \ \\ &f^{\prime \prime}(x)=\frac{4}{x^{3}} \end{aligned}

For local maxima or minima, we have $f^{\prime}(x)=0$

$\begin{gathered} \frac{1}{2}-\frac{2}{x^{2}}=0 \\ x^{2}=4 \\ x=2 \text { or } x=-2 \end{gathered}$

since, $x>0$

$x = 2$

Now, at x=2 ,

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

Thus, its min. value at 2 is 2.