Need solution for R.D. Sharma maths class 12 chapter 17 Maxima and Minima exercise 17.3  question 1 sub question 4.

Point of local maxima is 2 and it’s local maximum value is $\frac{1}{2}$

Hint:

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

If $f''(c_1) >0$ then $c_1$ is point of local minima.

If $f"(c_2) <0$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:

$f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0$

Explanation:

We have,

$\begin{gathered} f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0 \\ \therefore f^{\prime}(x)=-\frac{2}{x^{2}}+\frac{4}{x^{3}} \\ f^{\prime \prime}(x)=\frac{4}{x^{2}}-\frac{12}{x^{4}} \end{gathered}$

For maxima and minima, f’(x)=0

$\begin{array}{r} -\frac{2}{x^{2}}+\frac{4}{x^{3}}=0 \\ -\frac{2(x-2)}{x^{3}}=0 \\ x=2 \end{array}$

At x=2,

\begin{aligned} f^{\prime \prime}(2) &=\frac{4}{(2)^{3}}-\frac{12}{(2)^{2}} \\ &=\frac{4}{8}-\frac{12}{16} \\ &=\frac{1}{2}-\frac{3}{4} \\ &=-\frac{1}{4}<0 \end{aligned}

So, x=2 is point of local maxima.

So, local max. value at x= 2 is

\begin{aligned} f(2) &=\frac{2}{2}-\frac{2}{(2)^{2}} \\ &=1-\frac{2}{4}=\frac{1}{2} \end{aligned}

Hence, point of local maxima is 2 & its max. value is $\frac{1}{2}$