#### Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 20 maths textbook solution

2

Hint:

For maxima or minima we must have ${f}'\left ( x \right )=0$

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

Solution:

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

For extremum${f}'\left ( x \right )=0$

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

$x=\underline{+}2$

${{f}'}'\left ( x \right )=\frac{4}{x^{3}}> 0$for$x=2$

$\therefore x=2$ is the point of minima

$\therefore$ Given function has local minima at $x=2$

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2

Hint:

For maxima or minima we must have ${f}'\left ( x \right )=0$

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

Solution:

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

For extremum${f}'\left ( x \right )=0$

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

$x=\underline{+}2$

${{f}'}'\left ( x \right )=\frac{4}{x^{3}}> 0$for$x=2$

$\therefore x=2$ is the point of minima

$\therefore$ Given function has local minima at $x=2$

#### Infoexpert

2

Hint:

For maxima or minima we must have ${f}'\left ( x \right )=0$

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

Solution:

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

For extremum${f}'\left ( x \right )=0$

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

$x=\underline{+}2$

${{f}'}'\left ( x \right )=\frac{4}{x^{3}}> 0$for$x=2$

$\therefore x=2$ is the point of minima

$\therefore$ Given function has local minima at $x=2$

#### Infoexpert

2

Hint:

For maxima or minima we must have ${f}'\left ( x \right )=0$

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

Solution:

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

For extremum${f}'\left ( x \right )=0$

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

$x=\underline{+}2$

${{f}'}'\left ( x \right )=\frac{4}{x^{3}}> 0$for$x=2$

$\therefore x=2$ is the point of minima

$\therefore$ Given function has local minima at $x=2$

#### Infoexpert

2

Hint:

For maxima or minima we must have ${f}'\left ( x \right )=0$

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

Solution:

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

For extremum${f}'\left ( x \right )=0$

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

$x=\underline{+}2$

${{f}'}'\left ( x \right )=\frac{4}{x^{3}}> 0$for$x=2$

$\therefore x=2$ is the point of minima

$\therefore$ Given function has local minima at $x=2$