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

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

Answers (5)

Answer:

              2

Hint:

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

Given: 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}}> 0forx=2

\therefore x=2 is the point of minima

\therefore Given function has local minima at x=2

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Answer:

              2

Hint:

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

Given: 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}}> 0forx=2

\therefore x=2 is the point of minima

\therefore Given function has local minima at x=2

Posted by

Infoexpert

View full answer

Answer:

              2

Hint:

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

Given: 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}}> 0forx=2

\therefore x=2 is the point of minima

\therefore Given function has local minima at x=2

Posted by

Infoexpert

View full answer

Answer:

              2

Hint:

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

Given: 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}}> 0forx=2

\therefore x=2 is the point of minima

\therefore Given function has local minima at x=2

Posted by

Infoexpert

View full answer

Answer:

              2

Hint:

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

Given: 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}}> 0forx=2

\therefore x=2 is the point of minima

\therefore Given function has local minima at x=2

Posted by

Infoexpert

View full answer

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