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  • Please solve RD Sharma class 12 chapter 17 Maxima and Minima exercise Very short answer type question 5 maths textbook solution

Please solve RD Sharma class 12 chapter 17 Maxima and Minima exercise Very short answer type question  5 maths textbook solution

Answers (1)

Answer: -2

Given: f(x)=x+\frac{1}{x}

Solution:

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

For a local maxima or a local minima we must have

        \begin{aligned} &f^{\prime}(x)=0 \\\\ &1-\frac{1}{x^{2}}=0 \\\\ &x^{2}-1=0 \\\\ &x^{2}=1 \end{aligned}

        \begin{aligned} &x=1,-1 \\\\ &x<0 \\\\ &x=-1 \end{aligned}

Now

        \begin{aligned} &f^{\prime \prime}(x)=\frac{-2}{x^{3}} \\\\ &\text { At } x=-1 \\\\ &f^{\prime \prime}(-1)=\frac{2}{-1^{3}}=-2<0 \end{aligned}

So,x = -1 is a point of local maximum. Thus the local maximum value is given by

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

 

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