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Explain solution RD Sharma class 12 chapter 17 Maxima and Minima exercise Very short answer type question  4 maths

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Answer: 2


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

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 \end{aligned}

        \begin{aligned} &x^{2}=1 \\\\ &x=1,-1 \end{aligned}

\begin{aligned} &\text { But } x>0 \\\\ &\qquad x=1 \\\\ &\text { Now } f^{\prime \prime}(x)=\frac{2}{x^{3}} \\\\ &\text { At } x=1 \end{aligned}

        f^{\prime \prime}(1)=\frac{2}{1^{3}}=2>0

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



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