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Please solve RD Sharma class 12 chapter 17 Maxima and Minima exercise Very short answer type question  9 maths textbook solution

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Answer: e^{\frac{1}{e}}

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



Taking log on both sides we get

        \log \mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}} \log \mathrm{x}

Differentiating with respect to x , we get

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

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

For  a local maxima or a local minima we must have

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


        f^{\prime \prime}(x)=x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)^{2}+x^{\frac{1}{x}}\left(-\frac{2}{x^{3}}+\frac{2}{x^{3}} \log x-\frac{1}{x^{3}}\right)

                    =x^{\frac{1}{x}}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)^{2}+x^{\frac{1}{x}}\left(-\frac{3}{x^{3}}+\frac{2}{x^{3}} \log x\right)

        \begin{aligned} &\text { At } x=e \\ &f^{\prime \prime}(e)=e^{\frac{1}{e}}\left(\frac{1}{e^{2}}-\frac{1}{e^{2}} \text { loge }\right)^{2}+e^{\frac{1}{e}}\left(-\frac{3}{e^{3}}+\frac{2}{e^{3}} \text { loge }\right) \end{aligned}


So x = e is a point of local maximum.

Thus the maximum value is given by



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