in Calculus edited by
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in Calculus edited by
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$\\y=x^{1/x}\\ taking\ ln\ on\ both\ sides\\ ln\ y=\dfrac{1}{x}ln\ x\\ y=e^{\dfrac{lnx}{x}}\\ x^{1/x}=e^{\dfrac{lnx}{x}}\\ (e^x\ is\ max\ when\ x\ is\ max)\\ x^{1/x}\ is\ max\ when\ \dfrac{lnx}{x}\ is\ max\\ f(x)=\dfrac{lnx}{x}\\ f'(x)=\dfrac{1-lnx}{x^2}=0\\ so\ x=e\\ f''(x)=\dfrac{-3x+2xlnx}{x^4}\\ f''(e)=\dfrac{-e}{e^4}<0\\ \\ \therefore x^{1/x}\ is\ max\ at\ x=e$