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Question-120243




Question Number 120243 by benjo_mathlover last updated on 30/Oct/20
Answered by bemath last updated on 30/Oct/20
let f(x) = x^(1/x^6 )  ⇔ ln f(x)= ((ln (x))/x^6 )  differentiating both sides gives  ((f ′(x))/(f(x))) = ((x^5 −6x^5 ln (x))/x^(12) ) ; f ′(x)= ((x^(1/x^6 ) .x^5 (1−6ln (x)))/x^(12) )  f ′(x) = ((x^(1/x^6 ) (1−6ln x))/x^7 )  taking f ′(x)=0 , we get ln x=(1/6) or x=e^(1/6)   curve increasing on 0<x<e^(1/6)  and   decreasing on x >e^(1/6) . Thus maximum  value is (e^(1/6) )^(1/e) = e^(1/(6e))
$${let}\:{f}\left({x}\right)\:=\:{x}^{\frac{\mathrm{1}}{{x}^{\mathrm{6}} }} \:\Leftrightarrow\:\mathrm{ln}\:{f}\left({x}\right)=\:\frac{\mathrm{ln}\:\left({x}\right)}{{x}^{\mathrm{6}} } \\ $$$${differentiating}\:{both}\:{sides}\:{gives} \\ $$$$\frac{{f}\:'\left({x}\right)}{{f}\left({x}\right)}\:=\:\frac{{x}^{\mathrm{5}} −\mathrm{6}{x}^{\mathrm{5}} \mathrm{ln}\:\left({x}\right)}{{x}^{\mathrm{12}} }\:;\:{f}\:'\left({x}\right)=\:\frac{{x}^{\frac{\mathrm{1}}{{x}^{\mathrm{6}} }} .{x}^{\mathrm{5}} \left(\mathrm{1}−\mathrm{6ln}\:\left({x}\right)\right)}{{x}^{\mathrm{12}} } \\ $$$${f}\:'\left({x}\right)\:=\:\frac{{x}^{\frac{\mathrm{1}}{{x}^{\mathrm{6}} }} \left(\mathrm{1}−\mathrm{6ln}\:{x}\right)}{{x}^{\mathrm{7}} } \\ $$$${taking}\:{f}\:'\left({x}\right)=\mathrm{0}\:,\:{we}\:{get}\:\mathrm{ln}\:{x}=\frac{\mathrm{1}}{\mathrm{6}}\:{or}\:{x}={e}^{\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$${curve}\:{increasing}\:{on}\:\mathrm{0}<{x}<{e}^{\frac{\mathrm{1}}{\mathrm{6}}} \:{and}\: \\ $$$${decreasing}\:{on}\:{x}\:>{e}^{\frac{\mathrm{1}}{\mathrm{6}}} .\:{Thus}\:{maximum} \\ $$$${value}\:{is}\:\left({e}^{\frac{\mathrm{1}}{\mathrm{6}}} \right)^{\frac{\mathrm{1}}{{e}}} =\:{e}^{\frac{\mathrm{1}}{\mathrm{6}{e}}} \\ $$

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