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A-question-related-to-Q-15184-Find-the-maximum-of-f-x-ln-x-1-x-




Question Number 15234 by mrW1 last updated on 08/Jun/17
A question related to Q.15184  Find the maximum of f(x)=(ln x)^(1/x)
$$\mathrm{A}\:\mathrm{question}\:\mathrm{related}\:\mathrm{to}\:\mathrm{Q}.\mathrm{15184} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{ln}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$
Commented by mrW1 last updated on 09/Jun/17
the domain of f(x)=(ln x)^(1/x)  is [1,+∞)  let t=(1/x), t∈(0,1]  y=(ln (1/t))^t =(−ln t)^t   ln y=tln (−ln t)  (1/y)×(dy/dt)=ln (−ln t)+t×(1/(−ln t))×((−1)/t)=ln (−ln t)+(1/(ln t))  (dy/dt)=(−ln t)^t [ln (−ln t)+(1/(ln t))]=0  ⇒ln (−ln t)+(1/(ln t))=0  ⇒t=0.1714907907 (through graph)  ⇒y_(max) =1.102147
$$\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{ln}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:\mathrm{is}\:\left[\mathrm{1},+\infty\right) \\ $$$$\mathrm{let}\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{x}},\:\mathrm{t}\in\left(\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{y}=\left(\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{t}} =\left(−\mathrm{ln}\:\mathrm{t}\right)^{\mathrm{t}} \\ $$$$\mathrm{ln}\:\mathrm{y}=\mathrm{tln}\:\left(−\mathrm{ln}\:\mathrm{t}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{y}}×\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{ln}\:\left(−\mathrm{ln}\:\mathrm{t}\right)+\mathrm{t}×\frac{\mathrm{1}}{−\mathrm{ln}\:\mathrm{t}}×\frac{−\mathrm{1}}{\mathrm{t}}=\mathrm{ln}\:\left(−\mathrm{ln}\:\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{t}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dt}}=\left(−\mathrm{ln}\:\mathrm{t}\right)^{\mathrm{t}} \left[\mathrm{ln}\:\left(−\mathrm{ln}\:\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{t}}\right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{ln}\:\left(−\mathrm{ln}\:\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{t}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{t}=\mathrm{0}.\mathrm{1714907907}\:\left(\mathrm{through}\:\mathrm{graph}\right) \\ $$$$\Rightarrow\mathrm{y}_{\mathrm{max}} =\mathrm{1}.\mathrm{102147} \\ $$
Commented by mrW1 last updated on 09/Jun/17

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