Question

show that the maximum value of (1/x)x is e1/e

Answer 1

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Answer 2

$y_{max} = e^{\dfrac{1}{e}}$

Step-by-step explanation:

Let $y = (\frac{1}{x} )^{x}$

Taking ln both sides, we get

$\ln y = x \ln\frac{1}{x} = x\ln (x)^{-1} = - x \ln x$ {Since, $\ln x^{a} = a \ln x$ }

Now, differentiating both sides with respect to x we get,

$\frac{1}{y} \frac{dy}{dx} = - 1 - \ln x$

⇒ $\frac{dy}{dx} = (\frac{1}{x} )^{x}[- 1 - \ln x]$ = 0 {Condition for y to be maximum}

⇒ $\ln x = - 1$ {Since, $(\frac{1}{x}) ^{x} \neq 0$}

⇒ x = $e^{-1}$ {Converting logarithmic equation to exponential equation}

⇒ $x = \frac{1}{e}$

So, $y_{max} = e^{\frac{1}{e}}$ (Proved)