Find the maximum of the function $$f(x)=x^{1/x}$$ and the value of $x$ which gives the maximum value?
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2$\begingroup$ Like every other maximum problem: differentiate, set the derivative to 0, solve for $x$, check whether any of the solutions happen to be maxima. $\endgroup$ – hmakholm left over Monica Mar 3 '12 at 21:32
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3$\begingroup$ It's a bit easier if you consider $\ln f(x)$ instead. $\endgroup$ – Harald Hanche-Olsen Mar 3 '12 at 21:34
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$\begingroup$ @HaraldHanche-Olsen: How does taking the logarithm make it easier? $\endgroup$ – celtschk Jul 9 '12 at 10:40
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$\begingroup$ @celtschk: Just take a look at J.D.'s answer. $\endgroup$ – Harald Hanche-Olsen Jul 10 '12 at 19:28
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2$\begingroup$ I'm looking. But I still don't see how it's easier than just deriving $x^{1/x}$ directly: $(x^{1/x})' = (1/x)x^{1/x-1} + x^{1/x}\ln x \cdot (-1/x^2) = x^{1/x-2}(1-\ln x)$ $\endgroup$ – celtschk Jul 10 '12 at 19:43
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Let $y = x^{1/x}.$ So $$ \ln y = (1/x)\ln x. $$ Differentiate both sides w.r.t $x$, we get $$ y'/y = (1/x)(1/x) + (-1/x^2) \ln x. $$ Rearranging, we have $$\dfrac{d}{dx}(x^{1/x}) = (1-\ln x)x^{1/x - 2} $$ Set $\dfrac{d}{dx}(x^{1/x}) = 0$ and work from there to get the maximum.
$$\text{(Hint: maximum occurs at x = e.)}$$
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2$\begingroup$ I differentiated both sides of the equation $\ln y = (1/x) \ln x$. $\dfrac{d}{dx} \ln y = (\dfrac{d}{dx} y)/y$ by the rules of derivatives of logarithmic functions, and $y'$ is a shorthand for $\dfrac{d}{dx} y$. $\endgroup$ – user2468 Mar 3 '12 at 22:00
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This is not original with me.
If we know that $e^x \ge 1+x$ with equality only when $x = 0$, $e^{(x-e)/e} \ge 1 + (x-e)/e = x/e$ or $e^{x/e} \ge x$ or $e^{1/e} \ge x^{1/x}$ with equality only if $x = e$.
Ta-dah!
At no time do the fingers leave the hands!
answered Mar 3 '12 at 22:48
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1$\begingroup$ +1 for presentation! I watch them magic tricks close though, I do. $\endgroup$ – Eugene Shvarts Jul 9 '12 at 8:28
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$\begingroup$ I like this! It is so smooth just like exp itself. $\endgroup$ – Philimathmuse Apr 9 '18 at 21:09
