Find the maximum of the function $f(x)=x^{1/x}$ and the value of $x$ which gives the maximum value?
Find the maximum of $f(x)=x^{1/x}$
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$\begingroup$
calculus
optimization
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0@celtschk: No, if you know the rule you just used, it's not a great improvement. But unless you have learned a specific rule for differentiationg $f(x)^{g(x)}$, this relies on a modest bit of multivariable calculus, using the partial derivatives of $x^y$ and the multivariable chain rule. But it's not really a big deal (which is why I put the words “a bit” in my comment). – 2012-07-11
2 Answers
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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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2I 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 $\dfrac{d}{dx} y$. – 2012-03-03
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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!
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0I like this! It is so smooth just like exp itself. – 2018-04-09