7

Find the maximum of the function $$f(x)=x^{1/x}$$ and the value of $x$ which gives the maximum value?

J. M. ain't a mathematician
  • 75,051
Victor
  • 8,372
  • 2
    Like every other maximum problem: differentiate, set the derivative to 0, solve for $x$, check whether any of the solutions happen to be maxima. – hmakholm left over Monica Mar 03 '12 at 21:32
  • 3
    It's a bit easier if you consider $\ln f(x)$ instead. – Harald Hanche-Olsen Mar 03 '12 at 21:34
  • @HaraldHanche-Olsen: How does taking the logarithm make it easier? – celtschk Jul 09 '12 at 10:40
  • @celtschk: Just take a look at J.D.'s answer. – Harald Hanche-Olsen Jul 10 '12 at 19:28
  • 2
    I'm looking. But I still don't see how it's easier than just deriving $x^{1/x}$ directly: $(x^{1/x})' = (1/x)x^{1/x-1} + x^{1/x}\ln x \cdot (-1/x^2) = x^{1/x-2}(1-\ln x)$ – celtschk Jul 10 '12 at 19:43
  • @HaraldHanche-Olsen: Sorry, forgot the at-addressing. – celtschk Jul 10 '12 at 20:00
  • @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). – Harald Hanche-Olsen Jul 11 '12 at 12:52

2 Answers2

24

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!

marty cohen
  • 107,799
15

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.

$$\text{(Hint: maximum occurs at x = e.)}$$

Soham
  • 9,990
  • What is y'/y? i don't get it – Victor Mar 03 '12 at 21:57
  • 2
    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 $\dfrac{d}{dx} y$. –  Mar 03 '12 at 22:00