7
$\begingroup$

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

$\endgroup$
7
  • 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
  • 3
    $\begingroup$ It's a bit easier if you consider $\ln f(x)$ instead. $\endgroup$ – Harald Hanche-Olsen Mar 3 '12 at 21:34
  • $\begingroup$ @HaraldHanche-Olsen: How does taking the logarithm make it easier? $\endgroup$ – celtschk Jul 9 '12 at 10:40
  • $\begingroup$ @celtschk: Just take a look at J.D.'s answer. $\endgroup$ – Harald Hanche-Olsen Jul 10 '12 at 19:28
  • 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
14
$\begingroup$

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.

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

$\endgroup$
2
  • $\begingroup$ What is y'/y? i don't get it $\endgroup$ – Victor Mar 3 '12 at 21:57
  • 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
21
$\begingroup$

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!

$\endgroup$
2
  • 1
    $\begingroup$ +1 for presentation! I watch them magic tricks close though, I do. $\endgroup$ – Eugene Shvarts Jul 9 '12 at 8:28
  • $\begingroup$ I like this! It is so smooth just like exp itself. $\endgroup$ – Philimathmuse Apr 9 '18 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.