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How to find the "dip point" in the graph of $y=x^{x}$ graph?

I was exploring various graphs on Desmos, and I stumbled upon the graph of $x^x$.enter image description here

Now I tried putting various values to examine the nature of this graph between $0$ and $1$ but how do I find that turning point of this graph where it starts rising up? Is there any elementary method to find it? Thank you

轻型八神
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Derivative of $x^x$ is $x^x(1+\ln(x))$. Solving $1+\ln(x) = 0$ gives us $x = e^{-1}$, which is the point you are looking for. Of course having derivative equal to $0$ isn't a sufficient condition, but you can easily check that in fact there is a local minimum there.

mowzorn
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  • isn't a sufficient condition --- Actually it's immediate from the first derivative test, which is almost trivial to apply in this case. Since the $x^x$ factor is positive for all $x >0,$ it can be ignored. Now note that $1 + \ln x$ is negative when $x$ is less than $e^{-1}$ and positive when $x$ is greater than $e^{-1}.$ – Dave L. Renfro Dec 27 '22 at 20:57
  • I agree. I just assumed that the author of the question might not be very familiar with calculus, and I added this part so that they don't think that in general having derivative equal to $0$ means that there is a local extremum. So when I wrote 'you can easily check that...', I meant the first derivative test which you described. – mowzorn Dec 27 '22 at 21:05
  • OK, I thought that might be the case, but I wasn't sure. I'm used to people jumping right to the second derivative test for verification of max/min (which you didn't do, but I thought this is what you might have been implying) when the much more informative first derivative test (especially for non-local behavior, needed for example when trying to come up with a rough graph) is often no more difficult, and sometimes even easier (when the second derivative gets a bit messy to carry out). Oh, and regarding my comment to Bob Dobbs, I gave you an upvote when I wrote my earlier comment. – Dave L. Renfro Dec 28 '22 at 09:49
  • New contributors deserve upvote when they are hungry for math. I upvothed "this thread" too. – Bob Dobbs Jan 01 '23 at 12:58
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Since $y=x^x=e^{x\ln x}$, it is enough to find the minumum of $y=x\ln x$, $0<x<1$. Now, let $x=e^{-z}$, $0<z<\infty$. Then, the problem is finding the minumum of $y=-ze^{-z}$ and thus finding the maximum of $y=\frac{z}{e^z}$.

We claim that maximum of $y=\frac{z}{e^z}$ occurs at $z=1$ and it is $\frac{1}{e}$. If we increase $z=1$ to $z=1+\epsilon$, $\epsilon>0$, the the inequality $\frac{1+\epsilon}{e^{1+\epsilon}}<\frac{1}{e}$ gives $1+\epsilon<e^\epsilon$ which is true. Similarly for the decrease from $z=1$ to $z=1-\epsilon$. Since, this also gives a similar thing.

Hence, the maximum of $y=x^x$ occurs at $x=e^{-z}=e^{-1}=\frac{1}{e}$ and it is $\frac{1}{\sqrt[e]{e}}$.

Bob Dobbs
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    (+1) This seems to be OK, although I haven't super-carefully checked all the details (just did a quick read through). However, it's worth pointing out to others that some details might be needed, depending on one's "mathematical maturity" level. For example, the part "gives $1+\epsilon<e^\epsilon$ which is true" is technically the wrong implication direction, but can be fixed (in the reader's mind, if appropriately "mathematically mature") since the inequality operation performed is reversible. (continued) – Dave L. Renfro Dec 28 '22 at 09:41
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    And then there's the issue of why $\epsilon > 0$ implies $1+\epsilon<e^\epsilon,$ which depends on one's definition of the exponential function for why this holds. And while I'm here, if anyone is interested in slightly less trivial max/min problems using the first derivative test (slightly less trivial means by U.S. introductory calculus standards) and done entirely without the aid of calculators, see my 29 October 2006 sci.math post Sign charts and the first derivative test. – Dave L. Renfro Dec 28 '22 at 09:42
  • By the way, to see the sign charts properly in that sci.math post, you'll have to copy the post somewhere that allows you to use fixed width font. Google used to use this for their sci.math archive (and formerly the standard for all USENET groups), but at some later time someone at google -- who apparently didn't know better and presumably thought they were adding value to the archived posts when in fact they were massively subtracting value in many cases -- retroactively changed everything in google's archive to a variable font width. (The same thing happened at Math Forum a little earlier.) – Dave L. Renfro Dec 28 '22 at 09:55
  • @DaveL.Renfro Will these post take likes? They are so ancient. You know clicks and likes are everything in modern world. – Bob Dobbs Dec 28 '22 at 12:08
  • @DaveL.Renfro Yes epsilon part is "I don't wanna talk about it anymore." part of the answer. Together with delta they are worse. – Bob Dobbs Dec 28 '22 at 12:09
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    The reason I explicitly said I upvoted is that I didn't want you to think that I didn't because of the comments I made. Then I realized that I should also mention this for the other answerer, since mentioning it [= an upvote] here and not mentioning it there might lead one to suspect I didn't upvote there, when in fact I had actually already upvoted there. Ordinarily I only mention upvoting when I think something is especially good, but in this case I wanted to avoid the perception that I didn't upvote. – Dave L. Renfro Dec 28 '22 at 12:25
  • @DaveL.Renfro If it is mathematecillay OK no problem. Upvoting is nice joke. Down voting is bad joke. That's all. – Bob Dobbs Dec 28 '22 at 12:36