High School How Can We Prove That x^x Reaches a Minimum at x = e^-1?

[cmb]

I've noticed that x^x is a minimum for x = e^-1

I put it as a high school problem because I presume it's one of those simple differential proofs/identities, but I can't really see how to get to e^-1. Too long since I did any calculus. Can someone please show me how to arrive at that?

How about (x^x)^x is a minimum for x = e^-(1/2)?

and ((x^x)^x)^x is a minimum for x = e^-(1/3)

I presume the pattern goes on.

 

[fresh_42]

I've noticed that x^x is a minimum for x = e^-1

I put it as a high school problem because I presume it's one of those simple differential proofs/identities, but I can't really see how to get to e^-1. Too long since I did any calculus. Can someone please show me how to arrive at that?

How about (x^x)^x is a minimum for x = e^-(1/2)?

and ((x^x)^x)^x is a minimum for x = e^-(1/3)

I presume the pattern goes on.

How did you find those minima? And what do you remember about the properties of local extrema?

 

[cmb]

If you're asking me to differentiate and find the turning point, sure, I just don't know how to do that.

I did this with a calculator.

In the meantime, it is curious that;-

(x^x) is NOT a minimum for the lowest possible x
BUT
if y=(x^x) then; (y^y) IS a minimum for the lowest possible y

(i.e. (x^x)^(x^x) IS a minimum for the lowest possible (x^x) but NOT the lowest possible x)

Can calculus show that too? I'm just no good at calculus.

 

[PeroK]

If you're asking me to differentiate and find the turning point, sure, I just don't know how to do that.

I did this with a calculator.

In the meantime, it is curious that;-

(x^x) is NOT a minimum for the lowest possible x
BUT
if y=(x^x) then; (y^y) IS a minimum for the lowest possible y

(i.e. (x^x)^(x^x) IS a minimum for the lowest possible (x^x) but NOT the lowest possible x)

Can calculus show that too? I'm just no good at calculus.

You can find the method for differentiating $x^x$ on line. There are several videos showing how it's done.

The minimum is when $x = 1/e$.

If $y = x^x$ then the minimum of $y^y$ is when $y = 1/e$, hence $x^x = 1/e$.

This can be simplified to $x \ln x = -1$, which is a transcendental equation and can ot be solved numerically.

 

[PeroK]

I've noticed that x^x is a minimum for x = e^-1

I put it as a high school problem because I presume it's one of those simple differential proofs/identities, but I can't really see how to get to e^-1. Too long since I did any calculus. Can someone please show me how to arrive at that?

How about (x^x)^x is a minimum for x = e^-(1/2)?

and ((x^x)^x)^x is a minimum for x = e^-(1/3)

I presume the pattern goes on.

Yes, this is not too hard to prove using the same technique as minimizing $x^x$.