Study of the function $f(x)=x^{\frac{1}{x}}$ with $x\geq 1$ without derivative and using of Lambert's function

Question

Well we want to study the function : $$f(x)=x^{\frac{1}{x}}$$ with $x\geq 1$

Using the definition of an increasing/decreasing function we have :

$$x^{\frac{1}{x}}\leq^{?} y^{\frac{1}{y}}$$

Or $$\Big(\frac{1}{x}\Big)^{\frac{1}{x}}\geq^{?} \Big(\frac{1}{y}\Big)^{\frac{1}{y}}$$

Putting $a=\frac{1}{x}$ and $b=\frac{1}{y}$ we get :

$$a^a\geq^{?}b^b$$

Or $$a\ln(a)\geq^{?}b\ln(b)$$

Now we apply the Lambert's function :

$$\operatorname{W}(a\ln(a))\geq^{?}\operatorname{W}(b\ln(b))$$

Or :$$\ln(a)\geq^{?}\ln(b)$$

As the Lambert's function have two branchs the inequality can be reversed at the point $x=\frac{1}{e}$ . We deduce that :$$g(x)=\Big(\frac{1}{x}\Big)^{\frac{1}{x}}$$ have an extrema at $x=e$

So we deduce that $f(x)$ have an extrema at $x=e$

My question :

Is it right ?

The use of the Lambert's function need the use of derivatives ?

Have you some advices for me ?

Thanks a lot for all your contributions.

Ps:if it's a duplicate feel free to correct me.

Answer 1

It's easier to prove that $x = e$ is an extreme of that function using derivatives:

Let's use logarithmic differentiation:

$$\begin{align} &f = x^{\frac{1}{x}} \\ \\ &\ln f= \ln (x^{\frac{1}{x}}) \\ \\ &\frac{d}{dx} \ln f = \frac{d}{dx} \frac{\ln x}{x} \\ \\ &\frac{f'}{f}= \frac{1 - \ln x}{x^2} \\ \\ &f' = f \frac{1 - \ln x}{x^2} \\ \\ &f' = x^{\frac{1}{x}}\frac{1 - \ln x}{x^2} \end{align}$$

We can further simplify this to:

$$f' = x^{\frac{1}{x} - 2}({1 - \ln x})$$

Now we just simply solve the equation:

$$\begin{align} &x^{\frac{1}{x} - 2}({1 - \ln x}) = 0 \\ \\ & x^{\frac{1}{x} - 2} = 0 \vee ({1 - \ln x}) = 0 \end{align}$$

because $x \geq 1$, then $x^{\frac{1}{x} - 2}$ is never 0. This leaves us with:

$$\begin{align} &{1 - \ln x} = 0 \\ \\ & x = e \end{align}$$

Here's a few other things that you can say about the function you want to study:

• $f$ is a continuous function because it is the composition of continuous function

• $f$ has an horizontal asymptote with equation: $y = 1$

• $f$ is differentiable in it's domain because it's the composition of differentiable functions

• $\max \{f(x), x \geq 1\} = e$