Determine whether $x^x$ function is injective or surjective $?$

Question

Let $f$ be a functions from $\mathbb{R^+}$ to $\mathbb{R^+}$ defined by $f(x)=x^x$

Determine whether the function $f$ injective or surjective ?

I saw here similar questions regarding this function but this differ from other problem, I will tell you what I tried

Attempt:

$\underline{Surjective}$

for $f(x) \in \mathbb{R^+}, $ no element $x \in \mathbb{R^+} $ such that $f(x)=0 \Rightarrow x^x=0$

Thus $f$ is not onto

I have no idea how to prove it or disprove it as a surjective, Actually this problem is very strange to me, I referred this problem How can we describe the graph of $x^x$ for negative values? then I thought my problem would be wrong, Any help would be greatly appreciated regarding this problem.

@lulu I tried to use it but I saw somewhere $f(0)=0^0$ is undefined — DARK, Aug 08 '20 at 17:25
Ok, so then argue that $x^x$ approaches $1$ from below if $x$ approaches $0$ from the right or if $x$ approaches $1$ from the left. — lulu, Aug 08 '20 at 17:27
Take the derivative, show the function has a local min (that's the approach the posted solutions take). — lulu, Aug 08 '20 at 17:30
It cannot be surjective because $f (x)\gt 0$ and it is not injective because the derivative vanishes at a point $x_0$ between $0$ and $1$ then $f$ has an extreme at $x_0$ (it is not hard to verify this and that in effect a minimum). — Piquito, Aug 08 '20 at 17:33

score 4 · Answer 1 · answered Aug 08 '20 at 18:28

4

One can see that it's not injective since $f(1/2)=(1/2)^{1/2}=1/\sqrt{2}=(1/4)^{1/4}=f(1/4)$

answered Aug 08 '20 at 18:28

jjagmath

18,214

score 3 · Accepted Answer · edited Aug 08 '20 at 17:44

3

$$(x^x)'=\left(e^{x\ln{x}}\right)'=x^x(\ln{x}+1),$$ which says $f$ is injective on $[\frac{1}{e},+\infty)$ and on $(0,\frac{1}{e}]$.

By the way, $f$ is a surjection $$f:(0,+\infty)\rightarrow\left[\frac{1}{e^{\frac{1}{e}}},+\infty\right)$$

edited Aug 08 '20 at 17:44

Bernard

175,478

answered Aug 08 '20 at 17:27

Michael Rozenberg

194,933

can you please explain this? – DARK Aug 08 '20 at 17:41
@Akalanka Which step? – Michael Rozenberg Aug 08 '20 at 17:45
I am not familiar with this method can you please tell what is the method that you used? – DARK Aug 08 '20 at 18:00
1

@Akalanka I find where $f$ increase and where $f$ decreases, which gave the minimum point: $\left(\frac{1}{e},\frac{1}{e^{\frac{1}{e}}}\right)$. All these gives when $f$ is injective. Also, we see when $f$ is a surjection. – Michael Rozenberg Aug 08 '20 at 18:04
but there is a comment $f(x)$ is surjective – DARK Aug 08 '20 at 18:07
@Akalanka See a last part of my post. – Michael Rozenberg Aug 08 '20 at 18:08
1

got it. I will try to undestand this. Thank you – DARK Aug 08 '20 at 18:10

score 0 · Answer 3 · answered Aug 08 '20 at 17:24

0

$f'(x) = x^x (\ln(x)+1)$. So attains a maximum or minimum at $e^{-1}$. Then you have equal values of $f$ for some 2 numbers $x,y$ where $x<e^{-1}$ and $y>e^{-1}$. This means $f$ is not 1-1 on the positive reals.

answered Aug 08 '20 at 17:24

Jared S.

98

Determine whether $x^x$ function is injective or surjective $?$

3 Answers3