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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.

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  • $\begingroup$ For injectivity...what's $f(0)$? what's $f(1)$? $\endgroup$ Commented Aug 8, 2020 at 17:21
  • $\begingroup$ is $f(0)=0^0$ undefined? $\endgroup$ Commented Aug 8, 2020 at 17:23
  • $\begingroup$ @lulu I tried to use it but I saw somewhere $f(0)=0^0$ is undefined $\endgroup$ Commented Aug 8, 2020 at 17:25
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    $\begingroup$ 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. $\endgroup$ Commented Aug 8, 2020 at 17:27
  • $\begingroup$ @lulu got it. is there any easy method? $\endgroup$ Commented Aug 8, 2020 at 17:29

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

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

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

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  • $\begingroup$ can you please explain this? $\endgroup$ Commented Aug 8, 2020 at 17:41
  • $\begingroup$ @Akalanka Which step? $\endgroup$ Commented Aug 8, 2020 at 17:45
  • $\begingroup$ I am not familiar with this method can you please tell what is the method that you used? $\endgroup$ Commented Aug 8, 2020 at 18:00
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    $\begingroup$ @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. $\endgroup$ Commented Aug 8, 2020 at 18:04
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    $\begingroup$ got it. I will try to undestand this. Thank you $\endgroup$ Commented Aug 8, 2020 at 18:10
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$f'(x) = x^x (\ln(x)+1)$. 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.

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