If $\;f(x)+\frac{1}{x}=f^{-1}(\frac{1}{f(x)})\;$ and $f^{-1}(\frac{1}{f(x)})>0$. Then $f(x)$ is?

If $$f'(x)>0\; \forall \;x\in \mathbb R^+\;$$ where $$f:\mathbb R^+\rightarrow \mathbb R\;$$ and $$\;f(x)+\dfrac{1}{x}=f^{-1}\bigg(\dfrac{1}{f(x)}\bigg)\;$$ and $$f^{-1}\bigg(\dfrac{1}{f(x)}\bigg)>0$$. Then $$f(x)$$ is?.

My Approach:

I replaced $$f(x)$$ by $$\dfrac{1}{f(x)}\implies x\rightarrow f^{-1}\bigg(\dfrac{1}{f(x)}\bigg)$$.

$$\implies \; \dfrac{1}{f(x)}+\dfrac{1}{f^{-1}\bigg(\dfrac{1}{f(x)}\bigg)}=f^{-1}(f(x))$$

$$\implies\;$$ $$\dfrac{1}{f(x)}+\dfrac{x}{xf(x)+1}=x$$

$$\implies\; \dfrac{2xf(x)+1}{f(x)(xf(x)+1)}=x\; \implies \{xf(x)\}^2-\{xf(x)\}-1=0$$

$$\implies\;$$ $$xf(x)=\dfrac{1-\sqrt{5}}{2}$$ and $$xf(x)=\dfrac{1+\sqrt{5}}{2}$$.

But answer given is $$f(x)=\dfrac{\sqrt{5}-1}{2x}$$.

My Doubts:

$$1.$$ What is wrong in my method?

$$2.$$ Did I make any mistake by putting $$f(x)\rightarrow \dfrac{1}{f(x)}?$$

• You've got your domains mixed up, if f is only defined on $\mathbb{R}^+$, how can $f' > 0$ on $\mathbb{R}$? Assuming the domain is $\mathbb{R}^+$, then to replace $f(x)$ by $\frac{1}{f(x)}$, you need to show the following property: $\forall x \exists t_x \: f(t_x) = \frac{1}{f(x)}$ - I'm not sure how to check this. Now as you have that $f' > 0$, you only know that $f$ is strictly increasing (monotone), you don't know if it is onto or not. Then this also means that $f^{-1} \circ f(x) = x$ iff $x \in \text{range}(f)$, so that step needs explanation too.
– Anon
Commented Jun 1, 2022 at 20:27
• Could be missing something, but by switching $f$ and $1/f$, you will also change the meaning of $f^{-1}$. Indeed, it will change to $x \mapsto f(1/x)$, which will mean that the right hand side shouldn’t change! Commented Jun 1, 2022 at 21:29
• @TheoBendit Then how can i solve this problem? Commented Jun 2, 2022 at 4:35

The given answer is wrong, as can be verified in WA. OP's results are correct (also in WA), but only the negative solution satisfies the monotonicity condition $$\,f'(x) \gt 0\,$$ on $$\,\mathbb R^+\,$$.

Below is an alternate proof with the steps and substitutions made more explicit.

$$f(x)+\dfrac{1}{x}=f^{-1}\left(\dfrac{1}{f(x)}\right)$$

Let $$\,y = f(x)\,$$, then $$\,x = f^{-1}(y) = h(y)\,$$ and the equation becomes $$\,y + \dfrac{1}{h(y)}=h\left(\dfrac{1}{y}\right)\,$$, or:

$$y \, h(y) + 1 = h(y)\,h\left(\dfrac{1}{y}\right) \tag{1}$$

Writing $$\,(1)\,$$ for $$\,y \mapsto \dfrac{1}{y}\,$$:

$$\dfrac{1}{y} \,h\left(\dfrac{1}{y}\right) + 1 = h\left(\dfrac{1}{y}\right)\,h(y) \;\;\iff\;\; h\left(\dfrac{1}{y}\right)\,\big(y\,h(y) - 1\big) = y \tag{2}$$

Multiplying $$\,(1) \times(2)\,$$ and canceling out $$\,h\left(\dfrac{1}{y}\right) \gt 0\,$$:

$$\require{cancel} \big(y \, h(y) + 1\big)\,\big(y \, h(y) - 1\big)\,\cancel{h\left(\dfrac{1}{y}\right)} = y\,h(y)\,\cancel{h\left(\dfrac{1}{y}\right)} \\ \iff\quad y^2\,h^2(y) - y\,h(y) - 1 = 0 \quad\iff\quad h(y) = \dfrac{1 \pm \sqrt{5}}{2y}$$

It is easily verified that $$\,h\big(h(y)\big) = y\,$$, so $$\,f(x)=f^{-1}(x)=h(x)=\dfrac{1 \pm \sqrt{5}}{2x}\,$$.

Both solutions satisfy the additional constraint $$\,f^{-1}\left(\dfrac{1}{f(x)}\right) \gt 0\,$$ for $$\,x \in \mathbb R^+\,$$, because $$\,f\,$$ is of the form $$\,f(x)=f^{-1}(x) = \dfrac{\lambda}{x}\,$$, and therefore $$\,f^{-1}\left(\dfrac{1}{f(x)}\right) = \dfrac{\lambda^2}{x} \gt 0\,$$ when $$\,x \gt 0\,$$.

Only the negative solution satisfies the condition $$\,f'(x) \gt 0\,$$, so in the end $$\,f(x) = \dfrac{1-\sqrt{5}}{2x}\,$$.

• But $f'(x) > 0 \; \forall \; x \in \mathbb{R}^+$ so shouldn't ${1 - \sqrt{5} \over 2x}$ be the only solution? Commented Jun 2, 2022 at 6:26
• @VictorSouza Thanks, I completely missed that condition somehow. I corrected the answer now.
– dxiv
Commented Jun 2, 2022 at 6:42