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While looking at the previou questions, I came upon this If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$

Here, in the answer, it is,used that $f^{-1}(x)=1$ and thus, $x=f(1)$.

I could not understand that. please somebody help me understand that.

thanks

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    When we write $f^{-1}$, we mean a function with $f^{-1}(f(a)) = a$ for every $a$ and $f(f^{-1}(b)) = b$ for every $b$. So Since $f^{-1}(x)$ and $1$ are the same number, when we apply $f$ to both, we get the same thing, i.e., we get $f(f^{-1}(x)) = f(1)$. But by the second rule for inverses, applied to the case $b = x$, the left hand side is just $x$. – 2017-02-07
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    To understand the previous Question, one statement is "general", true for all values $x$, namely the claim about $2f(x) + 3f(\frac{1}{x})$, and the other statement holds only for the "specific" value of $x$ that we are asked to find ("find the value of $x$"), namely $f^{-1}(x)=1$. It would have been better to use a different symbol, e.g. $b$ rather than $x$, in the second statement and say "find $b$", as @JohnHughes suggests. – 2017-02-07

3 Answers 3

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Apply f on both sides, we have -

$f(f^{-1}(x)) = f(1)$

Then,

$x = f(1)$

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    I also like to try that question. – 2017-02-07
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    @ Kanwaljit Singh, which question are you talking about? – 2017-02-07
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    Link you sent to Arnaldo – 2017-02-07
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    @ Kanwaljit Singh, Then did you see that? – 2017-02-07
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If $f$ is a function and $f^{-1}$ it's inverse, it holds that $$f[f^{-1}(x)]=x$$ Thus if $f^{-1}(x)=1$ it follows that $f[f^{-1}(x)]=f(1)=x$

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The definition of inverse function:

$$f^{-1}(f(x))=f(f^{-1}(x))=x \quad\text{for all $x$ in the domain}$$

so, in your case,

$$f^{-1}(x)=1 \to f(f^{-1}(x))=f(1) \to x=f(1)$$

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    I have a question. Could you please help me? – 2017-02-07
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    @tRiGoNoMeTrY: of course, go ahead – 2017-02-07
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    @ Arnaldo, I am sending the link. math.stackexchange.com/questions/2132698/assign-to-each-side-b-of-a-convex-polygon-p-the-maximum-area-of-a-triangle – 2017-02-07
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    @tRiGoNoMeTrY: ok, I will try that one! – 2017-02-07
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    @ Arnaldo, Ok. Thank u sir... – 2017-02-07
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    @ Arnaldo, Could you please help me with this problem: math.stackexchange.com/questions/2135188/find-the-gcd-of .. – 2017-02-08
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    @Euler: Please, take a look there! – 2017-02-08