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If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$.

My Attempt: $$2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$$ $$2f(x)+3f(\frac {1}{x})=4x + \frac {6}{x}$$

At this point, I couldn't get other idea except comparing the corresponding terms. please provide any other method...

3 Answers 3

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We are trying to find the following:

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

Plugging $x=1$ into the original functional equation...

$$2f(1)+3f(1)=\frac{4+6}1$$

$$5f(1)=10\implies x=f(1)=2$$

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    why is $x=f(1)$? How is it? – 2017-01-27
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    @occasionaluser what does it mean to say $f^{-1}(x)$? – 2017-01-27
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    It means replacing $y$ in the place of $x$? – 2017-01-27
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    @occasionaluser which is the same as getting $f(1)=x$ – 2017-01-27
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    and, why did you put $x=1$ in the given function? – 2017-01-27
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    @occasionaluser so that I could find $f(1)$ – 2017-01-27
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    Can we write like this, $$f^{-1}(x)=1$$ $$\frac {1}{f} (x)=1$$ $$x=f(1)$$?? – 2017-01-27
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    @occasionaluser no, but like this:$$f^{-1}(x)=1\\f(f^{-1}(x))=f(1)\\x=f(1)$$ – 2017-01-27
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You can solve the following system.

$2f(x)+3f\left(\frac{1}{x}\right)=\frac{4x^2+6}{x}$ and $2f\left(\frac{1}{x}\right)+3f(x)=\frac{\frac{4}{x^2}+6}{\frac{1}{x}}$,

which gives $f(x)=2x$.

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Replace $x$ by $\frac1x$ in the original identity and you get another identity: $$ \textstyle 2f(\frac1x) + 2f(x) = \frac4x + 6x $$ Now you have two equations for two unknowns -- namely, $f(x)$ and $f(\frac1x)$. Solve these and obtain $$ f(x)=2x. $$ Finally, if $f^{-1}(x)=1$, then $x=f(1)=2$.

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    what is the main objective of replacing $x$ by $\frac {1}{x}$??? – 2017-01-27
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    This is one way to get an explicit form for $f(x)$. The original equation involves both $f(x)$ and $f(\frac1x)$, so this new equation allows you to eliminate $f(\frac1x)$. – 2017-01-27