<s>[INST] If $2f(x)+3f(\frac {1}{x})=\frac {4x^2+6}{x}$ and $f^{-1}(x)=1$ then find the value of $x$ 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... 
 [/INST] 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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