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

Question

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...

Answer 1

We are trying to find the following:

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

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

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

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

Answer 2

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 $f(x)=2x$.

Answer 3

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 $x=f(1)=2$.