 John Landry

2022-07-19

If $2f\left(x\right)+3f\left(\frac{1}{x}\right)=\frac{4{x}^{2}+6}{x}$ and ${f}^{-1}\left(x\right)=1$ then find the value of $x$
If $2f\left(x\right)+3f\left(\frac{1}{x}\right)=\frac{4{x}^{2}+6}{x}$ and ${f}^{-1}\left(x\right)=1$ then find the value of $x$
My Attempt:
$2f\left(x\right)+3f\left(\frac{1}{x}\right)=\frac{4{x}^{2}+6}{x}$
$2f\left(x\right)+3f\left(\frac{1}{x}\right)=4x+\frac{6}{x}$
At this point, I couldn't get other idea except comparing the corresponding terms. please provide any other method... dasse9

Expert

We are trying to find the following:
${f}^{-1}\left(x\right)=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=f\left(1\right)$
Plugging $x=1$ into the original functional equation...
$2f\left(1\right)+3f\left(1\right)=\frac{4+6}{1}$
$5f\left(1\right)=10\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=f\left(1\right)=2$ Brenton Dixon

Expert

You can solve the following system.
$2f\left(x\right)+3f\left(\frac{1}{x}\right)=\frac{4{x}^{2}+6}{x}$ and $2f\left(\frac{1}{x}\right)+3f\left(x\right)=\frac{\frac{4}{{x}^{2}}+6}{\frac{1}{x}}$
which gives $f\left(x\right)=2x$

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