John Landry

Answered

2022-07-19

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

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

My Attempt:

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

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

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

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

My Attempt:

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

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

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

Answer & Explanation

dasse9

Expert

2022-07-20Added 12 answers

We are trying to find the following:

${f}^{-1}(x)=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=f(1)$

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

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

$5f(1)=10\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=f(1)=2$

${f}^{-1}(x)=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=f(1)$

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

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

$5f(1)=10\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=f(1)=2$

Brenton Dixon

Expert

2022-07-21Added 5 answers

You can solve the following system.

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

which gives $f(x)=2x$

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