Question

If $f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right)f\left(\frac{1}{x}\right)$ and $f\left(3\right)=28$ find $f\left(\frac{1}{3}\right)$

Answer 1

$f\left(x\right)+f\left(\frac{1}{x}\right)=f\left(x\right)f\left(\frac{1}{x}\right)$

$\Rightarrow f\left(3\right)+f\left(\frac{1}{3}\right)=f\left(3\right)f\left(\frac{1}{3}\right)$

$\Rightarrow 28+f\left(\frac{1}{3}\right)=28f\left(\frac{1}{3}\right)$

$\Rightarrow f\left(\frac{1}{3}\right)=28f\left(\frac{1}{3}\right)-28$

$\Rightarrow f\left(\frac{1}{3}\right)=28(f\left(\frac{1}{3}\right)-1)$

Let $t=f\left(\frac{1}{3}\right)$

$\Rightarrow t=28(t-1)$

$\Rightarrow t=28t-28$

$\Rightarrow 28t-t=28$

$\Rightarrow 27t=28$

$\Rightarrow t=\frac{28}{27}$

$\therefore f\left(\frac{1}{3}\right)=\frac{28}{27}$