Let $f:R\to R$ be a function defined by $f\left(x\right)=\frac{4^x}{4^x+2}$. what is the value of $f\left(\frac{1}{4}\right)+2f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)=$?

Given, $f\left(x\right)=\frac{4^x}{4^x+2}$

We will show that for the given function $f(1-x)=1-f\left(x\right)$

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

$=\frac{4·4^{-x}}{4·4^{-x}+2}$

$=\frac{4}{4+2·4^x}$

$=\frac{4^x+2-4^x}{2+4^x}$

$=\frac{4^x+2}{2+4^x}-\frac{4^x}{2+4^x}$

$=1-\frac{4^x}{2+4^x}$

$=1-f\left(x\right)$

$\Rightarrow f\left(\frac{1}{4}\right)=f\left(1-\frac{3}{4}\right)=1-f\left(\frac{3}{4}\right)$

Now, $f\left(\frac{1}{4}\right)+2f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)=1-f\left(\frac{3}{4}\right)+2f\left(\frac{1}{2}\right)+f\left(\frac{3}{4}\right)$

$=1+2f\left(\frac{1}{2}\right)$

$=1+2\left[\frac{4^{{\displaystyle \frac{1}{2}}}}{4^{{\displaystyle \frac{1}{2}}}+2}\right]$

$=1+2\times \frac{2}{4}$

$=2$