Question

Let f(x) be a function given by $f:[0,2]\to [\frac{1}{7},\frac{2}{7}]\cup [1,4)$ and satisfies 3x -f(x) +1 = 0 for $0\le x\le 1$ and x - 7 f(x) = 0 for $1\le x\le 2$, there the sum of solutions of the equation $f\left(x\right)=f^{-1}\left(x\right)$ is

Answer 1

$f\left(x\right)=\left\{\begin{array}{cc}3x+1,& 0\le x<1\\ \frac{x}{7},& 1\le x\le 2\end{array}\right\}{f}^{-1}\left(x\right)=\left\{\begin{array}{cc}7x& \frac{1}{7}\le x\le \frac{2}{7}\\ \frac{x-1}{3}& 1\le x<4\end{array}\right\}$
For solution $\begin{array}{cc}3x+1=7x& \frac{x}{7}=\frac{x-1}{3}\\ x=\frac{1}{4}& 3x=7x-7\\ & 4x=7\to x=\frac{7}{4}\end{array}$

So sum $=\frac{1}{4}+\frac{7}{4}=2$