Найти обратную функцию: f(x)= 4 - 3*2^( (1+x)/(2-x) )

[latex]y= 4 - 3*2^{\frac{ 1+x}{2-x} } \\ \\ x= 4 - 3*2^{\frac{ 1+y}{2-y} } \\ \\ 3*2^{\frac{ 1+y}{2-y} } =4-x \\ \\2^{\frac{ 1+y}{2-y} } =\frac{4-x}{3} \\ \\ log_2(2^{\frac{ 1+y}{2-y} } )=log_2(\frac{4-x}{3}) \\ \\ \frac{ 1+y}{2-y} log_22=log_2(\frac{4-x}{3}) \\ \\ \frac{ 1+y}{2-y} =log_2(\frac{4-x}{3}) \\ \\ 1+y=(2-y)log_2(\frac{4-x}{3}) \\ \\ 1+y=2log_2(\frac{4-x}{3}) -ylog_2(\frac{4-x}{3}) \\ \\ y+ylog_2(\frac{4-x}{3}) =2log_2(\frac{4-x}{3}) -1 \\ \\ [/latex] [latex]y(1+log_2(\frac{4-x}{3}) )=2log_2(\frac{4-x}{3}) -1 \\ \\ y= \frac{2log_2(\frac{4-x}{3}) -1}{1-log_2(\frac{4-x}{3})} =\frac{log_2(\frac{4-x}{3})^2 -log_22}{log_22+log_2(\frac{4-x}{3})} = \frac{log_2 \frac{(4-x)^2}{9*2} }{log_2(2* \frac{4-x}{3}) } = \frac{log_2 \frac{(4-x)^2}{18} }{log_2( \frac{8-2x}{3}) } = \\ \\ log_{ \frac{8-2x}{3}} \frac{16-8x+x^2}{18} \\ \\ \\ f^{-1}(x)=log_{ \frac{8-2x}{3}} \frac{x^2-8x+16}{18} [/latex]