Given the following question:
[tex]\begin{gathered} f(x)=\frac{x-1}{x+5},x\ne-5 \\ f(x)=\frac{x-1}{x+5},x-5=y=\frac{x-1}{x+5},x-5 \\ \begin{equation*} y=\frac{x-1}{x+5},x\ne-5 \end{equation*} \\ x=\frac{y-1}{y+5},y-5=xy+5x=y-1 \\ \begin{equation*} xy+5x=y-1 \end{equation*} \\ y-y=y\times(1-x)=(5x+1)=y=\frac{\left[5x+1\right]}{\left[1-x\right]} \\ f\mleft(x\mright)-1=\frac{\left[5x+1\right]}{\left[1-x\right]} \end{gathered}[/tex]First option isn't the same
[tex]\begin{gathered} g(x)=\frac{x-2}{x-1},x\ne1 \\ y=\frac{\left[x-2\right]}{\left[x-1\right]} \\ y=\frac{\left[x-2\right]}{\left[x-1\right]}=x=\frac{\left[y-2\right]}{\left[y-1\right]}=xy-x=y-2 \\ xy-x=y-2 \\ xy-y=-2+x \\ y*\left[x-1\right]=\left[x-2\right]=y=\frac{\left[x-2\right]}{\left[x-1\right]} \\ g\mleft(x\mright)-1=\frac{\left[x-2\right]}{\left[x-1\right]} \end{gathered}[/tex]Second option is the same
[tex]\begin{gathered} h\mleft(x\mright)=\frac{\left[x+3\right]}{\left[x-2\right]} \\ y=\frac{\left[x+3\right]}{\left[x-2\right]} \\ y=\frac{\left[x+3\right]}{\left[x-2\right]}=x=\frac{\left[y+3\right]}{\left[y-2\right]} \\ x=\frac{\left[y+3\right]}{\left[y-2\right]} \\ x\times\left[y-2\right]=\left[y+3\right]=xy-2x=y+3 \\ xy-2x=y+3 \\ xy-2x=y+3-=xy-y=3+2x \\ xy-y=3+2x \\ y\times\left[x-1\right]=\left[2x+3\right] \\ y=\frac{\left[2x+3\right]}{\left[x-1\right]} \\ y=\frac{\left[2x+3\right]}{\left[x-1\right]}=h\mleft(x\mright)-1=\frac{\left[2x+3\right]}{\left[x-1\right]} \\ h\mleft(x\mright)-1=\frac{\left[2x+3\right]}{\left[x-1\right]} \end{gathered}[/tex]Third option isn't the same
[tex]\begin{gathered} k\left(x\right)=\frac{\left[x+1\right]}{\left[x-1\right]} \\ y=\frac{\left[x+1\right]}{\left[x-1\right]} \\ y=\frac{\left[x+1\right]}{\left[x-1\right]}=x=\frac{\left[y+1\right]}{\left[y-1\right]} \\ x=\frac{\left[y+1\right]}{\left[y-1\right]} \\ x\times\left[y-1\right]=\left[y+1\right] \\ xy-x=y+1 \\ xy-x=y+1=xy-y=x+1 \\ xy-y=x+1=y\times\left[x-1\right]=\left[x+1\right] \\ y=\frac{\left[x+1\right]}{\left[x-1\right]} \\ k\left(x\right)-1=\frac{\left[x+1\right]}{\left[x-1\right]} \end{gathered}[/tex]Fourth option is the same.
