Question

Which functions are the same as their inverse functions? multiple answers.

Answer 1

Given the following question:

`\begin{gathered} f(x)=(x-1)/(x+5),x\\e-5 \\ f(x)=(x-1)/(x+5),x-5=y=(x-1)/(x+5),x-5 \\ \begin{equation*} y=(x-1)/(x+5),x\\e-5 \end{equation*} \\ x=(y-1)/(y+5),y-5=xy+5x=y-1 \\ \begin{equation*} xy+5x=y-1 \end{equation*} \\ y-y=y*(1-x)=(5x+1)=y=(\left[5x+1\right])/(\left[1-x\right]) \\ f\mleft(x\mright)-1=(\left[5x+1\right])/(\left[1-x\right]) \end{gathered}`

First option isn't the same

`\begin{gathered} g(x)=(x-2)/(x-1),x\\e1 \\ y=(\left[x-2\right])/(\left[x-1\right]) \\ y=(\left[x-2\right])/(\left[x-1\right])=x=(\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=(\left[x-2\right])/(\left[x-1\right]) \\ g\mleft(x\mright)-1=(\left[x-2\right])/(\left[x-1\right]) \end{gathered}`

Second option is the same

`\begin{gathered} h\mleft(x\mright)=(\left[x+3\right])/(\left[x-2\right]) \\ y=(\left[x+3\right])/(\left[x-2\right]) \\ y=(\left[x+3\right])/(\left[x-2\right])=x=(\left[y+3\right])/(\left[y-2\right]) \\ x=(\left[y+3\right])/(\left[y-2\right]) \\ x*\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*\left[x-1\right]=\left[2x+3\right] \\ y=(\left[2x+3\right])/(\left[x-1\right]) \\ y=(\left[2x+3\right])/(\left[x-1\right])=h\mleft(x\mright)-1=(\left[2x+3\right])/(\left[x-1\right]) \\ h\mleft(x\mright)-1=(\left[2x+3\right])/(\left[x-1\right]) \end{gathered}`

Third option isn't the same

`\begin{gathered} k\left(x\right)=(\left[x+1\right])/(\left[x-1\right]) \\ y=(\left[x+1\right])/(\left[x-1\right]) \\ y=(\left[x+1\right])/(\left[x-1\right])=x=(\left[y+1\right])/(\left[y-1\right]) \\ x=(\left[y+1\right])/(\left[y-1\right]) \\ x*\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*\left[x-1\right]=\left[x+1\right] \\ y=(\left[x+1\right])/(\left[x-1\right]) \\ k\left(x\right)-1=(\left[x+1\right])/(\left[x-1\right]) \end{gathered}`

Fourth option is the same.