Respuesta :
The functions given are
[tex]\begin{gathered} f(x)=7x+1 \\ g(x)=\frac{x-1}{7} \end{gathered}[/tex]We are asked to prove that
[tex]\begin{gathered} f^{-1}(x)=g(x) \\ g^{-1}(x)=f(x) \end{gathered}[/tex]Let
[tex]\begin{gathered} f(x)=y=7x+1 \\ y=7x+1 \end{gathered}[/tex]Concept: To get the inverse of a function, we will make x the subject of the formula and then replace the value of y with x
[tex]\begin{gathered} y=7x+1 \\ \text{substract 1 from both sides} \\ y-1=7x+1-1 \\ y-1=7x \end{gathered}[/tex]Divide both sides by 7
[tex]\begin{gathered} 7x=y-1 \\ \frac{7x}{7}=\frac{y-1}{7} \\ x=\frac{y-1}{7} \\ \text{replace the value of y with x} \\ f^{-1}(x)=\frac{x-1}{7}=g(x)(\text{PROVED)} \end{gathered}[/tex]Alternatively,
let
[tex]\begin{gathered} g(x)=y=\frac{x-1}{7} \\ y=\frac{x-1}{7} \end{gathered}[/tex]To get the inverse of a function, we will make x the subject of the formula and then replace the value of y with x
[tex]\begin{gathered} y=\frac{x-1}{7} \\ \text{Cross multiply} \\ 7y=x-1 \\ \text{add 1 to both sides} \\ 7y+1=x-1+1 \\ 7y+1=x \\ x=7y+1 \\ \text{replace y with x therefore, we will have} \\ g^{-1}(x)=7x+1=f(x)(\text{PROVED)} \end{gathered}[/tex]