let f(x) = y
To find the inverse of f(x), we would interchange x and y:
[tex]\begin{gathered} y\text{ = }\sqrt[]{5x\text{ - 10}} \\ \text{Interchanging:} \\ x\text{ = }\sqrt[]{5y\text{ - 10}} \end{gathered}[/tex]Then we would make the subject of formula:
[tex]\begin{gathered} \text{square both sides:} \\ x^2\text{ = (}\sqrt[]{5y-10)^2} \\ x^2\text{ = 5y - 10} \end{gathered}[/tex][tex]\begin{gathered} \text{Add 5 to both sides:} \\ x^2+10\text{ = 5y} \\ y\text{ = }\frac{x^2+10}{5} \\ \text{The result above is }f^{\mleft\{-1\mright\}}\mleft(x\mright) \end{gathered}[/tex][tex]\begin{gathered} f^{\mleft\{-1\mright\}}\mleft(x\mright)\text{ = }\frac{x^2+10}{5} \\ The\text{ domain of the inverse is all real numbers} \\ \text{That is from negative infinity to positive infinity} \end{gathered}[/tex]In interval notation:
[tex]\begin{gathered} \text{Domain = (-}\infty,\text{ }\infty) \\ f^{\{-1\}}(x)\text{ = }\frac{x^2+10}{5}\text{for domain (-}\infty,\text{ }\infty) \end{gathered}[/tex]