Question

Consider the function f(x)= square root 5x-10 for the domain [2, +infinity). find f^-1(x), where f^-1 is the inverse of f. also state the domain of f^-1 in interval notation.edit: PLEASE DOUBLE CHECK ANSWERS.

Answer 1

`f^{\{-1\}}(x)\text{ = }(x^2+10)/(5)\text{for domain (-}\infty,\text{ }\infty)`Step-by-step explanation:
`\begin{gathered} f(x)\text{ = }\sqrt[]{5x\text{ - 10}} \\ \text{Domain = \lbrack{}2, }\infty) \end{gathered}`

let f(x) = y

To find the inverse of f(x), we would interchange x and y:

`\begin{gathered} y\text{ = }\sqrt[]{5x\text{ - 10}} \\ \text{Interchanging:} \\ x\text{ = }\sqrt[]{5y\text{ - 10}} \end{gathered}`

Then we would make the subject of formula:

`\begin{gathered} \text{square both sides:} \\ x^2\text{ = (}\sqrt[]{5y-10)^2} \\ x^2\text{ = 5y - 10} \end{gathered}`
`\begin{gathered} \text{Add 5 to both sides:} \\ x^2+10\text{ = 5y} \\ y\text{ = }(x^2+10)/(5) \\ \text{The result above is }f^{\mleft\{-1\mright\}}\mleft(x\mright) \end{gathered}`
`\begin{gathered} f^{\mleft\{-1\mright\}}\mleft(x\mright)\text{ = }(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}`

In interval notation:

`\begin{gathered} \text{Domain = (-}\infty,\text{ }\infty) \\ f^{\{-1\}}(x)\text{ = }(x^2+10)/(5)\text{for domain (-}\infty,\text{ }\infty) \end{gathered}`