Question

Please help
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Answer 1

first option

Explanation:

To find the inverse
`f^(-1)` (x)

let f(x) = y and rearrange , making x the subject

given

f(x) =
`√(x-2)` + 5 , then

y =
`√(x-2)` + 5 ( subtract 5 from both sides )

y - 5 =
`√(x-2)` ( square both sides to clear the radical )

(y - 5)² = (
`√(x-2)` )² , that is

(y - 5)² = x - 2 ( add 2 to both sides )

(y - 5)² + 2 = x

change y back into terms of x, with x being
`f^(-1)` (x) , then

`f^(-1)` (x) = (x - 5)² + 2

Answer 2

`\textsf{A)} \quad f^(-1)(x)=(x-5)^2+2,\qquad x\geq 5`

Explanation:

The inverse of a function is the reflection of the original function across the line y = x where:

• The domain of the original function corresponds to the range of the inverse function.
• The range of the original function corresponds to the domain of the inverse function.

As the domain of given function f(x) is restricted to x ≥ 2, its range is restricted to f(x) ≥ 5. Therefore, the domain of the inverse function f⁻¹(x) is restricted to x ≥ 5.

To find the inverse of function f(x), begin by interchanging the x and y values:

`x=√(y-2)+5`

Solve the equation for y:

`\begin{aligned}x-5&=√(y-2)+5-5\\\\x-5&=√(y-2)\\\\(x-5)^2&=\left(√(y-2)\right)^2\\\\(x-5)^2&=y-2\\\\(x-5)^2+2&=y-2+2\\\\y&=(x-5)^2+2\end{aligned}`

Swap the y for f⁻¹(x):

`f^(-1)(x)=(x-5)^2+2`

Include the restricted domain:

`f^(-1)(x)=(x-5)^2+2,\qquad x\geq 5`