Find the domain and range of `y=sqrt(x-2)+1` . And find the inverse equation and its domain and range.
(I) To determine the domain of a function that has a radical, consider the expression inside the square root which is x - 2.
Note that in square root, a negative value inside it is not allowed. So, get the domain, set x - 2 greater than or equal to 0.
`x - 2gt=2`
And, solve for x.
Hence, the domain of `y =sqrt(x-2)+1` is `[2,+oo)` .
(II) To determine the range, consider the smallest value of x which is 2. Plug-in this value to the function.
`y=sqrt(x-2)+1 = sqrt(2-2)+1=sqrt0+1=0+1=1`
Since the operation between the radical `sqrt(x-2)` and the constant 1 is addition, it indicates that y=1 is the smallest value of y.
Hence, the range of `y=sqrt(x-2)+1` is `[1,+oo)` .
(III) To determine the inverse function of `y=sqrt(x-2) + 1` , replace y with x and replace x with y.
Then, solve for y.
And, replace y with `f^(-1)(x)` to indicate that it is the inverse of the given function.
Hence, the inverse function is `f^(-1)(x)=(x-1)^2+2` .
(IV) To determine the domain and range of the inverse function, take note that its domain y-values of the original function. And its range is the x-values of the original function too.
Thus, domain of `f^(-1)(x) ` is `[1,+oo)` and its range is `[2,+oo)` .