Find the domain and range of `y=sqrt(x-2)+1` . And find the inverse equation and its domain and range.

Expert Answers

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(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)` .

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