Question

f(x)=sqrt{x}-2. find the inverse of f(x) and its domain.
a. f^{-1}(x)=(x - 2)^2;xgeq2
b. f^{-1}(x)=(x - 2)^2;xgeq0
c. f^{-1}(x)=x^2-2;xgeq0
d. f^{-1}(x)=x^2-2;xgeq2

Explanation:

Step1: Let \(y = f(x)=\sqrt{x}-2\).

First, solve for \(x\) in terms of \(y\). Add 2 to both sides: \(y + 2=\sqrt{x}\).

Step2: Square both sides.

We get \((y + 2)^2=x\). Then, swap \(x\) and \(y\) to find the inverse function. So \(f^{-1}(x)=(x + 2)^2\).
For the domain of \(f(x)=\sqrt{x}-2\), the expression under the square - root must be non - negative, so \(x\geq0\). The range of \(f(x)\) is \(y\geq - 2\). For the inverse function \(f^{-1}(x)=(x + 2)^2\), the domain is the range of the original function, so \(x\geq - 2\).

Answer:

Assuming the options are meant to be:
A. \(f^{-1}(x)=(x - 2)^2,x\geq2\)
B. \(f^{-1}(x)=(x - 2)^2,x\geq0\)
C. \(f^{-1}(x)=x^2-2,x\geq0\)
D. \(f^{-1}(x)=x^2 - 2,x\geq2\)
None of the given options are correct. The correct inverse function is \(f^{-1}(x)=(x + 2)^2,x\geq - 2\)