For the following function defined that is one-to-one, write an equation for the inverse function in the form $y=f^{-1}(x)$, and then graph $f$ and $f^{-1}$ on the same axes. Give the domain and range of $f$ and $f^{-1}$. If the function is not one-to-one, say so.
\[
f(x)=\sqrt{-4+x}, x \geq 4
\]
Find the equation of the inverse function. Select the correct choice below, and fill in the answer box if necessary.
A. $f^{-1}(x)=\square, x \leq 0$
B. $f^{-1}(x)=\square, x \geq 0$
C. The function is not one-to-one.
Answer
Thus, the correct choice is B. \(f^{-1}(x)=x^2+4, x \geq 0\).
Step 1 ：First, we need to determine if the function is one-to-one. The function \(f(x)=\sqrt{-4+x}\) is decreasing for \(x \geq 4\), so it is one-to-one.
Step 2 ：To find the inverse function, we first replace \(f(x)\) with \(y\), so we get \(y=\sqrt{-4+x}\).
Step 3 ：Next, we swap \(x\) and \(y\), so we get \(x=\sqrt{-4+y}\).
Step 4 ：Squaring both sides, we get \(x^2=-4+y\).
Step 5 ：Solving for \(y\), we get \(y=x^2+4\).
Step 6 ：So, the inverse function is \(f^{-1}(x)=x^2+4\), and its domain is \(x \geq 0\).
Step 7 ：Thus, the correct choice is B. \(f^{-1}(x)=x^2+4, x \geq 0\).
