Question

13
select the correct answer.
which function is the inverse of $f(x) = x^2 - 16$ if the domain of $f(x)$ is $x \geq 0$?
\bigcirc a. $f^{-1}(x) = \sqrt{x + 16}$
\bigcirc b. $f^{-1}(x) = \sqrt{x} + 4$
\bigcirc c. $f^{-1}(x) = \sqrt{x - 16}$
\bigcirc d. $f^{-1}(x) = \sqrt{x} - 4$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x) = x^2 - 16 \). Replace \( f(x) \) with \( y \), so we have \( y = x^2 - 16 \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = y^2 - 16 \).

Step3: Solve for \( y \)

Now, we solve the equation \( x = y^2 - 16 \) for \( y \). First, add 16 to both sides: \( x + 16 = y^2 \). Then, take the square root of both sides. Since the original domain of \( f(x) \) is \( x \geq 0 \), the range of \( f(x) \) will be \( y \geq - 16 \), and for the inverse function, the domain of the inverse will correspond to the range of the original function, and the range of the inverse will correspond to the domain of the original function ( \( y \geq 0 \) ). So we take the positive square root: \( y=\sqrt{x + 16} \). We can write this as \( f^{-1}(x)=\sqrt{x + 16} \).

Answer:

A. \( f^{-1}(x)=\sqrt{x + 16} \)