Question

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 \( f(x) = x^2 - 16 \), so we write \( y = x^2 - 16 \).

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

To find the inverse, we swap \( x \) and \( y \), getting \( x = y^2 - 16 \).

Step3: Solve for \( y \)

First, we add 16 to both sides of the equation: \( x + 16 = y^2 \).
Then, since the domain of \( f(x) \) is \( x \geq 0 \), the range of \( f(x) \) (and thus the domain of \( f^{-1}(x) \)) will result in \( y \) being non - negative when we solve for \( y \). So we take the square root of both sides: \( y=\sqrt{x + 16}\) (we take the positive square root because the original function \( f(x)=x^2 - 16\) with \( x\geq0\) is one - to - one and the inverse should also be a function, so we take the non - negative root).
So the inverse function \( f^{-1}(x)=\sqrt{x + 16}\)

Answer:

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