Question

select the correct answer. which function is the inverse of $f(x)=x^2 - 16$ if the domain of $f(x)$ is $x\geq0$? \\(\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 \( y \)

First, we add 16 to both sides of the equation to isolate the \( y^2 \) term. So, \( x + 16 = y^2 \). Then, since the domain of \( f(x) \) is \( x \geq 0 \), the range of \( f(x) \) (which becomes the domain of the inverse function) will be such that \( y \) (the original function's output) will be \( y \geq - 16 \), but when finding the inverse, we need to consider the domain of the original function to determine the correct inverse. Since the original function has \( x \geq 0 \), the inverse function should have a range of \( y \geq 0 \). Taking the square root of both sides, we get \( y=\sqrt{x + 16}\) (we take the positive square root because the original function's domain is \( x\geq0 \), so the inverse function should have a range of \( y\geq0 \)). So the inverse function \( f^{-1}(x)=\sqrt{x + 16}\).

Answer:

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