Question

find the inverse $f^{-1}(x)$ of $f(x)$ restricted to the domain $x \geq 0$.
$f(x) = \frac{1}{\sqrt{x^2 + 5}}$
\bigcirc $f^{-1}(x) = \sqrt{\frac{5 - 5x^2}{x^2}}$
\bigcirc $f^{-1}(x) = \sqrt{\frac{1 - 5x^2}{x^2}}$
\bigcirc $f^{-1}(x) = -\sqrt{\frac{1 - 5x^2}{x^2}}$
\bigcirc $f^{-1}(x) = \sqrt{\frac{1 - x^2}{5x^2}}$

Explanation:

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

Let \( y = \frac{1}{\sqrt{x^2 + 5}} \), where \( x \geq 0 \).

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

We get \( x = \frac{1}{\sqrt{y^2 + 5}} \), with the domain of the inverse function corresponding to the range of \( f(x) \) for \( x \geq 0 \).

Step3: Solve for \( y \)

First, square both sides: \( x^2 = \frac{1}{y^2 + 5} \).
Then, take reciprocals: \( \frac{1}{x^2} = y^2 + 5 \).
Subtract 5: \( y^2 = \frac{1}{x^2} - 5=\frac{1 - 5x^2}{x^2} \).
Since the original function has \( x \geq 0 \), the inverse function (for the restricted domain) should have \( y \geq 0 \), so take the positive square root: \( y = \sqrt{\frac{1 - 5x^2}{x^2}} \) (we need to ensure the expression inside the square root is non - negative, i.e., \( 1-5x^{2}\geq0\) and \(x
eq0\), which is consistent with the domain and range considerations for the inverse function of \(f(x)\) with \(x\geq0\)).

Answer:

\(f^{-1}(x)=\sqrt{\frac{1 - 5x^2}{x^2}}\) (corresponding to the second option: \(f^{-1}(x)=\sqrt{\frac{1 - 5x^2}{x^2}}\))