The given problem asks us to determine a domain for the function $f(x) = \frac{1}{\sqrt{x^2 + 5}}$ on which it is one-to-one, and then find the formula for the inverse of $f$ when restricted to that domain.

Step 1: Analyze the function $f(x)$

1. The domain of $f(x) = \frac{1}{\sqrt{x^2 + 5}}$:

   • The square root $\sqrt{x^2 + 5}$ is well-defined for all $x$, because $x^2 + 5 > 0$ for all real numbers $x$.
   • Thus, the function is defined for all real numbers: $x \in (-\infty, \infty)$.

2. Behavior of $f(x)$:

   • $f(x)$ is positive because the denominator $\sqrt{x^2 + 5}$ is always positive.
   • As $|x|$ increases, $\sqrt{x^2 + 5}$ grows, making $f(x)$ decrease.
   • $f(x)$ has a horizontal symmetry because $x^2 = (-x)^2$, so $f(x)$ is even: $f(-x) = f(x)$.

3. $f(x)$ is not one-to-one on its natural domain $(-\infty, \infty)$ because it has the same value for $x$ and $-x$.

Step 2: Restrict the domain to make $f(x)$ one-to-one

To make $f(x)$ one-to-one, we can restrict the domain to either:

• $[0, \infty)$, or
• $(-\infty, 0]$.

Let us choose the domain $[0, \infty)$ (we could also choose $(-\infty, 0]$ and solve similarly).

Step 3: Find the inverse of $f(x)$

To find the inverse, follow these steps:

1. Start with $y = f(x) = \frac{1}{\sqrt{x^2 + 5}}$.

2. Solve for $x$ in terms of $y$:

   • Multiply both sides by $\sqrt{x^2 + 5}$:
     $y \cdot \sqrt{x^2 + 5} = 1$
   • Divide by $y$:
     $\sqrt{x^2 + 5} = \frac{1}{y}$
   • Square both sides:
     $x^2 + 5 = \frac{1}{y^2}$
   • Subtract 5:
     $x^2 = \frac{1}{y^2} - 5$
   • Take the square root:
     $x = \sqrt{\frac{1}{y^2} - 5}.$

3. Since we restricted the domain to $[0, \infty)$, $x \geq 0$, so we take the positive square root: $x = \sqrt{\frac{1}{y^2} - 5}.$

Thus, the inverse function is: $f^{-1}(y) = \sqrt{\frac{1}{y^2} - 5}, \quad y > 0.$

Final Answer:

1. Restricted domain: $[0, \infty)$.
2. Inverse function: $f^{-1}(y) = \sqrt{\frac{1}{y^2} - 5}$ for $y > 0$.

Would you like a detailed graph or further explanation about one-to-one functions or inverses? Let me know!

Related Questions:

1. How can we verify that $f^{-1}(f(x)) = x$ for the restricted domain?
2. What happens if we restrict the domain to $(-\infty, 0]$?
3. Why does the inverse function require the condition $y > 0$?
4. Can $f(x)$ have an inverse over its full domain $(-\infty, \infty)$? Why or why not?
5. How does the symmetry of $f(x)$ affect its one-to-one property?

Tip:

When finding the inverse of a function, always check if the function is one-to-one first, as a non-one-to-one function does not have a proper inverse over its full domain!