Step 1: Analyze the Function $g(x)$

We are given the function:

$g(x) = \frac{1}{\sqrt{x^2 + 5}}$

This is a rational function where the denominator is the square root of $x^2 + 5$. Let's first consider the properties of this function to determine its domain and where it may be one-to-one.

Domain:

The function is defined for all real values of $x$ because the expression inside the square root, $x^2 + 5$, is always positive for any real number (since $x^2 \geq 0$, so $x^2 + 5 \geq 5$).

Thus, the domain of $g(x)$ is all real numbers:

$\text{Domain of } g(x): (-\infty, \infty)$

One-to-One:

To find the largest interval where the function is one-to-one, we check if the function is strictly increasing or decreasing on certain intervals.

We know that $g(x)$ involves the square root of $x^2 + 5$, which is symmetric about the $y$-axis. The function will decrease as $x$ moves away from 0 in both directions. For a function to be one-to-one, it must either strictly increase or strictly decrease. Since $g(x)$ decreases symmetrically on both sides of the origin, we can consider restricting $g(x)$ to either the left or right side of the $y$-axis to ensure one-to-one behavior.

Step 2: Choose an Interval

We can restrict the domain of $g(x)$ to either $[0, \infty)$ or $(-\infty, 0]$, as the function is decreasing in both intervals, making it one-to-one on either.

Let’s choose the interval $[0, \infty)$ for simplicity.

Step 3: Inverse of the Restricted Function

Now, we need to find the inverse of the function $g(x) = \frac{1}{\sqrt{x^2 + 5}}$ when restricted to $[0, \infty)$.

1. Start with the equation $y = \frac{1}{\sqrt{x^2 + 5}}$.

2. Solve for $x$:

$y = \frac{1}{\sqrt{x^2 + 5}} \implies \sqrt{x^2 + 5} = \frac{1}{y}$

3. Square both sides:

$x^2 + 5 = \frac{1}{y^2}$

4. Solve for $x^2$:

$x^2 = \frac{1}{y^2} - 5$

5. Since we are restricting the domain to $[0, \infty)$, we only take the positive square root:

$x = \sqrt{\frac{1}{y^2} - 5}$

Thus, the inverse function is:

$g^{-1}(y) = \sqrt{\frac{1}{y^2} - 5}, \quad y \in (0, \frac{1}{\sqrt{5}}]$

Final Answer:

• The largest domain over which $g(x)$ is one-to-one is $[0, \infty)$.
• The inverse function for this restricted domain is:

$g^{-1}(y) = \sqrt{\frac{1}{y^2} - 5}, \quad y \in (0, \frac{1}{\sqrt{5}}]$

Would you like more details on any of these steps? Here are five related questions:

1. How can you verify if a function is one-to-one?
2. What is the significance of restricting the domain of a function?
3. How can you graphically interpret the inverse of a function?
4. How does the domain of a function affect its inverse?
5. What are the properties of rational functions involving square roots?

Tip: Always check if a function is strictly increasing or decreasing to determine where it might be one-to-one.