using intervral noation determin the largets domian over which the given function is one-to-one. Then provide the ecuations for the inverse of the function that is restricted to that domain. If two equally large domain exits over which the given function is one-to-one, you may use either domain. However certain that the equations for the inverse function you submit is apporparite for the pirtucyal domain you chose.  g(x)=1/squrtroot x^2+5

### 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.

using interval notation determine the largest domain over which the given function is one-to-one. Then provide the equations for the inverse of the function that is restricted to that domain. If two equally large domains exist over which the given function is one-to-one, you may use either domain. However, certain that the equations for the inverse function you submit are appropriate for the particular domain you chose. g(x)=1/sqrt(x^2 + 5)

Learn how to determine the largest domain over which g(x) = 1/sqrt(x^2 + 5) is one-to-one, and find the inverse of the function restricted to that domain. This detailed solution explores domain restrictions, inverse functions, and key algebraic concepts suitable for students in Grades 10-12.

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