Let's go through each part of your request:

1. Inverse of $g(x) = \sqrt{x+2}$

To find the inverse, we swap $x$ and $y$ in the equation and solve for $y$: $g(x) = \sqrt{x + 2}$ Let $y = \sqrt{x + 2}$. To find the inverse: $x = \sqrt{y + 2}$ Square both sides: $x^2 = y + 2$ Now solve for $y$: $y = x^2 - 2$ Thus, the inverse function is: $g^{-1}(x) = x^2 - 2$

2. Domain of $g^{-1}(x)$

The domain of $g^{-1}(x)$ is the range of $g(x)$. The function $g(x) = \sqrt{x+2}$ is defined for $x \geq -2$, and since square roots are non-negative, $g(x) \geq 0$. Therefore, the range of $g(x)$ is $[0, \infty)$.

Hence, the domain of $g^{-1}(x) = x^2 - 2$ is $[0, \infty)$.

3. Graph of $g(x)$ and $g^{-1}(x)$

We need to graph both $g(x) = \sqrt{x+2}$ and $g^{-1}(x) = x^2 - 2$. The graph of an inverse function is a reflection of the original function over the line $y = x$.

4. Solve the equation

The equation to solve is: $\sqrt{x^2} - 5 = x + 1$ First, simplify $\sqrt{x^2}$. Since $\sqrt{x^2} = |x|$, the equation becomes: $|x| - 5 = x + 1$

Now, solve this equation for two cases:

Case 1: $x \geq 0$

In this case, $|x| = x$, so the equation becomes: $x - 5 = x + 1$ Subtract $x$ from both sides: $-5 = 1$ This is a contradiction, so there are no solutions in this case.

Case 2: $x < 0$

In this case, $|x| = -x$, so the equation becomes: $-x - 5 = x + 1$ Add $x$ to both sides: $-5 = 2x + 1$ Subtract 1 from both sides: $-6 = 2x$ Now, divide by 2: $x = -3$

Thus, the solution to the equation is $x = -3$.

Would you like me to plot the graphs or explain any part in more detail?

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Follow-up Questions

1. How do we verify the inverse function algebraically?
2. Why is $g^{-1}(x)$ only defined for $x \geq 0$?
3. Can the solution $x = -3$ be verified by substitution?
4. What happens if you try to solve the equation $|x| - 5 = x + 1$ by squaring both sides directly?
5. How does the line $y = x$ help in finding inverse function graphs?

Tip:

When solving absolute value equations, always consider both the positive and negative cases separately.