To find the inverse of the function $g(x) = -4\sqrt{5 - 2x} + 1$, we need to follow these steps:

Step 1: Replace $g(x)$ with $y$

$y = -4\sqrt{5 - 2x} + 1$

Step 2: Swap $x$ and $y$ to find the inverse

$x = -4\sqrt{5 - 2y} + 1$

Step 3: Solve for $y$ (the inverse function)

1. Subtract 1 from both sides: $x - 1 = -4\sqrt{5 - 2y}$

2. Divide by -4: $\frac{x - 1}{-4} = \sqrt{5 - 2y}$

3. Square both sides to eliminate the square root: $\left(\frac{x - 1}{-4}\right)^2 = 5 - 2y$

4. Simplify the left side: $\frac{(x - 1)^2}{16} = 5 - 2y$

5. Subtract 5 from both sides: $\frac{(x - 1)^2}{16} - 5 = -2y$

6. Multiply by -1: $5 - \frac{(x - 1)^2}{16} = 2y$

7. Divide by 2: $y = \frac{5 - \frac{(x - 1)^2}{16}}{2}$

Thus, the inverse function $g^{-1}(x)$ is: $g^{-1}(x) = \frac{5}{2} - \frac{(x - 1)^2}{32}$

Step 4: Find the domain and range of $g^{-1}(x)$

1. Domain of $g^{-1}(x)$: The domain of $g^{-1}(x)$ is the range of the original function $g(x)$. Since $g(x) = -4\sqrt{5 - 2x} + 1$, the square root must be non-negative, meaning $5 - 2x \geq 0$, or $x \leq \frac{5}{2}$. Therefore, the domain of $g(x)$ is $(-\infty, \frac{5}{2}]$. The range of $g(x)$, and hence the domain of $g^{-1}(x)$, is $(-\infty, 1]$.

2. Range of $g^{-1}(x)$: The range of $g^{-1}(x)$ is the domain of $g(x)$, which is $(-\infty, \frac{5}{2}]$.

Final Answer:

• The inverse function is $g^{-1}(x) = \frac{5}{2} - \frac{(x - 1)^2}{32}$.
• Domain of $g^{-1}(x)$: $(-\infty, 1]$
• Range of $g^{-1}(x)$: $(-\infty, \frac{5}{2}]$

Would you like more details or have any questions?

Here are five related questions:

1. How do you verify if a function and its inverse are correct?
2. Can the inverse function have a different range than the original?
3. What happens if the function includes complex numbers?
4. How do the transformations in $g(x)$ affect its inverse?
5. How does the domain restriction influence the inverse function?

Tip: The domain of a function restricts the allowable values of $x$, while the range reflects the possible outputs or values of $y$. Always check both when finding inverses.