To find the inverse of the function $y = -\frac{7}{3}x^2 + 4$, follow these steps:

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

We start with: $y = -\frac{7}{3}x^2 + 4$ Replace $y$ with $f(x)$, which is the same thing: $f(x) = -\frac{7}{3}x^2 + 4$

Step 2: Swap $x$ and $y$

To find the inverse, switch $x$ and $y$: $x = -\frac{7}{3}y^2 + 4$

Step 3: Solve for $y$

Now, solve for $y$.

1. Subtract 4 from both sides: $x - 4 = -\frac{7}{3}y^2$

2. Multiply both sides by $\frac{3}{-7}$ to isolate $y^2$: $\frac{3(x - 4)}{-7} = y^2$

3. Take the square root of both sides: $y = \pm \sqrt{\frac{3(x - 4)}{-7}}$

Since the square root gives both positive and negative values, you must determine which branch of the function you're dealing with, depending on the context or constraints of the problem. Thus, the inverse function is: $f^{-1}(x) = \pm \sqrt{\frac{3(x - 4)}{-7}}$

Final Answer:

$f^{-1}(x) = \pm \sqrt{\frac{3(4 - x)}{7}}$

Would you like more details or have any questions?

Here are 5 follow-up questions for further understanding:

1. What happens when the coefficient of $x^2$ is positive instead of negative?
2. How would you restrict the domain to ensure the inverse is a function?
3. Can you find the inverse of a linear function? How is it different?
4. What role does the square root play in determining the nature of the inverse?
5. How does the sign of the coefficient $-\frac{7}{3}$ affect the shape of the graph?

Tip: When dealing with square roots in inverse functions, always check if the function needs to be restricted to keep the inverse valid.