Let's find the inverse functions for the given problems.

Problem 4: $g(x) = \frac{12x - 8}{2x + 3}$

To find the inverse, follow these steps:

1. Replace $g(x)$ with $y$: $y = \frac{12x - 8}{2x + 3}$

2. Swap $x$ and $y$: $x = \frac{12y - 8}{2y + 3}$

3. Solve for $y$: $x(2y + 3) = 12y - 8$ $2xy + 3x = 12y - 8$ $2xy - 12y = -3x - 8$ $y(2x - 12) = -3x - 8$ $y = \frac{-3x - 8}{2x - 12}$ $y = \frac{-(3x + 8)}{2(x - 6)}$

So, the inverse function is: $g^{-1}(x) = \frac{-(3x + 8)}{2(x - 6)}$

Problem 5: $g(x) = \sqrt[3]{5x^2 + 2} - 20$

To find the inverse:

1. Replace $g(x)$ with $y$: $y = \sqrt[3]{5x^2 + 2} - 20$

2. Swap $x$ and $y$: $x = \sqrt[3]{5y^2 + 2} - 20$

3. Solve for $y$:

   First, isolate the cube root term: $x + 20 = \sqrt[3]{5y^2 + 2}$

   Next, cube both sides: $(x + 20)^3 = 5y^2 + 2$

   Then, solve for $y^2$: $5y^2 = (x + 20)^3 - 2$ $y^2 = \frac{(x + 20)^3 - 2}{5}$

   Finally, take the square root: $y = \pm \sqrt{\frac{(x + 20)^3 - 2}{5}}$

So, the inverse function is: $g^{-1}(x) = \pm \sqrt{\frac{(x + 20)^3 - 2}{5}}$

Summary of Inverse Functions:

4. $g^{-1}(x) = \frac{-(3x + 8)}{2(x - 6)}$
5. $g^{-1}(x) = \pm \sqrt{\frac{(x + 20)^3 - 2}{5}}$

Would you like more details on any step or have any questions?

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Here are some related questions to deepen your understanding:

1. How can the domain and range affect the inverse of these functions?
2. Why do we sometimes end up with a $\pm$ sign when finding an inverse?
3. Can a cubic root function have an inverse if we restrict its domain?
4. How would you apply these inverse functions in real-life problems?
5. What steps would you follow to verify that the functions derived are indeed the correct inverses?

Tip: When dealing with roots and powers while finding inverse functions, always consider the implications for the domain and range of the function. This ensures the inverse is valid over the correct interval.