Let us analyze and solve the provided questions step-by-step.

Problem 1:

The table gives some function values for the continuous function $g(x)$.
We are asked whether the Intermediate Value Theorem (IVT) applies to show that there exists some $c$ in the interval $(0, 2)$ for which $g(c) = 5$.

Given Information:

• $g(0) = 3$
• $g(2) = 8$

Solution:

1. Verify continuity:
   The problem explicitly states that $g(x)$ is continuous. This satisfies the first condition of the IVT, which requires continuity on the closed interval $[0, 2]$.

2. Check for $g(c) = 5$ within the interval:
   The function value $g(x)$ changes from $g(0) = 3$ to $g(2) = 8$.
   Since $g(x)$ is continuous, it must take on every value between 3 and 8 as $x$ moves from 0 to 2.
   Specifically, $5$ lies in the interval $[3, 8]$, so there must exist some $c \in (0, 2)$ such that $g(c) = 5$.

Conclusion:

IVT applies.
This is because $g(x)$ is continuous on $[0, 2]$, and $g(0) < 5 < g(2)$.

Problem 2:

Let $k(x) = \frac{x^2 - x - 2}{x^2 - 4}$. We are tasked with determining if there must exist some $x$ between 0 and 2 for which $k(x) = -4$, based on the IVT.

Given Information:

• $k(x) = \frac{x^2 - x - 2}{x^2 - 4}$
• Interval: $(0, 2)$

Solution:

1. Check continuity of $k(x)$:
   To use IVT, $k(x)$ must be continuous on the interval $(0, 2)$.

   • Analyze the denominator $x^2 - 4$:
     $x^2 - 4 = (x - 2)(x + 2).$ The denominator equals zero at $x = 2$ and $x = -2$.
     Thus, $k(x)$ is not defined at $x = 2$, meaning $k(x)$ is not continuous on $(0, 2)$.

2. Conclude based on continuity:
   Since $k(x)$ is not continuous on $(0, 2)$, the IVT does not apply, and we cannot guarantee the existence of some $c \in (0, 2)$ for which $k(c) = -4$.

Conclusion:

IVT does not apply.
This is because $k(x)$ is not continuous at $x = 2$.

Final Answer Summary:

1. Problem 1: IVT applies because $g(x)$ is continuous, and $5$ is between $g(0) = 3$ and $g(2) = 8$.
2. Problem 2: IVT does not apply because $k(x)$ is not continuous at $x = 2$.

Let me know if you'd like further clarifications or details!
Here are 5 follow-up questions for deeper understanding:

1. What does it mean for a function to be continuous on an interval?
2. Why is the Intermediate Value Theorem dependent on continuity?
3. Can a function that has a removable discontinuity satisfy the IVT?
4. How can you rewrite $k(x) = \frac{x^2 - x - 2}{x^2 - 4}$ by factoring both the numerator and denominator?
5. What happens if we change the interval for $k(x)$ to $(0, 1.9)$? Does the IVT apply in this case?

Tip: Always check for continuity before applying the IVT. If a function is discontinuous on the interval, the theorem cannot guarantee intermediate values.