Let's focus on solving part b of the problem. To do that, we need to apply the Intermediate Value Theorem (IVT).

Restating Part B:

• We are given $f(x)$ in two pieces:
  • $f(x) = \sqrt{x - 1}$ for $1 \leq x \leq 5$
  • $f(x) = \frac{3x - 5}{x^3 + 1}$ for $5 < x \leq 10$
• The goal is to find whether there exists some $c \in [2, 8]$ such that $f(c) = \frac{4}{3}$, using the Intermediate Value Theorem (IVT).

Steps:

1. Check if $f(x)$ is continuous on the interval [2, 8]: The Intermediate Value Theorem requires that the function is continuous on the closed interval. We know that $f(x)$ is composed of two parts, each defined on different intervals:

   • $f(x) = \sqrt{x - 1}$ is continuous for $x \geq 1$, including on $[2, 5]$.
   • $f(x) = \frac{3x - 5}{x^3 + 1}$ is also continuous for $x \in (5, 10]$. Therefore, $f(x)$ is continuous on $[2, 8]$ because there are no discontinuities at $x = 5$ (based on part a).

2. Evaluate the function at the endpoints: We need to check the values of $f(2)$ and $f(8)$ to see if $f(c) = \frac{4}{3}$ lies between these values.

   • For $x = 2$, $f(2) = \sqrt{2 - 1} = \sqrt{1} = 1$.
   • For $x = 8$, $f(8) = \frac{3(8) - 5}{8^3 + 1} = \frac{24 - 5}{512 + 1} = \frac{19}{513} \approx 0.037$.

3. Check if $f(c) = \frac{4}{3}$ is between $f(2)$ and $f(8)$: We see that:

   • $f(2) = 1$
   • $f(8) \approx 0.037$

   Since $\frac{4}{3} \approx 1.333$, it is not between $f(2) = 1$ and $f(8) \approx 0.037$. Therefore, by the Intermediate Value Theorem, there is no such $c \in [2, 8]$ such that $f(c) = \frac{4}{3}$.

Conclusion:

There does not exist a value $c \in [2, 8]$ such that $f(c) = \frac{4}{3}$ based on the given function.

Do you have any questions or want further details on this solution? Here are 5 questions you could explore:

1. How would the IVT apply if the value $\frac{4}{3}$ were within the range of $f(x)$ on the interval?
2. How can you prove that $f(x)$ is continuous at $x = 5$?
3. What happens to the function as $x$ approaches values near 1 or near 10?
4. Can you calculate the precise value of $f(x)$ for $x = 8$ using a calculator?
5. What other intervals might satisfy the Intermediate Value Theorem for different target values of $f(x)$?

Tip: Always check the continuity of the function first when applying the Intermediate Value Theorem!