Question: The function f(x) is continuous on its domain of $[-9, 9]$ and is plotted below such that the portion of the graph on the interval $(-3, 2)$ is hidden from view. Given that $f(-3) = -2$ and $f(2) = -7$, determine what conclusions can be drawn based on the Intermediate Value Theorem on the interval $(-3, 2)$. Must there exist a value of $c$ in the interval $(?, ?)$ where $f(c) = -1$?

1. **State the problem:** We have a continuous function $f(x)$ on $[-9,9]$ with $f(-3) = -2$ and $f(2) = -7$. The graph between $x=-3$ and $x=2$ is hidden. We want to determine if there must exist a value $c$ in $(-3,2)$ such that $f(c) = -1$. 2. **Recall the Intermediate Value Theorem (IVT):** If a function $f$ is continuous on a closed interval $[a,b]$ and $k$ is any value between $f(a)$ and $f(b)$, then there exists at least one $c$ in $(a,b)$ such that $f(c) = k$. 3. **Apply IVT to this problem:** Here, $a = -3$, $b = 2$, $f(a) = -2$, and $f(b) = -7$. The value $k = -1$ lies between $-2$ and $-7$? 4. **Check if $k = -1$ is between $f(-3)$ and $f(2)$:** Since $-1 > -2$ and $-1 > -7$, $-1$ is not between $-2$ and $-7$. The values $-2$ and $-7$ are both less than $-1$, so $-1$ is not between them. 5. **Conclusion:** Because $-1$ is not between $f(-3)$ and $f(2)$, the Intermediate Value Theorem does not guarantee the existence of a $c$ in $(-3,2)$ such that $f(c) = -1$. 6. **Final answer:** There does not have to be a value $c$ in $(-3,2)$ with $f(c) = -1$ based on the IVT. **Summary:** The IVT guarantees a value $c$ for any $k$ between $-2$ and $-7$, but since $-1$ is not between these values, no guarantee exists for $f(c) = -1$ on $(-3,2)$.