We are given that the function $f(x)$ is continuous on the interval $[-9, 9]$, and we are interested in applying the Intermediate Value Theorem (IVT) on the interval $(-4, 3)$.

Restating the problem:

• $f(-4) = -5$
• $f(3) = -2$
• The portion of the graph between $x = -4$ and $x = 3$ is hidden.

The Intermediate Value Theorem states that if a function $f$ is continuous on a closed interval $[a, b]$ and $f(a) \neq f(b)$, then for any value $N$ between $f(a)$ and $f(b)$, there exists some $c \in (a, b)$ such that $f(c) = N$.

Applying IVT:

• The function $f(x)$ is continuous on the interval $(-4, 3)$, which implies that for any value $N$ between $f(-4) = -5$ and $f(3) = -2$, there must exist some $c \in (-4, 3)$ such that $f(c) = N$.
• Since $f(-4) = -5$ and $f(3) = -2$, the IVT guarantees that the function $f(x)$ must take every value between -5 and -2 somewhere on the interval $(-4, 3)$.

Conclusion:

From the Intermediate Value Theorem, we can conclude that there exists at least one value $c \in (-4, 3)$ for each of the following:

1. $f(c) = -4$
2. $f(c) = -3$
3. $f(c) = -2.5$ (or any other value between -5 and -2)

In summary, the IVT guarantees that the function takes on every value between $-5$ and $-2$ on the hidden interval $(-4, 3)$, even though the exact behavior of the function in this interval is not visible.

Would you like further details or have any questions?

Here are 5 related questions:

1. What conditions are required for the Intermediate Value Theorem to apply?
2. Can the Intermediate Value Theorem tell us how many times $f(x)$ takes a particular value?
3. How does continuity of $f(x)$ affect its graph on the interval $(-4, 3)$?
4. What if the function were not continuous on $(-4, 3)$? Would the IVT still apply?
5. How does the IVT relate to finding roots of an equation?

Tip: When applying the IVT, always check that the function is continuous on the given interval.