To determine the validity of the statement using the Intermediate Value Theorem (IVT), let’s analyze the problem step by step.

Given:

• $f(x)$ is continuous on the interval $[0, 4]$.
• $f(0) = -4$, $f(3) = 2$, $f(4) = 8$.
• We need to check if $f(x) = 0$ has exactly 1 solution in $[0, 4]$.

Step 1: Apply the Intermediate Value Theorem

The IVT states that if $f$ is continuous on $[a, b]$, and $f(a)$ and $f(b)$ have opposite signs, then there is at least one $c \in (a, b)$ such that $f(c) = 0$.

1. At $x = 0$, $f(0) = -4$ (negative).
2. At $x = 3$, $f(3) = 2$ (positive).

Since $f$ is continuous, and $f(0) < 0 < f(3)$, the IVT guarantees that there is at least one solution $c \in (0, 3)$ such that $f(c) = 0$.

Step 2: Check for Additional Roots

• Beyond $x = 3$, we observe $f(3) = 2$ and $f(4) = 8$. Both values are positive, and there is no sign change between $x = 3$ and $x = 4$.
• Therefore, $f(x) = 0$ cannot have another root in $(3, 4]$.

Conclusion:

1. The IVT ensures at least one root in $(0, 3)$.
2. There are no roots in $(3, 4]$, as the function remains positive there.

Thus, there is exactly one solution in the interval $[0, 4]$, and the statement is true.

Final Answer:

• The statement must be true.

Additional Questions:

1. What is the Intermediate Value Theorem, and how does it apply to continuous functions?
2. Can the IVT guarantee multiple roots within an interval?
3. How would the solution change if $f(x)$ were not continuous?
4. What happens if additional points within the interval are given?
5. How can we confirm the exact location of the root?

Tip: The IVT guarantees at least one solution but does not give its exact value; further analysis (e.g., bisection) can help pinpoint the root.