1. The problem asks us to explain why Rolle's theorem does not apply to the function $f(x) = \frac{1}{x} - 3$ on the interval $[-4,4]$. 2. Rolle's theorem states that if a function $f$ is continuous on a closed interval $[a,b]$, differentiable on the open interval $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$. 3. First, check if $f$ is continuous on $[-4,4]$. The function $f(x) = \frac{1}{x} - 3$ is undefined at $x=0$, which lies inside the interval $[-4,4]$. Therefore, $f$ is not continuous on the entire interval. 4. Since continuity on the closed interval is a requirement for Rolle's theorem, the theorem does not apply here. 5. Additionally, check the values at the endpoints: $f(-4) = \frac{1}{-4} - 3 = -\frac{1}{4} - 3 = -3.25$ and $f(4) = \frac{1}{4} - 3 = 0.25 - 3 = -2.75$. Since $f(-4) \neq f(4)$, another condition of Rolle's theorem fails. 6. In summary, the function $f(x) = \frac{1}{x} - 3$ is not continuous on $[-4,4]$ due to the discontinuity at $x=0$, and the endpoint values are not equal, so Rolle's theorem does not hold on this interval.