Applying the IVT to show a root exists

First evaluate the function at the endpoints: $f(0) = \tan(0) - 0 - 1 = 0 - 0 - 1 = -1$. Next, $f(\pi/3) = \tan(\pi/3) - \pi/3 - 1 = \sqrt{3} - \pi/3 - 1 \approx 1.732 - 1.047 - 1 = 0.685$. Since $f(0) = -1 < 0$ and $f(\pi/3) \approx 0.685 > 0$, the function values have opposite signs at the endpoints. By the Intermediate Value Theorem, since $f$ is continuous on the closed interval $[0, \pi/3]$ and $f(0) < 0 < f(\pi/3)$, there must exist at least one value $c$ in the open interval $(0, \pi/3)$ where $f(c) = 0$. That is, there exists $c \in (0, \pi/3)$ such that $\tan(c) - c - 1 = 0$, or equivalently, $\tan(c) = c + 1$.

The function must be continuous on the closed interval, here from zero to $\pi/3$. The Intermediate Value Theorem requires continuity to guarantee that every value between $f(0)$ and $f(\pi/3)$, including zero, is actually attained somewhere inside the interval. Without continuity, the function could skip over the value zero even if the endpoint values have opposite signs. This is why the problem first confirms tangent minus $x$ minus one is continuous on the given interval before applying the theorem.