Applying the IVT to show a value is attained

Since $\arctan(x)$ and $x$ are both continuous on all of $\mathbb{R}$, their sum $f(x) = \arctan(x) + x$ is continuous everywhere, and in particular on the closed interval $[-1, 1]$. This satisfies the continuity hypothesis required by the Intermediate Value Theorem.

Evaluate the endpoints: $f(-1) = \arctan(-1) + (-1) = -\frac{\pi}{4} - 1 \approx -1.785$, and $f(1) = \arctan(1) + 1 = \frac{\pi}{4} + 1 \approx 1.785$. Since $f(-1) \approx -1.785 < 0.5 < 1.785 \approx f(1)$, the value $N = 0.5$ lies strictly between $f(-1)$ and $f(1)$.

Because $f$ is continuous on $[-1,1]$ and $0.5$ is between $f(-1)$ and $f(1)$, the Intermediate Value Theorem guarantees the existence of at least one $c \in (-1, 1)$ such that $f(c) = 0.5$.