Applying the IVT to prove a root exists

Since $\ln(x)$ is continuous for all $x>0$ and $[1,4] \subset (0,\infty)$, and since polynomials like $-x+2$ are continuous everywhere, $f(x) = \ln(x)-x+2$ is a sum of functions continuous on $[1,4]$, hence continuous on $[1,4]$.

Evaluate the endpoints: $f(1) = \ln(1) - 1 + 2 = 0 - 1 + 2 = 1 > 0$, and $f(4) = \ln(4) - 4 + 2 = \ln(4) - 2 \approx 1.386 - 2 = -0.614 < 0$.

Since $f$ is continuous on $[1,4]$ and $f(1) > 0$ while $f(4) < 0$, the value $N = 0$ lies strictly between $f(1)$ and $f(4)$. By the Intermediate Value Theorem, there exists at least one $c \in (1,4)$ with $f(c) = 0$, so $f$ has a root in $(1,4)$.