Applying the IVT to prove existence of a root

The function $f(x) = \sqrt{x} - \dfrac{2}{x}$ is continuous on $[1,4]$ because $\sqrt{x}$ is continuous for $x \ge 0$ and $2/x$ is continuous for $x \ne 0$; since $[1,4]$ contains neither a negative number nor $0$, $f$ is a difference of two functions continuous on all of $[1,4]$, hence continuous there.

Evaluate the endpoints: $f(1) = \sqrt{1} - \dfrac{2}{1} = 1 - 2 = -1$, and $f(4) = \sqrt{4} - \dfrac{2}{4} = 2 - 0.5 = 1.5$. Since $f(1) = -1 < 0$ and $f(4) = 1.5 > 0$, the value $N=0$ lies strictly between $f(1)$ and $f(4)$.

Because $f$ is continuous on $[1,4]$ and $0$ is between $f(1)$ and $f(4)$, the Intermediate Value Theorem guarantees at least one $c$ in $(1,4)$ with $f(c) = 0$.