Applying the Intermediate Value Theorem to show a root exists

Since $f(x) = x^2 - \sin(x) - 1$ is a difference of a polynomial and $\sin(x)$, both of which are continuous on all real numbers, $f$ is continuous everywhere, and in particular on the closed interval $[0,2]$. This satisfies the continuity hypothesis required by the Intermediate Value Theorem.

Now evaluate the endpoints: $f(0) = 0^2 - \sin(0) - 1 = -1$, which is negative. And $f(2) = 4 - \sin(2) - 1 = 3 - \sin(2) \approx 3 - 0.909 = 2.091$, which is positive.

Since $f$ is continuous on $[0,2]$ and $f(0) < 0 < f(2)$, the Intermediate Value Theorem guarantees that there exists at least one $c$ in $(0,2)$ such that $f(c) = 0$. This shows the equation $x^2 - \sin(x) - 1 = 0$ has a root in $(0,2)$.