If $f$ is continuous on the closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there is at least one $c$ in $(a, b)$ with $f(c) = N$. Read aloud: f of c equals N for some c between a and b. Here $a$ and $b$ are the ends of the interval, $f(a)$ and $f(b)$ the heights there, $N$ a target height between them, and $c$ the input the theorem promises but does not locate. A continuous graph cannot get from one height to another without passing through every height in between: that is the whole idea, and it is why the pencil-never-lifts picture matters.

The exam version takes $N = 0$: if $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f$ has a root in $(a, b)$. For $f(x) = x^3 - 3x + 1$: $f(0) = 1$ and $f(1) = -1$, a polynomial is continuous, so a root lies strictly between 0 and 1. At least one: the theorem counts nothing and locates nothing. To close in, check the midpoint, keep the half where the sign still changes, and repeat; that is bisection, and it is how a calculator finds roots too.