Applying the Intermediate Value Theorem to prove a root exists

Since $f(x) = x^3 - 4x - 1$ is a polynomial, it is continuous for all real numbers, and in particular on the closed interval $[-1, 3]$. This satisfies the continuity hypothesis required by the Intermediate Value Theorem.

Evaluate the function at the endpoints: $f(-1) = (-1)^3 - 4(-1) - 1 = -1 + 4 - 1 = 2$, and $f(3) = 3^3 - 4(3) - 1 = 27 - 12 - 1 = 14$. Both endpoint values are positive, so evaluating $f$ only at the endpoints does not produce a sign change. To apply IVT to guarantee a root at $f(c) = 0$, we need to find where $f$ changes sign. Evaluating at the interior point $x = 1$: $f(1) = 1^3 - 4(1) - 1 = 1 - 4 - 1 = -4$. Now we have $f(-1) = 2 > 0$ and $f(1) = -4 < 0$, which shows a sign change.

Since $f$ is continuous on $[-1, 1] \subset [-1, 3]$ and $f(-1) = 2 > 0$ while $f(1) = -4 < 0$, the value $0$ lies between $f(-1)$ and $f(1)$. By the Intermediate Value Theorem, there exists at least one $c \in (-1, 1)$ such that $f(c) = 0$. Therefore, $f$ has a root in the interval $[-1, 3]$.

The answer is one. The solution evaluated f at x equals one, getting negative four, since the endpoints negative one and three both gave positive results. This interior value provided the needed sign change with f of negative one equal to two. That sign change let the Intermediate Value Theorem guarantee a root between negative one and one.