## Calculus (3rd Edition)

$[-\frac{1}{2},0]$
One can see that the function $f(x)=x^3+2x+1$ is continuous on $[-\frac{1}{2},0]$ and $f(-\frac{1}{2})=-\frac{1}{8}\neq f(0)=1$ and $0$ is between $f(\frac{1}{2})$ and $f(0)$. Then, by the Intermediate Value Theorem, the function $f(x)$ takes on the value $0$ for some $x\in (-\frac{1}{2},0)$. Hence, $f(x)$ has a root in the interval $[-\frac{1}{2},0]$ which has length $\frac{1}{2}$.