## Calculus 10th Edition

Since $f(x)$ is a polynomial, it is continuous for all values of $x.$ $f(0)=(0)^3+5(0)-3=-3\to f(0)\lt0.$ $f(1)=(1)^3+5(1)-3=3\to f(1)\gt0$ Since $f(x)$ is continuous over the interval $[0, 1]$ and the sign of $f(x)$ changes over the interval $[0, 1]$, then the Intermediate Value Theorem guarantees at least one root in the interval.