## Calculus 10th Edition

Since $f(x)$ is a polynomial, it is continuous for all values of $x.$ $f(1)=2(1)^3-3=-1\to f(1)\lt0.$ $f(2)=2(2)^3-3=13\to f(2)\gt0.$ Since $f(x)$ is continuous over the interval $[1, 2]$ and the sign of $f(x)$ changes over the interval $[1, 2]$, then the Intermediate Value Theorem guarantees at least one zero in the interval.
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