Answer
$\frac{1}{3}$ is not a zero of the function
Work Step by Step
$f(\frac{1}{3})=4(\frac{1}{3})^3-5(\frac{1}{3})^2-3(\frac{1}{3})+1=\frac{4}{27}-\frac{5}{9}-\frac{3}{3}+1=-.407\ne0$
Therefore $\frac{1}{3}$ is not a zero of the function
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