Answer
\[f\left( x \right)={{x}^{3}}-4x+2\] has a zero between 0 and 1.
Work Step by Step
Find $f\left( 0 \right)$ and $f\left( 1 \right)$. If the sign changes in the values of $f\left( 0 \right)$ and $f\left( 1 \right)$, then there must be at least one zero between 1 and 2.
That is.,
$\begin{align}
& f\left( x \right)={{x}^{3}}-4x+2 \\
& f\left( 0 \right)=0-4\times 0+2 \\
& f\left( 0 \right)=2, \\
\end{align}$
Which is a positive.
And
$\begin{align}
& f\left( x \right)={{x}^{3}}-4x+2 \\
& f\left( 1 \right)=1-4+2 \\
& f\left( 1 \right)=-1, \\
\end{align}$
Which is a negative.
Thus, there is a sign change in the value of $f\left( 0 \right)$ and $f\left( 1 \right)$ , and there is at least one zero between 0 and 1.
