Answer
By the intermediate value theorem the function $f\left( x \right)={{x}^{4}}+6{{x}^{3}}-18{{x}^{2}}$ has a real zero between integers 2 and 3.
Work Step by Step
According to the intermediate value theorem if $g\left( a \right)$ and $g\left( b \right)$ are of different signs, that is $g\left( a \right)g\left( b \right)<0$ , then there will exist at least one point ‘c’ such that $g\left( c \right)=0$.
The value of the function $f$ at 2 and 3 is:
$\begin{align}
& f\left( 2 \right)={{\left( 2 \right)}^{4}}+6{{\left( 2 \right)}^{3}}-18{{\left( 2 \right)}^{2}} \\
& =16+48-72 \\
& =-8
\end{align}$
$\begin{align}
& f\left( 3 \right)={{\left( 3 \right)}^{4}}+6{{\left( 3 \right)}^{3}}-18{{\left( 3 \right)}^{2}} \\
& =81+162-162 \\
& =81
\end{align}$
Since $f\left( 2 \right)=-8$ and $f\left( 3 \right)=81$ have opposite signs, by the intermediate value theorem the function $f\left( x \right)={{x}^{4}}+6{{x}^{3}}-18{{x}^{2}}$ has a real zero between integers 2 and 3.
