## Precalculus (6th Edition) Blitzer

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.
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.