Precalculus (6th Edition) Blitzer

By the intermediate value theorem the function $f$ has a real zero between integers −3 and −2.
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 a point ‘c’ such that $g\left( c \right)=0$. So, to the value of the function $f$ at −3 and −2 is: \begin{align} & f\left( -3 \right)=3{{\left( -3 \right)}^{3}}-10\left( -3 \right)+9 \\ & =-81+30+9 \\ & =-42 \end{align} \begin{align} & f\left( -2 \right)=3{{\left( -2 \right)}^{3}}-10\left( -2 \right)+9 \\ & =-24+20+9 \\ & =5 \end{align} Since, $f\left( -3 \right)=-42$ and $f\left( -2 \right)=5$ have opposite signs, by the intermediate value theorem the function $f$ has a real zero between integers −3 and −2.