## Precalculus (6th Edition) Blitzer

The value of $\left( fog \right)\left( x \right)$ by the use of the functions $f\left( x \right)={{x}^{2}}-x-4$ and $g\left( x \right)=2x-6$ is $4{{x}^{2}}-26x+38$.
We know that the function $\left( fog \right)\left( x \right)$ can also be defined as $f\left( g\left( x \right) \right)$. Therefore, $\left( fog \right)\left( x \right)=f\left( 2x-6 \right).$ Now, substitute $2x-6$ in place of x -- that is, $x\to 2x-6$ \begin{align} & \left( fog \right)\left( x \right)={{\left( 2x-6 \right)}^{2}}-\left( 2x-6 \right)-4 \\ & =4{{x}^{2}}-24x+36-2x+6-4 \end{align} $=4{{x}^{2}}-26x+38$ Hence, $\left( fog \right)\left( x \right)=4{{x}^{2}}-26x+38$. Hence, the value of $\left( fog \right)\left( x \right)$ by the use of the functions $f\left( x \right)={{x}^{2}}-x-4$ and $g\left( x \right)=2x-6$ is $4{{x}^{2}}-26x+38$.