## Precalculus (6th Edition) Blitzer

$\left( f\circ g \right)\left( x \right)=2{{x}^{2}}-3x$ and $\left( g\circ f \right)\left( x \right)=2-2{{x}^{2}}+x$.
We have $f\left( x \right)=2{{x}^{2}}-x-1$ and $g\left( x \right)=1-x$: Therefore, $\left( f\circ g \right)\left( x \right)$ can be computed as: \begin{align} & \left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right) \\ & =2{{\left( 1-x \right)}^{2}}-\left( 1-x \right)-1 \\ & =2\left( 1+{{x}^{2}}-2x \right)-\left( 1-x \right)-1 \\ & =2{{x}^{2}}-3x \end{align} And, $\left( g\circ f \right)\left( x \right)$ is computed as: \begin{align} & \left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right) \\ & =1-\left( 2{{x}^{2}}-x-1 \right) \\ & =2-2{{x}^{2}}+x \end{align} Thus, $\left( f\circ g \right)\left( x \right)=2{{x}^{2}}-3x$ and $\left( g\circ f \right)\left( x \right)=2-2{{x}^{2}}+x$.