Answer
$\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 $.
Work Step by Step
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 $.
