Answer
The value of $\left( f\circ g \right)\left( x \right)=-{{x}^{2}}+2$ and $\left( g\circ f \right)\left( x \right)=-{{x}^{2}}-2x $.
Work Step by Step
We know that $\left( f\circ g \right)\left( x \right)$ is written as $ f\left( g\left( x \right) \right)$:
Therefore, $\left( f\circ g \right)\left( x \right)$ = $ f\left( g\left( x \right) \right)$ and $\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)$.
Now, for $\left( f\circ g \right)\left( x \right)$
$\begin{align}
& f\left( g\left( x \right) \right)=-{{\left\{ g\left( x \right) \right\}}^{2}}-2\left\{ g\left( x \right) \right\}+1 \\
& =-{{\left( x-1 \right)}^{2}}-2\left( x-1 \right)+1
\end{align}$
And simplify it further, to get
$\begin{align}
& =-\left( {{x}^{2}}-2x+1 \right)-2x+2+1 \\
& =-{{x}^{2}}+2x-1-2x+2+1 \\
& =-{{x}^{2}}+2
\end{align}$
Similarly, for $\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)$
$\begin{align}
& g\left( f\left( x \right) \right)=f\left( x \right)-1 \\
& =-{{x}^{2}}-2x+1-1 \\
& =-{{x}^{2}}-2x
\end{align}$
Thus, the value of $\left( f\circ g \right)\left( x \right)=-{{x}^{2}}+2$ and $\left( g\circ f \right)\left( x \right)=-{{x}^{2}}-2x $.
