#### Answer

See below:

#### Work Step by Step

Consider the provided function,
$ f\left( x \right)=\left\{ \begin{align}
& -x+1\ \ \ \text{ for }-1\le x<1 \\
& 2\ \ \ \ \text{ for }x=1 \\
& {{x}^{2}}\ \ \text{ for }x>\text{1} \\
\end{align} \right.$
Here, the function is a combination of three sub-functions, $ f\left( x \right)=-x+1\text{ for }-1\le x<1$, $ f\left( x \right)=2\text{ for }x=1$ and $ f\left( x \right)={{x}^{2}}\text{ for }x>1$
Consider the first rule $ f\left( x \right)=-x+1\text{ for }-1\le x<1$, which defines a straight line.
The line should be graphed only for $-1\le x<1$ -- that is, to the left of $ x=1$ and to the right of $ x=-1$
Substitute some values of x in $ f\left( x \right)=-x+1\text{ for }-1\le x<1$
Put $ x=-1$
$\begin{align}
& f\left( -1 \right)=-\left( -1 \right)+1 \\
& =1+1 \\
& =2
\end{align}$
The point $\left( -1,2 \right)$ is graphed as a closed dot as the point is a part of the function.
And, put $ x=1$
$\begin{align}
& f\left( 1 \right)=-\left( 1 \right)+1 \\
& =-1+1 \\
& =0
\end{align}$
The point $\left( 1,0 \right)$ is graphed as open dot as the point is not the part of the function.
Consider the second rule $ f\left( x \right)=2\text{ for }x=1$ which defines a single point to be plotted only for $ x=1$.
Now consider the third rule $ f\left( x \right)={{x}^{2}}\text{ for }x>1$
The line should be graphed only for $ x>1$