## Precalculus (6th Edition) Blitzer

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 $x>1$