MathematicsPreCalculus Mathematics at Nebraska

Think of the plane as divided into two regions by the vertical line \(x = 1\text{,}\) as shown in Figure141. In the left-hand region (\(x \le 1\)), we graph the line \(y = x + 1\text{.}\) (The fastest way to graph the line is to plot its intercepts, \((-1, 0)\) and \((0, 1)\text{.}\))

Notice that the value \(x = 1\) is included in the first region, so \(f (1) = 1 + 1 = 2\text{,}\) and the point \((1, 2)\) is included on the graph. We indicate this with a solid dot at the point \((1, 2)\text{.}\)

In the right-hand region (\(x \gt 1\)), we graph the horizontal line \(y = 3\text{.}\) The value \(x = 1\) is not included in the second region, so the point \((1, 3)\) is not part of the graph. We indicate this with an open circle at the point \((1, 3)\text{.}\)

1. In the definition for \(\abs{x}\text{,}\) we replace \(x\) by \(x - 3\) to get

   \begin{equation*} g(x) =\abs{x-3} = \begin{cases} x-3 \amp \text{if }~ x-3\ge 0\\ -(x-3) \amp \text{if }~ -(x-3)\lt 0 \end{cases}. \end{equation*}

   We can simplify this expression to

   \begin{equation*} g(x) =\abs{x-3} = \begin{cases} x-3 \amp \text{if }~ x\ge 3\\ -x+3 \amp \text{if }~ x\lt 3 \end{cases}. \end{equation*}

2. In the first region, \(x \ge 3\text{,}\) we graph the line \(y = x - 3\text{.}\) Because \(x = 3\) is included in this region, the endpoint of this portion of the graph, \((3, 0)\text{,}\) is included, too.

   In the second region, \(x \lt 3\text{,}\) we graph the line \(y = -x + 3\text{.}\) Note that the two pieces of the graph meet at the point \((0, 3)\text{,}\) as shown in Figure144.