consists of four line segments and give the equation of each segment. Sketch the graph.

Let’s look at the four regions described.

#### Region 1: \(x < 0\)

In this case, \(x\), \(x-1\) and \(x-4\) are all negative so \[\begin{equation*} y = \big|x\big| + \big|x-1\big| + \big|x-4\big| = -x - (x-1) - (x-4) = -3x + 5. \end{equation*}\]#### Region 2: \(0 \le x < 1\)

When we are in this region, \(x\) is positive but \(x-1\) and \(x-4\) are negative so \[\begin{equation*} y = \big|x\big| + \big|x-1\big| + \big|x-4\big| = x - (x-1) - (x-4) = -x + 5. \end{equation*}\]#### Region 3: \(1 \le x < 4\)

In this case, \[\begin{equation*} y = \big|x\big| + \big|x-1\big| + \big|x-4\big| = x + (x-1) - (x-4) = x + 3. \end{equation*}\]#### Region 4: \(x \ge 4\)

Finally, in this region, \[\begin{equation*} y = \big|x\big| + \big|x-1\big| + \big|x-4\big| = x + (x-1) + (x-4) = 3x - 5. \end{equation*}\]Thus, the graph of \(y\) consists of four line segments. This leads to the following sketch.

Notice that the line segments join up at their ends. Why must this be the case for a function like this?

Calculate the roots of the equations

- \(\big|x\big| + \big|x-1\big| + \big|x-4\big| = 6 - \dfrac{1}{3}x\),

If we add the line \(y = 6 - \dfrac{1}{3}x\) to our sketch in our mind’s eye, we can see that we expect two solutions, one negative, one positive.

#### Region 1: \(x < 0\)

In this region, the equation to solve becomes \[\begin{equation*} -3x + 5 = 6 - \frac{x}{3} \iff \frac{8x}{3} = -1 \iff x = -\frac{3}{8}. \end{equation*}\]As this lies in the region \(x < 0\), this is a valid solution.

#### Region 2: \(0 \le x < 1\)

The equation to solve can be written as \[\begin{equation*} -x + 5 = 6 - \frac{x}{3} \iff \frac{2x}{3} = -1 \iff x = -\frac{3}{2}. \end{equation*}\]As this does *not* fall in the region under consideration (\(0 \le x < 1\)), there is no solution in this region.

#### Region 3: \(1 \le x < 4\)

In this region, the equation to solve becomes \[\begin{equation*} x + 3 = 6 - \frac{x}{3} \iff \frac{4x}{3} = 3 \iff x = \frac{9}{4}. \end{equation*}\]As this is part of the region \(1 \le x < 4\), this is the second solution to the original equation.

#### Region 4: \(x \ge 4\)

Finally, in this region, we are to solve \[\begin{equation*} 3x - 5 = 6 - \frac{x}{3} \iff \frac{10x}{3} = 11 \iff x = \frac{33}{10}. \end{equation*}\]As \(\dfrac{33}{10} < 4\), the solution does not fall in the region under consideration, and hence this is not a solution to the equation.

Thus, the two solutions to the equation are \[\begin{equation*} x = -\frac{3}{8} \quad\text{and}\quad x = \frac{9}{4}. \end{equation*}\]By sketching the line \(y = 6 - \frac{1}{3}x\) we could probably have seen that the solutions had to fall in regions 1 and 3 and saved ourselves some work.

- \(\big|x\big| + \big|x-1\big| + \big|x-4\big| = \dfrac{1}{x}\).

Once again, if we imagine adding the line \(y = \dfrac{1}{x}\) to our sketch, we can see that we expect one positive solution to this.

#### Region 1: \(x < 0\)

It’s clear there are no solutions in this region, since \(\dfrac{1}{x}\) is negative here.

#### Region 2: \(0 \le x < 1\)

The value \(x = 0\) cannot be a solution. When \(0 < x < 1\), the equation becomes \[\begin{equation*} -x + 5 = \frac{1}{x} \iff -x^2 + 5x - 1 = 0. \end{equation*}\] The quadratic formula implies that \[\begin{equation*} x = \frac{-5 \pm \sqrt{5^2 - 4 \times (-1) \times (-1)}}{2 \times (-1)} = \frac{-5 \pm \sqrt{25 - 4}}{-2} = \frac{5 \mp \sqrt{21}}{2}. \end{equation*}\] It is clear that \[\begin{equation*} \frac{5 + \sqrt{21}}{2} > \frac{5}{2} > 1 \end{equation*}\] so that one of the two solutions does not fall into the region of interest. As for the other, \[\begin{equation*} \frac{5 - \sqrt{21}}{2} \le \frac{5 - 4}{2} = \frac{1}{2} < 1 \end{equation*}\] and \[\begin{equation*} \frac{5 - \sqrt{21}}{2} > \frac{5 - 5}{2} = 0 \end{equation*}\]and so this falls into the region of interest and it is a valid solution.

From a sketch, we would not expect to find any more solutions, but to be sure we’ll check.

#### Region 3: \(1 \le x < 4\)

The equation becomes \[\begin{equation*} x + 3 = \frac{1}{x} \iff x^2 + 3x - 1 = 0 \end{equation*}\] and so \[\begin{equation*} x = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-1)}}{2} = \frac{-3 \pm \sqrt{9 + 4}}{2} = \frac{-3 \pm \sqrt{13}}{2}. \end{equation*}\]As \(3 < \sqrt{13} < 4\), it is plain that neither of these values falls into the region under consideration.

#### Region 4: \(x \ge 4\)

Finally, in the fourth region, the equation can be written as \[\begin{equation*} 3x - 5 = \frac{1}{x} \iff 3x^2 - 5x - 1 = 0 \end{equation*}\] and so \[\begin{equation*} x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 3 \times (-1)}}{2 \times 3} = \frac{5 \pm \sqrt{25 + 12}}{6} = \frac{5 \pm \sqrt{37}}{6}. \end{equation*}\] As \(\sqrt{37} > 6\), it follows that \[\begin{equation*} \frac{5 - \sqrt{37}}{6} < \frac{5 - 6}{6} < 0 \end{equation*}\] so that this possible solution falls outside of the region of interest. For the other, as \(\sqrt{37} < 7\), \[\begin{equation*} \frac{5 + \sqrt{37}}{6} < \frac{5 + 7}{6} = 2 \end{equation*}\]the solution also falls outside of the region of interest.

Thus there is just a single solution to the equation, namely \[\begin{equation*} x = \frac{5 - \sqrt{21}}{2}. \end{equation*}\]