Example 3
Completion requirements
| Example 3 |
Express \(y = \left| 5x + 1 \right|\) as a piecewise function.
The function will remain the same when \(5x + 1\) is positive, and will become negative when \(5x + 1\) is negative. The zeros of \(y = 5x + 1\) can be used to split the function into these two regions.
The expression \(5x + 1\) is positive for \(x \gt -\frac{1}{5}\) and negative for \(x \lt -\frac{1}{5}\), so
\[\begin{align}
0 &= 5x + 1 \\
- 1 &= 5x \\
- \frac{1}{5} &= x
\end{align}\]
0 &= 5x + 1 \\
- 1 &= 5x \\
- \frac{1}{5} &= x
\end{align}\]
The expression \(5x + 1\) is positive for \(x \gt -\frac{1}{5}\) and negative for \(x \lt -\frac{1}{5}\), so
\[y = \left\{ \begin{align}
&5x + 1{\rm \thinspace { for }}\thinspace x \ge - \frac{1}{5} \\
-&\left({5x + 1} \right){\rm \thinspace { for }} \thinspace x < - \frac{1}{5} \\
\end{align} \right.\]
&5x + 1{\rm \thinspace { for }}\thinspace x \ge - \frac{1}{5} \\
-&\left({5x + 1} \right){\rm \thinspace { for }} \thinspace x < - \frac{1}{5} \\
\end{align} \right.\]