Simplify. Consider all the cases.

Example:
[tex]\[|x-5|=\left\{\begin{array}{l}x-5, \text{ if } x\ \textgreater \ 5 \\ 0, \text{ if } x=5 \\ 5-x, \text{ if } x\ \textless \ 5\end{array}\right.\][/tex]

(c) \(|2x+7|\)

Answer:

If \(x > -\frac{7}{2}\), then \(|2x+7| = 2x + 7\)

If \(x = -\frac{7}{2}\), then \(|2x+7| = 0\)

If [tex]\(x \ \textless \ -\frac{7}{2}\)[/tex], then [tex]\(|2x+7| = -(2x + 7)\)[/tex]



Answer :

To simplify the expression \(|2x + 7|\), we need to consider the properties of the absolute value function. The absolute value \(|u|\) of a function \(u\) is defined as:

[tex]\[ |u| = \begin{cases} u & \text{if } u \geq 0 \\ -u & \text{if } u < 0 \end{cases} \][/tex]

In this case, \(u = 2x + 7\). We need to determine the break point where \(2x + 7 = 0\), as this divides the cases to consider. Solving for \(x\):

[tex]\[ 2x + 7 = 0 \\ 2x = -7 \\ x = -\frac{7}{2} \][/tex]

So,
- If \(x > -\frac{7}{2}\), \(2x + 7 > 0\).
- If \(x = -\frac{7}{2}\), \(2x + 7 = 0\).
- If \(x < -\frac{7}{2}\), \(2x + 7 < 0\).

Now, we can write the absolute value expression in its different cases:

1. If \(x > -\frac{7}{2}\):
- In this region, \(2x + 7\) is positive, so \(|2x + 7| = 2x + 7\).

2. If \(x = -\frac{7}{2}\):
- Here, \(2x + 7 = 0\), so \(|2x + 7| = 0\).

3. If \(x < -\frac{7}{2}\):
- In this region, \(2x + 7\) is negative, so \(|2x + 7| = -(2x + 7)\).

So the expression for \(|2x + 7|\) fully simplified considering all cases is:

[tex]\[ |2x + 7| = \begin{cases} 2x + 7 & \text{if } x > -\frac{7}{2} \\ 0 & \text{if } x = -\frac{7}{2} \\ -(2x + 7) = -2x - 7 & \text{if } x < -\frac{7}{2} \end{cases} \][/tex]

Therefore, filling in the blanks:

[tex]\[ \begin{matrix} \text{If } x > & -\frac{7}{2} & \text{, then } |2x + 7| = & 2x + 7 \\ \text{If } x = & -\frac{7}{2} & \text{, then } |2x + 7| = & 0 \\ \text{If } x < & -\frac{7}{2} & \text{, then } |2x + 7| = & -2x - 7 \end{matrix} \][/tex]