Section 1.8 - Absolute Value Function#
The absolute value of a number \(a\), denoted \(|a|\), gives the undirected distance from \(a\) to \(0\) on a number line. The absolute value function is defined as
\[\begin{split}|x|=\begin{cases}
x & \text{if }x\ge0\\
-x & \text{if }x<0
\end{cases}.\end{split}\]
Eq or Ineq |
Equivalent Form |
Solution Set |
|---|---|---|
Case 1: \(|x|=k\) |
\(x=k\) or \(x=-k\) |
\(\{-k,k\}\) |
Case 2: \(0\le|x|<k\) |
\(-k<x<k\) |
\((-k,k)\) |
Case 3: \(|x|>k\ge0\) |
\(x<-k\) or \(x>k\) |
\((-\infty,-k)\cup(k,\infty)\) |
Theorem 2
The equation \(|x|=a\) if and only if \(x=a\) or \(x=-a\). Furthermore, \(|x|=|a|\) if and only if \(x=a\) or \(x=-a\).
Example 22
Solve
\[|9-4x|=7\]
Solution:
Since \(|x|=a\) if and only if \(x=a\) or \(x=-a\) we have
\[\begin{split}\begin{aligned}
9-4x & =7 & 9-4x & =-7\\
-4x & =-2 & -4x & =-16\\
x & =\frac{-2}{-4} & x & =\frac{-16}{-4}\\
& =\frac{1}{2} & & =4\end{aligned}\end{split}\]
Therefore, the solution set is
\[x=\{\frac{1}{2},4\}\]
Example 23
Solve
\[|3x+2|=|x-5|\]
Since \(\|x\|=\|a\|\) if and only if \(x=a\) or \(x=-a\) we have
\[\begin{split}\begin{aligned}
3x+2 & =x-5 & 3x+2 & =-(x-5)\\
3x-x & =-5-2 & 3x+2 & =-x+5\\
2x & =-7 & 3x+x & =5-2\\
x & =\frac{-7}{2} & 4x & =3\\
& & x & =\frac{3}{4}\end{aligned}\end{split}\]
Therefore, the solution set is
\[x=\{-\frac{7}{2},\frac{3}{4}\}.\]