Absolute Value (Modulus) Function

Absolute value function (modulus function). Definition: $f(x)=|x|$ is the piecewise function $f(x)=\begin{cases}x, & x\ge0\\ -x, & x<0\end{cases}$ (Fig. 6.39). Domain: $R$ (or $(-\infty,\infty)$); Range: $[0,\infty)$.

Properties: (1) The graph of $f(x)=|x|$ is the union of the line $y=x$ from quadrant I with the line $y=-x$ from quadrant II. Since the origin marks where the two lines' directions change, we call it a critical point. (2) The graph is symmetric about the $Y$-axis. (3) The graph of $f(x)=|x-3|$ is the graph of $|x|$ shifted 3 units to the right, with the critical point now at $(3,0)$. (4) $f(x)=|x|$ represents the distance of $x$ from the origin. (5) If $|x|=m$, this represents every $x$ whose distance from the origin is $m$, that is $x=+m$ or $x=-m$ (Fig. 6.40). (6) If $|x|<m$, this represents every $x$ whose distance from the origin is less than $m$: $0\le x<m$ and $0\ge x>-m$, that is $-m<x<m$, i.e. $x\in(-m,m)$ (Fig. 6.41). (7) If $|x|\ge m$, this represents every $x$ whose distance from the origin is greater than or equal to $m$: $x\ge m$ and $x\le-m$, i.e. $x\in(-\infty,-m]\cup[m,\infty)$ (Fig. 6.42). (8) If $m<|x|<n$, this represents every $x$ whose distance from the origin is greater than $m$ but less than $n$, that is $x\in(-n,-m)\cup(m,n)$ (Fig. 6.43). (9) Triangle inequality: $|x+y|\le|x|+|y|$ (verify by trying different positive and negative values of $x,y$). (10) $|x|$ can also be defined as $|x|=\sqrt{x^2}=\max\{x,-x\}$.

Ex. 10: Solve $|4x-5|\le3$.

Solution: Using $|x|\le m \iff -m\le x\le m$: $-3\le4x-5\le3 \implies -3+5\le4x\le3+5 \implies2\le4x\le8 \implies\dfrac24\le x\le\dfrac84 \implies\dfrac12\le x\le2$.

Ex. 11: Find the domain of $f(x)=\dfrac{1}{\sqrt{\left||x|-1\right|-3}}$.

Solution: Since $f(x)$ sits over a square root in a denominator, it is defined only where the radicand is strictly positive: $\left||x|-1\right|-3>0$, therefore $\left||x|-1\right|>3$, so $|x|-1>3$ or $|x|-1<-3$, that is $|x|>4$ or $|x|<-2$. But $|x|<-2$ is impossible, since $|x|\ge0$ always — so only $|x|>4$ survives, giving $x>4$ or $x<-4$. Domain is $(-\infty,-4)\cup(4,\infty)$.

Ex. 12: Solve $|x-1|+|x+2|=8$.

Solution: Let $f(x)=|x-1|+|x+2|$. The critical points are $x=1$ and $x=-2$, dividing the number line into three regions (Fig. 6.44):

Region I ($x<-2$, test value $-3$): $(x-1)<0,\ (x+2)<0$, so $f(x)=-(x-1)-(x+2)=-2x-1$. …