Absolute value and step functions homework answer key
Evaluations, equivalent piecewise forms, solution sets for equations and inequalities, and graph features for absolute value functions and step functions in Algebra.
Conventions and notation
Absolute value satisfies \( |x| \ge 0 \) and represents distance from \(0\) on the number line. Step function is interpreted here as the greatest integer (floor) function \(\lfloor x \rfloor\), which returns the greatest integer less than or equal to \(x\). Graph endpoints use the standard convention: a filled (closed) point indicates inclusion; a hollow (open) point indicates exclusion.
Core algebra facts used throughout
Absolute value as a piecewise function: \[ |u| = \begin{cases} u, & u \ge 0,\\ -u, & u < 0. \end{cases} \] Absolute value equation form (with \(b \ge 0\)): \[ |u| = b \;\Longleftrightarrow\; u=b \text{ or } u=-b. \] Absolute value strict inequality form: \[ |u| < b \;\Longleftrightarrow\; -b < u < b. \]
Answer key
| Item | Given | Answer | Essential notes |
|---|---|---|---|
| 1 | \(f(x)=|x-3|\). Evaluate \(f(-2)\), \(f(0)\), \(f(5)\). | \(f(-2)=5\), \(f(0)=3\), \(f(5)=2\). | Distance interpretation: \( |x-3| \) is the distance from \(x\) to \(3\) on the number line. |
| 2 | \(g(x)=|2x+1|\). Equivalent piecewise form. | \[ g(x)= \begin{cases} -2x-1, & x<-\tfrac12,\\ 2x+1, & x\ge -\tfrac12. \end{cases} \] | The sign change occurs where \(2x+1=0\), i.e., \(x=-\tfrac12\). |
| 3 | Solve \( |3x-6| = 9 \). | \(x=5\) or \(x=-1\). | Two linear cases: \(3x-6=9\) and \(3x-6=-9\). |
| 4 | Solve \( |x+2| < 5 \). | \(-7 < x < 3\). | Equivalent compound inequality: \(-5<x+2<5\). |
| 5 | \(A(x)=-2|x-3|+6\). Vertex, intercepts, range. |
Vertex \((3,6)\). \(x\)-intercepts \((0,0)\) and \((6,0)\). Range \(y \le 6\). |
Negative leading factor produces an inverted “V”. Symmetry line at \(x=3\). |
| 6 | \(A(x)=-2|x-3|+6\). Equivalent piecewise form. | \[ A(x)= \begin{cases} 2x, & x<3,\\ -2x+12, & x\ge 3. \end{cases} \] | For \(x<3\), \(|x-3|=3-x\). For \(x\ge 3\), \(|x-3|=x-3\). |
| 7 | \(p(x)=\lfloor x \rfloor\). Evaluate \(p(-1.2)\), \(p(0)\), \(p(2.7)\), \(p(3)\). | \(p(-1.2)=-2\), \(p(0)=0\), \(p(2.7)=2\), \(p(3)=3\). | \(\lfloor x \rfloor\) is the greatest integer \(\le x\). |
| 8 | \(S(x)=2\lfloor x \rfloor -3\). Interval rule and endpoint convention. | For any integer \(k\): if \(k \le x < k+1\), then \(S(x)=2k-3\). | Each step is closed at \(x=k\) and open at \(x=k+1\) because \(\lfloor x \rfloor\) jumps at integers. |
| 9 | Solve \(\lfloor x \rfloor = -1\). | \(-1 \le x < 0\). | The output \(-1\) occurs exactly on the half-open interval \([-1,0)\). |
| 10 | \(T(x)=\lfloor |x-1| \rfloor\). Evaluate \(T(-2.7)\), \(T(-1)\), \(T(0.99)\), \(T(3.2)\). | \(T(-2.7)=3\), \(T(-1)=2\), \(T(0.99)=0\), \(T(3.2)=2\). | The absolute value produces a nonnegative input to the floor; small changes near an integer threshold can change the output abruptly. |
Derivation notes for representative items
Piecewise reduction and solution-set structure
Item 3 rests on the equivalence \( |u|=b \Longleftrightarrow u=b \text{ or } u=-b \) for \(b\ge 0\). With \(u=3x-6\) and \(b=9\):
\[ 3x-6=9 \;\Longrightarrow\; 3x=15 \;\Longrightarrow\; x=5, \] \[ 3x-6=-9 \;\Longrightarrow\; 3x=-3 \;\Longrightarrow\; x=-1. \]
Item 4 rests on the distance interpretation: \( |x+2|<5 \) means the distance from \(x\) to \(-2\) is less than \(5\), which is the open interval centered at \(-2\) with radius \(5\). Algebraically, \(-5<x+2<5\), hence \(-7<x<3\).
Common pitfalls
- Sign boundary errors: the split point for \(|ax+b|\) is \(ax+b=0\), not \(x=0\) unless \(b=0\).
- Endpoint confusion for step graphs: \(\lfloor x \rfloor\) is closed at integers on the left interval \([k,k+1)\) and open at \(k+1\).
- Inequality direction mistakes: \( |u| < b \) produces a compound inequality, while \( |u| > b \) produces a union of two rays.