📏 Calculus Deep Dives · Visual Study Guide

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Absolute Value: Equations & Inequalities

Every technique here — isolating, splitting into "and"/"or", checking candidates, taming degenerate right-hand sides — follows from one geometric sentence. Learn the picture and the algebra becomes obvious.

💡 In one sentence: $|x|$ is the distance from $x$ to $0$, so $|x-a|$ is the distance from $x$ to $a$. Then $|x-a|=c$ asks "which points lie exactly at distance $c$ from $a$?", $|x-a|<c$ asks for all points closer than $c$, and $|x-a|>c$ for all points farther than $c$.

📐 Distance-first🗂 15 sections✍ 17 canonical examples🧮 45+ practice problems💻 Python checks

🔬 Distance Lab

drag x around!

$|x-3|$ is the distance from $x$ to $3$. Move $x$ and watch the distance — it's never negative, and it hits $0$ only when $x=3$.

a (target)3

|x − a| distance3

The Idea First: Absolute Value Is Distance

Signs record direction; absolute value keeps only size.

The sign of a number records a direction (left/right, debt/credit); the absolute value keeps only the size: $|-7|=7$, $|7|=7$, $|0|=0$. On a number line, "size" has a direct geometric reading:

The one idea behind everything

$$|x| = \text{distance from } x \text{ to } 0, \qquad |x-a| = \text{distance from } x \text{ to } a.$$

An absolute value equation or inequality is never really about signs — it is a question about distance on the line.

The second identity is a shift of the first: instead of measuring from $0$, measure from $a$. For example $|7-2|=|5|=5$ (7 and 2 are five apart), and $|-5-3|=|-8|=8$ (−5 and 3 are eight apart).

Why $|x-a|$ is "distance from $x$ to $a$": slide the whole number line until the origin sits under $a$; the coordinate of $x$ becomes $x-a$, and its distance to the new origin is $|x-a|$. Distance is shift-invariant — that's why $|153-151|=|3-1|=2$.

A machined rod is acceptable if $|L-50|\le 0.5$ mm.
The temperature must satisfy $|T-68|\le 3$ °F.
In the $\varepsilon$–$\delta$ definition of limits, "close" is $|x-a|<\delta$ and $|f(x)-L|<\varepsilon$.

Absolute value inequalities are the arithmetic of closeness. For programmers: abs(x) is the 1-D distance function, and $|x-y|$ is the simplest distance between two data points.

Definition & Properties

The piecewise definition, the identities you'll use constantly, and two proofs worth knowing.

Definition of the absolute value

$$|x| = \begin{cases} x, & x \ge 0,\\ -x, & x<0. \end{cases}$$

The expression inside the bars is the argument.

Reading the minus correctly: in the second case, $-x$ means flip the sign, not "a negative number." When $x<0$, the value $-x$ is positive: $|-4|=-(-4)=4$. The bars are a sign-removal machine that always outputs a nonnegative number.

Property	Statement	Comment
Nonnegativity	$\|x\|\ge0$, $\|x\|=0\iff x=0$	outputs never negative
Symmetry	$\|-x\|=\|x\|$	left/right look the same
Multiplicative	$\|xy\|=\|x\|\|y\|$; $\left\|\frac{x}{y}\right\|=\frac{\|x\|}{\|y\|}$	pull bars past $\times,\div$
Square root	$\|x\|=\sqrt{x^2}$	the honest $\sqrt{x^2}$
Triangle	$\|x+y\|\le\|x\|+\|y\|$	equality iff $xy\ge0$
Reverse triangle	$\big\|\|x\|-\|y\|\big\|\le\|x-y\|$	used to bound errors
Distance	$\|x-y\|=\|y-x\|$	distance is symmetric

Theorem — Multiplicative property

For all real $x,y$: $|xy|=|x||y|$.

Using $|t|=\sqrt{t^2}$: $$|xy|=\sqrt{(xy)^2}=\sqrt{x^2y^2}=\sqrt{x^2}\sqrt{y^2}=|x||y|.$$

Theorem — Triangle inequality

For all real $x,y$: $|x+y|\le|x|+|y|$.

Both sides are $\ge0$, so compare squares: $$|x+y|^2=x^2+2xy+y^2\le x^2+2|xy|+y^2=(|x|+|y|)^2,$$ since $xy\le|xy|$, with equality iff $xy\ge0$. Taking square roots preserves the inequality.

The picture to remember: $-3$ and $3$ sit at the same distance, $3$, from the origin — exactly why $|-3|=|3|$. Opposite numbers share a distance from the origin: the geometric content of $|-x|=|x|$.

A Visual Dictionary of Solution Sets

Three pictures describe every elementary absolute-value problem.

An equation is a search for exact distances. The solutions of $|x-a|=c$ (with $c>0$) are the two points at distance $c$ from $a$: one left, one right — and $a$ is their midpoint.

Solving $|x-a|=c$ with $a=3$, $c=2$: the solutions are the two points at distance 2 from 3.

Inequalities are inside/outside statements. "Closer than $c$" carves out the interval between the two points; "farther than $c$" keeps everything outside them.

Shape	Equivalence ($c>0$)	Solution set
$\|X\|<c$	$-c<X<c$ ("and")	one interval
$\|X\|>c$	$X<-c$ or $X>c$ ("or")	two rays

The graph view. Plot $y=|x-2|$ (a V). A horizontal line $y=c$ cuts the V at the exact-distance solutions; where the V dips below the line solves $|x-2|<c$, and where it rises above solves $|x-2|>c$.

How to read any solution-set picture: endpoints come from the boundary equation $|X|=c$; strict inequalities leave endpoints out (open dots, parentheses); non-strict include them (solid dots, brackets). "Smaller than" = between the endpoints; "bigger than" = beyond them.

The Master Rules — everything in one table

Rows = shape of statement; columns = sign of $c$. Read "±" as "or."

Isolate the absolute value

Rearrange until the bars stand alone: $3|2x-1|=12 \Rightarrow |2x-1|=4$.

Classify the right-hand side $c$

$c<0$: no solution ($\varnothing$). $\;c=0$: solve $E=0$. $\;c>0$: split into $E=c$ or $E=-c$.

Solve each branch & present the set

Collect the surviving values into $\{\dots\}$ or interval notation.

Two absolute values: $|A|=|B|$

$$|A|=|B| \iff A=B \ \text{or}\ A=-B.$$

Both sides are nonnegative, so this is equivalent to $A^2=B^2$, i.e. $(A-B)(A+B)=0$. Squaring is legitimate only because both sides are $\ge0$.

Special case to memorize: $|x-a|=|x-b|$ means "equidistant from $a$ and $b$," so the solution is the midpoint $\dfrac{a+b}{2}$. If the right-hand side contains the unknown ($|E|=G(x)$), a solution must first satisfy $G(x)\ge0$, and every candidate must be checked.

Where the "and" and "or" come from

Because $|X|=\max(X,-X)$, the statement $|X|<c$ demands both $X<c$ and $-X<c$; the statement $|X|>c$ is satisfied when either holds:

$$|X|<c \iff X<c\ \text{and}\ -X<c \iff -c<X<c,$$

$$|X|>c \iff X>c\ \text{or}\ -X>c \iff X>c\ \text{or}\ X<-c.$$

"Size below $c$" is a sandwich around 0 (an and); "size above $c$" escapes the sandwich on either side (an or).

The master table

Statement	$c>0$	$c=0$	$c<0$
$\|X\|=c$	$X=\pm c$	$X=0$	no solution ($\varnothing$)
$\|X\|<c$	$-c<X<c$ (and)	$\varnothing$	$\varnothing$
$\|X\|\le c$	$-c\le X\le c$ (and)	$X=0$	$\varnothing$
$\|X\|>c$	$X<-c$ or $X>c$ (or)	$X\ne0$	all reals $\mathbb R$
$\|X\|\ge c$	$X\le-c$ or $X\ge c$ (or)	$\mathbb R$	$\mathbb R$

The argument $X$ can be anything: $3x-5$, $x^2-4$, $\sqrt{x}-2$. Isolate first, then substitute.

The degenerate cells aren't trivia: $|3x+4|\le-2$ is impossible; $|x+6|>0$ excludes a single point; $|x-4|\ge0$ is solved by every real.

Interval notation: $|X|<c\to(-c,c)$ · $|X|\le c\to[-c,c]$ · $|X|>c\to(-\infty,-c)\cup(c,\infty)$ · $|X|\ge c\to(-\infty,-c]\cup[c,\infty)$.

Canonical Examples: Equations

Each is a family representative. Classify RHS first, choose the template, check if needed.

5.1 · The three RHS cases in one breath

Solve (a) $|x|=8$, (b) $|x|=0$, (c) $|x|=-3$.

(a) $c=8>0$: split → $x=8$ or $x=-8$. Set $\{-8,8\}$.
(b) $c=0$: bars vanish, $x=0$. Set $\{0\}$.
(c) $c=-3<0$: bars never output a negative → no solution, $\varnothing$.

5.2 · Linear: $|2x+1|=7$

Split: $2x+1=7\Rightarrow x=3$, or $2x+1=-7\Rightarrow x=-4$.
Check: $|2(3)+1|=7$ ✓ and $|2(-4)+1|=|-7|=7$ ✓. Solution $\{-4,3\}$.

5.3 · Isolate first: $2|4x-1|=6$

Divide by 2: $|4x-1|=3$. Split: $x=1$ or $x=-\tfrac12$.
Solution $\{-\tfrac12,1\}$.

5.4 · Two bars: $|x-1|=|x-5|$

Branch $x-1=x-5$: $-1=-5$, impossible.
Branch $x-1=-(x-5)$: $2x=6\Rightarrow x=3$.
Solution $\{3\}$ — the midpoint of 1 and 5.

5.5 · A branch becomes an identity: $|x-4|=|4-x|$

$|4-x|=|x-4|$, so the equation says a quantity equals itself.
Branch 2 collapses to $0=0$ (an identity).
Solution: every real number $\mathbb R$.

5.6 · Quadratic inside: $|x^2-4|=3$

$x^2-4=3\Rightarrow x=\pm\sqrt7$; $x^2-4=-3\Rightarrow x=\pm1$.
Solution $\{-\sqrt7,-1,1,\sqrt7\}$.

5.7 · Variable RHS — check answers! $|x+3|=x^2-4x-3$

Branch 1: $x^2-5x-6=0\Rightarrow x=6,-1$.
Branch 2: $x^2-3x=0\Rightarrow x=0,3$.
Check all four candidates:

$x$	LHS $=\|x+3\|$	RHS $=x^2-4x-3$	Verdict
$-1$	2	2	valid
$6$	9	9	valid
$0$	3	$-3$	rejected
$3$	6	$-6$	rejected

Solution $\{-1,6\}$. The rejected candidates are not optional bookkeeping — they are the answer.

5.8 · Linear, variable on right: $|x-2|=2x-5$

Need $2x-5\ge0$, i.e. $x\ge\tfrac52$.
Branch 1: $x=3$ ✓. Branch 2: $x=\tfrac73<\tfrac52$ → rejected.
Solution $\{3\}$.

Canonical Examples: Inequalities

"and" sandwiches vs "or" splits — plus the four degenerate flavors.

6.1 · Basic "and": $|x|\le2$

$-2\le x\le2$, interval $[-2,2]$. Closed brackets because non-strict.

6.2 · Basic "or": $|x+3|>4$

$x+3>4\Rightarrow x>1$, or $x+3<-4\Rightarrow x<-7$.
Solution $(-\infty,-7)\cup(1,\infty)$.

6.3 · Isolate then sandwich: $3|4x-1|\le9$

$|4x-1|\le3\Rightarrow -3\le4x-1\le3\Rightarrow-\tfrac12\le x\le1$.
Interval $[-\tfrac12,1]$.

6.4 · Sidestep the sign flip: $|2-x|<8$

Use symmetry: $|2-x|=|x-2|$, so $-8<x-2<8\Rightarrow-6<x<10$.
Interval $(-6,10)$. The reflection trick avoids flipping signs.

6.5 · Degenerate RHS, four flavors

Solve (a) $2|3x+1|-4\ge2$, (b) $|3x+4|+5\le3$, (c) $|x+6|>0$, (d) $-10+|x-4|\ge-10$.

(a) $|3x+1|\ge3$: $x\ge\tfrac23$ or $x\le-\tfrac43$ → $(-\infty,-\tfrac43]\cup[\tfrac23,\infty)$.
(b) $|3x+4|\le-2$: $\varnothing$.
(c) every $x\ne-6$: $(-\infty,-6)\cup(-6,\infty)$.
(d) $|x-4|\ge0$: $\mathbb R$.

6.6 · Variable RHS inequality: $|x-1|<2x-3$

Need $2x-3>0\Rightarrow x>\tfrac32$. Sandwich: $-(2x-3)<x-1<2x-3$.
Right: $x>2$. Left: $x>\tfrac43$. Intersect with $x>\tfrac32$: $x>2$.
Solution $(2,\infty)$.

6.7 · Two bars, square then read geometry: $|x-3|<|x+1|$

Square: $(x-3)^2<(x+1)^2\Rightarrow-8x<-8\Rightarrow x>1$.
Solution $(1,\infty)$ — points closer to 3 than to $-1$.

The Key-Point Method — several absolute values

No single "±" split works — partition the line instead.

Find the key points

Solve "argument = 0" for every bar.

Cut the line into regions

In each region every argument has a fixed sign, so drop the bars.

Solve within each region

Keep only solutions that actually lie in that region.

Take the union over regions

Combine the survivors.

7.1 · $|x-1|+|x-2|=3$

Key points $1,2$ → three regions.
$x\le1$: $3-2x=3\Rightarrow x=0$ ✓.
$1\le x\le2$: $1=3$, impossible.
$x\ge2$: $2x-3=3\Rightarrow x=3$ ✓.
Union: $\{0,3\}$.

7.2 · $|x+1|-|x-2|>1$

Key points $-1,2$.
$x\le-1$: $-3>1$ false.
$-1\le x\le2$: $2x-1>1\Rightarrow x>1$ → $(1,2]$.
$x\ge2$: $3>1$ true → $[2,\infty)$.
Union: $(1,\infty)$.

Worked Examples — a full solution & tolerances

A template-by-template solution, then the physics of "within ε."

8.1 · Solve $|2x-5|\ge3$ and graph

Strategy: isolated, RHS the positive 3 → "or" shape. Endpoints from $|2x-5|=3$.
Unfold: $2x-5\ge3\Rightarrow x\ge4$, or $2x-5\le-3\Rightarrow x\le1$.
Endpoints: non-strict, so included: $|2(1)-5|=3$ ✓, $|2(4)-5|=3$ ✓.
Checks: $x=4$: $3\ge3$ ✓. Gap $x=2.5$: $0\not\ge3$ ✓. Far $x=0$: $5\ge3$ ✓.

Answer: $(-\infty,1]\cup[4,\infty)$ — two closed rays.

The tolerance template

$$|m-T|\le\varepsilon_t \iff T-\varepsilon_t \le m \le T+\varepsilon_t.$$

Center $T$ = target, half-width $\varepsilon_t$ = tolerance. Any band $[b,c]$ has $T=\frac{b+c}{2}$ and $\varepsilon_t=\frac{c-b}{2}$ — the same center–radius picture as $\varepsilon$–$\delta$ limits.

8.2 · Band ↔ inequality & proportional tightness

(a) Temperatures acceptable in $[64,72]^\circ$F — one absolute-value inequality.
(b) Bolt acceptable iff $|d-12|\le0.2$ mm — list diameters.
(c) Which is proportionally tighter: $\pm0.5$ on 50 mm, or $\pm0.2$ on 12 mm?

(a) Center 68, half-width 4: $|T_{\text{temp}}-68|\le4$.
(b) $[11.8,12.2]$ mm.
(c) $\frac{0.5}{50}=1\%$ vs $\frac{0.2}{12}\approx1.7\%$ — the bolt is proportionally tighter.

Common Misconceptions — eight traps, eight repairs

Most lost marks live here, not in the algebra.

"$\sqrt{x^2}=x$"

No: $\sqrt{x^2}=|x|$, always nonnegative. For $x=-5$, $\sqrt{(-5)^2}=5$.

"$|x|=-x$ means a negative answer"

No: for $x\le0$, the quantity $-x$ is nonnegative. $|-3|=-(-3)=3$.

"$|x|=-5$ has solutions $\pm5$"

No: bars never output a negative → $\varnothing$. Always classify the RHS first.

Swapping "and"/"or"

"Smaller" sandwiches ($-4<x<4$, one interval); "bigger" splits ($x<-4$ or $x>4$, two rays). Hook: less thAND · greatOR.

Distributing bars over sums

$|a+b|=|a|+|b|$ is false — only $\le$ (triangle inequality) holds. Counterexample: $a=1,b=-1$ gives 0 vs 2.

Forgetting the check when RHS has $x$

For $|E|=G(x)$, candidates must pass $G(x)\ge0$. The rejected candidates are part of the answer.

Squaring blindly

$|A|=|B|\iff A^2=B^2$ is safe (both $\ge0$); $|A|=B\iff A^2=B^2$ is not (it accepts $B<0$).

Dropping a sign flip

Multiplying a chain by $-1$ reverses both directions. Safer: use $|a-b|=|b-a|$ to avoid negatives.

The 5-Step Problem-Solving Algorithm

Steps 1–3 set up; step 4 produces candidates; step 5 audits.

Isolate

Get the absolute value alone: $|E|=c$, $|E|\lessgtr c$, or two bars.

Classify the right side

Positive, zero, negative — or contains $x$? Then checking is mandatory.

Unfold

"$=$" → ± split; "$<,\le$" → sandwich "and"; "$>,\ge$" → two-ray "or"; several bars → key-point method.

Solve each branch

Intersect the "and" pieces, union the "or" pieces, discard invalid candidates.

Verify

Graph it, write interval notation, plug endpoints into $|E|=c$, spot-check one interior point.

The two checks that pay for themselves: Endpoint check — substitute each endpoint into $|E|=c$. Spot check — pick one number inside the set and one outside; both must behave as claimed.

Computational Exploration

Draw the pictures in one line, and get an independent verification channel.

Computers know absolute value as abs(x), and symbolic engines solve it directly. All outputs below were produced with numpy, sympy and matplotlib.

11.1 · Seeing the solution set

import numpy as np, matplotlib.pyplot as plt
f = lambda x: np.abs(x - 2)              # the V-shaped function
x = np.linspace(-6, 10, 800); xs = np.linspace(-1, 5, 200)
plt.plot(x, f(x), lw=2, color="teal")
plt.axhline(3.0, color="tab:red", ls="--")      # y = 3
plt.fill_between(xs, f(xs), 3.0, color="tab:red", alpha=0.15)  # |x-2| < 3
plt.plot([-1,5],[3,3],"o", color="tab:red")        # |x-2| = 3
plt.show()

The dashed line $y=3$ meets the V at $(-1,3)$ and $(5,3)$; the shaded band marks $-1<x<5$.

11.2 · Truth tables without proofs

x = np.linspace(-10, 10, 2_000_001)
# Claim: |2x-5| >= 3  <=>  x <= 1 or x >= 4
lhs   = np.abs(2*x - 5) >= 3
claim = (x <= 1) | (x >= 4)
print("|2x-5| >= 3 mismatches:", np.count_nonzero(lhs != claim))

|2x-5| >= 3 mismatches: 0

Two million sample points, zero disagreements. Scanning isn't a proof, but it catches sign slips and flipped endpoints instantly.

11.3 · Symbolic solving & extraneous-filtering

import sympy as sp
x = sp.symbols("x", real=True)
print(sp.solveset(sp.Abs(x-3) - 2, x, domain=sp.S.Reals))           # {1, 5}
print(sp.solveset(sp.Abs(x-1) - sp.Abs(x-5), x, domain=sp.S.Reals)) # {3}
print(sp.solveset(sp.Abs(x+3) - (x**2-4*x-3), x, domain=sp.S.Reals))# {-1, 6}

The last line returns $\{-1,6\}$ — sympy silently discarded the extraneous candidates 0 and 3.

11.4 · "Check your answers" as code

candidates = [-1, 6, 0, 3]   # from |x+3| = x^2-4x-3
for x in candidates:
    lhs = abs(x + 3); rhs = x**2 - 4*x - 3
    print(x, lhs, rhs, lhs == rhs)

-1  2   2   True
6  9   9   True
0  3  -3  False
3  6  -6  False

Verification workflow: (1) solve symbolically; (2) confirm with a truth table or substitution; (3) draw the picture — if picture and symbols disagree, it's usually the symbols.

Connections — where absolute value lives

The same idea, wearing different clothes across mathematics.

📏 Distance geometry

$|x-y|$ is the distance between two points. $|x-a|=|x-b|$ gives the midpoint; in the plane it upgrades to the perpendicular bisector.

∞ Limits (ε–δ)

$\lim_{x\to a}f(x)=L$ is defined by $|x-a|<\delta$ and $|f(x)-L|<\varepsilon$ — the tolerance band with different letters.

🔬 Measurement science

Every tolerance, error bar, and QC limit is $|\text{measured}-\text{target}|\le\varepsilon$.

📊 Statistics

The absolute deviation $|x-\bar x|$ measures how far a point sits from the mean — the robust cousin of standard deviation.

📈 Functions & graphs

$y=|x|$ is the simplest piecewise function and the prototype V-shape; $y=a|x-h|+k$ has vertex $(h,k)$ and slopes $\pm a$.

🧭 Higher dimensions

$|x|$ is the 1-D case of a norm. For a complex number $z$, the modulus $|z|$ plays the role of absolute value.

💻 Programming

abs(x) is native in every language; the sample-compare-count truth-table style is how you test claims about signs and boundaries.

Practice Set — cover the answers, work on paper

Compare your checking, not just your final answer.

      Level 1 · Concepts
      Level 2 · Core skills
      Level 3 · Challenge
    

Level 1 — Concepts

C1 Explain in one sentence why $|x-a|<c$ describes an interval centered at $a$, and state its endpoints.concept

$|x-a|$ is the distance from $x$ to $a$; keeping it below $c$ keeps $x$ within $c$ of $a$. Endpoints: $a-c$ and $a+c$.

C2 True or false: (a) $|x|=-x$ for every $x\le0$; (b) $|a+b|=|a|+|b|$ for all reals; (c) $\sqrt{x^2}=|x|$ for all reals.true/false

(a) True. (b) False: $a=1,b=-1$ gives 0 versus 2. (c) True.

C3 Explain why $|x|=-4$ has no solutions while $|x|>-4$ holds for every real $x$.concept

$|x|\ge0$ always; a nonnegative quantity can never equal $-4$, but is always greater than $-4$.

C4 Without solving, how many solutions can $|2x-7|=5$ have? What about $=0$ and $=-1$?predict

Two; one ($x=7/2$); none.

C5 A student writes "$|x|>3$, so $x>\pm3$". Diagnose and repair.diagnose

"$x>\pm3$" is meaningless; correct is $x>3$ or $x<-3$.

C6 Describe the graph of $y=|2x-5|$: vertex, slopes, and where it lies on or above $y=3$.graph

V-shape with vertex $(\tfrac52,0)$ and slopes $\pm2$; meets $y=3$ at $x=1,4$ and lies above for $x\le1$ or $x\ge4$.

Level 2 — Core skills

S1 Solve $|x+7|=3$.equation

$\{-10,-4\}$.

S2 Solve $3|2x-4|=18$.equation

$|2x-4|=6$; $x=5$ or $x=-1$.

S3 Solve $|x^2-1|=3$.nonlinear

$x^2=4\Rightarrow x=\pm2$; $x^2=-2$: none. $\{-2,2\}$.

S4 Solve $|x-4|+2=1$.impossible

$|x-4|=-1$: $\varnothing$.

S5 Solve $-3|x-1|-6=3$.impossible

$|x-1|=-3$: $\varnothing$.

S6 Solve $|x-1|\le4$; give the interval.and

$[-3,5]$.

S7 Solve $|2x+6|>10$; give the interval.or

$(-\infty,-8)\cup(2,\infty)$.

S8 Solve $|5-x|\ge1$; give the interval.reflection

$(-\infty,4]\cup[6,\infty)$.

S9 Solve $2|x-3|<0$.impossible

$|x-3|\ge0$ always: $\varnothing$.

S10 Solve $|x-7|\ge-2$.all reals

$\mathbb R$.

S11 A bottling machine is acceptable when $|V-500|\le3$ (mL). Find the range.tolerance

$[497,503]$ mL.

Level 3 — Challenge

H1 Solve $|x-2|=3x+1$ (watch the RHS).variable RHS

Need $3x+1\ge0$. Branch $x=-\tfrac32$: rejected (RHS $<0$). Branch $x=\tfrac14$: valid. Solution $\{\tfrac14\}$.

H2 Solve $|x^2-1|=3$.nonlinear

$\{-2,2\}$.

H3 Solve $|x-3|+|x+1|=8$.key points

Key points $-1,3$. Solutions $\{-3,5\}$.

H4 Solve $|x-1|\ge|x+5|$.square

Square: $-12x\ge24\Rightarrow x\le-2$. Solution $(-\infty,-2]$.

H5 Prove the reverse triangle inequality $\big||x|-|y|\big|\le|x-y|$.proof

From $|x|\le|x-y|+|y|$ get $|x|-|y|\le|x-y|$; swapping gives $|y|-|x|\le|x-y|$. Together: $\big||x|-|y|\big|\le|x-y|$.

H6 Solve $\left|\tfrac{x}{3}+2\right|\le1$; give the interval.fraction

$[-9,-3]$.

Quick Drill — 25 mixed problems (answer key)

#	Problem	Answer	Type
1	$\|x\|=8$	$\{-8,8\}$	eq, +
2	$\|x-2\|=6$	$\{-4,8\}$	eq, +
3	$\|x+1\|=0$	$\{-1\}$	eq, zero
4	$\|x-4\|=-6$	No solution	eq, −
5	$\|3x+2\|=10$	$\{-4,\tfrac83\}$	eq, +
6	$\|2x+5\|+4=3$	No solution	eq, −
7	$2\|4x-1\|=6$	$\{-\tfrac12,1\}$	eq, +
8	$\tfrac12\|2x-6\|+1=2$	$\{2,4\}$	eq, +
9	$-3\|x-1\|-6=3$	No solution	eq, −
10	$\|x-7\|+2=2$	$\{7\}$	eq, zero
11	$\|3x+2\|=\|x-6\|$	$\{-4,1\}$	two bars
12	$\|x-4\|=\|4-x\|$	$\mathbb R$	two bars
13	$\|x\|\le2$	$[-2,2]$	ineq, +
14	$\|x+3\|>4$	$(-\infty,-7)\cup(1,\infty)$	ineq, +
15	$\|x+3\|<-6$	No solution	ineq, −
16	$3\|2x-4\|\ge-9$	$\mathbb R$	ineq, −
17	$2\|x-9\|+6>6$	$(-\infty,9)\cup(9,\infty)$	ineq, zero
18	$-4\|3x-1\|\ge8$	No solution	ineq, −
19	$-5\|2x+2\|-3\ge-3$	$\{-1\}$	ineq, zero
20	$-10+\tfrac56\|x-4\|\ge-10$	$\mathbb R$	ineq, zero
21	$3\|\tfrac56x+2\|+6<15$	$(-6,\tfrac65)$	ineq, +
22	$\|x^2-4\|=-4$	No solution	nonlinear
23	$\|x^2+9x+14\|=0$	$x=-2,-7$	nonlinear
24	$\|x^2+1\|=2x$	$x=1$	nonlinear
25	$\|x-1\|=x^2+4x-5$	$x=-6,1$	nonlinear

Study Plan & Exam Checklist

A schedule to own the chapter, then a checklist to run on every problem.

⏱ First 30 minutes

Read Sections 1–4 with a pencil. Redraw the equation/inequality pictures from memory, then write the master table once without looking. Do concept checks C1–C4.

⏱ Next 90 minutes

Work through Sections 5–7 actively: cover each solution, attempt it, then compare checks. Then do S1–S11 and H1–H3.

📅 Day two

Redo three problems you got wrong, from scratch. Draw the V-graph and key-point figures by hand. Run the four code listings (or read the outputs and match them to the worked examples).

🗓 End of week

Attempt H4–H6. Write the summary card from memory. Re-derive the "and"/"or" rules from $|X|=\max(X,-X)$ and the midpoint trick.

Exam Checklist — tick each before submitting

0 / 9 confirmed

The absolute value was isolated before unfolding. The right-hand side was classified: positive, zero, negative, or containing $x$. "$<,\le$" produced one interval ("and"); "$>,\ge$" produced two rays ("or"). Every candidate was checked whenever the right-hand side contained $x$. Endpoints were tested in $|E|=c$ and matched the strictness (open vs closed). The final answer is in interval notation (or a brace list for equations). A spot-check number inside and outside the set behaves correctly. Degenerate cases were spotted: $\varnothing$, a single point, or all of $\mathbb R$. No "$x>\pm c$" junk notation survived to the final line.

Summary Card & Further Resources

Everything on one card — reproduce it from memory and you own the chapter.

Absolute value in one card

Definition: $|x|=x$ if $x\ge0$; $|x|=-x$ if $x<0$; outputs are never negative.
One idea: $|x-a|$ is the distance from $x$ to $a$; everything else is geometry.
Equations: $|E|=c$ with $c>0$ splits $E=\pm c$; $c=0$ gives $E=0$; $c<0$ gives $\varnothing$.
Two bars: $|A|=|B|\iff A=\pm B$; special case $|x-a|=|x-b|$ → midpoint $\frac{a+b}{2}$.
Inequalities ($c>0$): $|E|<c$ sandwiches ("and"); $|E|>c$ splits ("or").
Degenerate: $|E|<0:\varnothing$; $|E|\le0:E=0$; $|E|>0$: all but $E=0$; $|E|\ge0$: all reals.
Variable right side: require the correct sign, then check every candidate.
Several bars: key points partition the line — unfold, solve per region, keep in-region, union.
Tolerance: $|m-T|\le\varepsilon_t$ is the band $[T-\varepsilon_t,\,T+\varepsilon_t]$ — the same center–radius picture as $\varepsilon$–$\delta$ limits.
Verify: endpoints in $|E|=c$, one spot value, and a truth-table or solveset run when a machine is available.

Further resources

📕

Stewart, Calculus — Appendix A (numbers, inequalities, absolute values): the same rules in textbook form.

📗

Spivak, Calculus, Ch. 1 — inequality exercises built almost entirely on absolute values; excellent for proof-flavored practice (H5 style).

📘

OpenStax, College Algebra 2e — equations & inequalities chapters, including graph interpretations.

📝

Paul's Online Math Notes — concise "Absolute Value Equations/Inequalities" pages with worked problem sets.

🎬

Khan Academy — "Absolute value equations" and "Absolute value functions" units for quick video review.

📈

Desmos / GeoGebra — type abs(x-2) and a slider for the right-hand side; watching the intersections slide builds the V-graph intuition permanently.

Built from "Absolute Value: Equations & Inequalities" (Calculus Deep Dives) & the companion worked-problem worksheet · Distance-first study guide