Exercises — Absolute value - modulus — definition, number line interpretation

Apply the definition: is the inside $\ge 0$ or $<0$?

• $|9|=9$ — already non-negative, keep it.
• $|-9|=-(-9)=9$ — negative, flip the sign.
• $|0|=0$ — zero is its own distance from zero.
• $\left|-\frac{3}{4}\right|=\frac{3}{4}$ — negative, flip.
• $|3-10|=|-7|=7$ — first simplify inside the bars ($3-10=-7$), then take absolute value.

The last one is the whole trap of Level 1: the bars act on the result, not on each piece.

• (a) True. Both branches of the definition output something non-negative.
• (b) False. Fails for negatives: $|-3|=3\ne -3$. It's only true when $x\ge 0$.
• (c) True. Reflecting across $0$ keeps the same distance. We must check both branches:
  • If $x\ge 0$: then $-x\le 0$, so by the definition $|-x|=-(-x)=x$. And $|x|=x$. Equal. ✓
  • If $x<0$: then $-x>0$, so $|-x|=-x$. And $|x|=-x$ (flip). Equal. ✓
  Both cases give $|-x|=|x|$, so it holds for every real $x$.
• (d) True. Only the point at distance $0$ from $0$ is $0$ itself.

Distance is $|b-a|$, and the order does not matter. $|11-(-6)|=|11+6|=|17|=17.$ $|-2-(-6)|=|-2+6|=|4|=4.$ Sanity check (first one): from $-6$ up to $0$ is $6$, from $0$ up to $11$ is $11$; total $6+11=17$. ✓

Read it as "the distance from $x$ to $4$ is $6$." You can sit $6$ units to the right of $4$ or $6$ units to the left.

• Right: $x-4=6\Rightarrow x=10$.
• Left: $x-4=-6\Rightarrow x=-2$.

Check: $|10-4|=|6|=6$ ✓ and $|-2-4|=|-6|=6$ ✓. See Solving absolute value equations.

The bars measure the distance of the whole quantity $2x+1$ from $0$; that distance is $9$, so $2x+1$ is $+9$ or $-9$.

• $2x+1=9\Rightarrow 2x=8\Rightarrow x=4$.
• $2x+1=-9\Rightarrow 2x=-10\Rightarrow x=-5$.

Check: $|2(4)+1|=|9|=9$ ✓, and $|2(-5)+1|=|-9|=9$ ✓.

"Distance from $x$ to $3$ is less than $5$" — so $x$ lives inside the band (the top strip of the figure), radius $5$ centred at $3$. Using the rule $|A|<b\iff -b<A<b$ with $A=x-3$ and $b=5$: $-5<x-3<5.$ Now add $3$ to all three parts of the chain (the same move applied everywhere keeps it valid): $-2<x<8.$ Interval: $(-2,8)$. The centre is $3$, the endpoints are $3\pm 5$ — exactly the two open circles in the figure. See Inequalities.

"Distance from $x$ to $-2$ is at least $4$" — so $x$ is outside (or on the edge of) the band of radius $4$ around $-2$; this is the two amber rays in the bottom strip of the figure. By the "outside" rule $|A|\ge b\iff A\ge b\text{ or }A\le -b$ with $A=x+2$, $b=4$: $x+2\ge 4\text{or}x+2\le -4.$ $x\ge 2\text{or}x\le -6.$ Solution: $(-\infty ,-6]\cup [2,\infty )$. Check an endpoint: $x=2$: $|2+2|=4\ge 4$ ✓. $x=-6$: $|-6+2|=|-4|=4\ge 4$ ✓.

Read geometrically: "$x$ is equally far from $1$ and from $-3$." The point equally distant from two points is their midpoint: $x=\frac{1+(-3)}{2}=\frac{-2}{2}=-1.$ Algebra confirms it. Squaring both sides (legal since both are $\ge 0$) removes the bars: $(x-1{)}^2=(x+3{)}^2\Rightarrow x^2-2x+1=x^2+6x+9\Rightarrow -8x=8\Rightarrow x=-1.$ Check: $|-1-1|=2$ and $|-1+3|=2$. ✓

The two bars "switch" at their inside-zeros: $x=2$ and $x=-1$. These cut the line into three regions. In each, decide the sign of each inside, then apply the definition. See Piecewise functions.

Region A: $x<-1$. Here $x-2<0$ and $x+1<0$, so flip both: $f=-(x-2)-(x+1)=(-x+2)+(-x-1)=-2x+1.$ Region B: $-1\le x<2$. Here $x-2<0$ (flip) but $x+1\ge 0$ (keep). Watch the $x$-terms cancel: $f=-(x-2)+(x+1)=(-x+2)+(x+1)=\underset{=0}{\underbrace{(-x+x)}}+(2+1)=3.$ The $-x$ from the flipped first bar exactly kills the $+x$ from the kept second bar — that is why the function is flat here, not just that it equals $3$. Region C: $x\ge 2$. Both non-negative, keep both: $f=(x-2)+(x+1)=2x-1.$ So $f(x)=\{\begin{array}{ll}-2x+1& x<-1\\ 3& -1\le x<2\\ 2x-1& x\ge 2.\end{array}$ Minimum: on region A, $-2x+1$ decreases toward $x=-1$, reaching $3$. On region C, $2x-1$ increases from $x=2$, starting at $3$. On region B it is constant at $3$. So the minimum value is $\boxed{3}$, achieved for every $x$ in $[-1,2]$.

Meaning: $f(x)$ is the total distance from $x$ to the two points $2$ and $-1$. That total is smallest — and equal to the gap between them, $|2-(-1)|=3$ — whenever $x$ sits between them.

$|x+y|=|4+(-9)|=|-5|=5,|x|+|y|=4+9=13.$ Indeed $5\le 13$. ✓ Why strict here: $x$ and $y$ point in opposite directions (one positive, one negative), so they partly cancel — you walked $4$ right then $9$ left, net displacement $5$ but total path $13$. Equality holds only when $x$ and $y$ have the same sign (or one is $0$): then nothing cancels, e.g. $x=4,y=9$ gives $|13|=13=4+9$.

Split on the sign of $x$, since $|x|$ has two faces.

• Case $x\ge 0$: $|x|=x$, so $x+x=8\Rightarrow 2x=8\Rightarrow x=4$. It satisfies $x\ge 0$ ✓.
• Case $x<0$: $|x|=-x$, so $-x+x=8\Rightarrow 0=8$, impossible. No solution here.

Only solution: $x=4$. Deeper reason: for $x<0$, $|x|=-x$ exactly cancels the $+x$, so the left side is always $0$ — it can never reach $8$. So negative inputs are structurally ruled out. Check: $|4|+4=8$ ✓.

Two exhaustive cases cover all reals.

• If $x\ge 0$: $|x|=x$, so $|x|\ge x$ holds with equality.
• If $x<0$: $|x|=-x$, which is positive, while $x$ is negative. A positive number is greater than a negative one, so $|x|>x$ (strict).

Together: $|x|\ge x$ for all real $x$, with equality iff $x\ge 0$. $■$

Handle each constraint as a set, then intersect. First: $|x-5|\le 3$ means distance from $5$ is at most $3$: $5-3\le x\le 5+3$, i.e. $[2,8]$. Second: $|x|>4$ means distance from $0$ exceeds $4$: $x<-4$ or $x>4$, i.e. $(-\infty ,-4)\cup (4,\infty )$. Intersect $[2,8]$ with that: the part of $[2,8]$ where also $x>4$ is $(4,8]$; the part where $x<-4$ is empty. Answer: $(4,8]$. Check the edges: $x=8$: $|8-5|=3\le 3$ ✓ and $|8|=8>4$ ✓. Just above $4$, say $x=4.1$: $|4.1-5|=0.9\le 3$ ✓ and $|4.1|>4$ ✓. At $x=4$ exactly, $|x|>4$ fails, so $4$ is excluded — hence the open bracket.