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Worked solution

Choose inside or outside intervals for a modulus comparison

Solve (a) ∣x−3∣<4|x-3|<4 and (b) ∣2x+1∣≥4|2x+1|\ge4.

CJC · Lecture · Example 8

(a) Method 1: remove modulus as a double inequality

∣x−3∣<4|x-3|<4 is equivalent to −4<x−3<4-4<x-3<4. Add 3 to every part: −1<x<7.

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(a) Method 2: compare two graphs

Sketch y=|x−3| and y=4. The V has vertex (3,0), y-intercept 3 and intersections at x=−1,7. Read where the V lies below the horizontal line: −1<x<7. Strict comparison excludes both crossings.

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(b) Method 1: split into two outside cases

∣2x+1∣≥4|2x+1|\ge4 means 2x+1≤−42x+1\le-4 or 2x+1≥42x+1\ge4. Hence x≤−5/2x\le-5/2 or x≥3/2x\ge3/2.

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(b) Method 2: compare two graphs

Sketch y=|2 x+1| and y=4. The vertex is (−1/2,0), the y-intercept is 1 and the intersection inputs are −5/2,3/2. Read where the V is on or above the horizontal line: x≤−5/2 or x≥3/2. Both crossings are included.

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Both original V graphs and level 4 are retained, including vertices 3 and −1/2, y-intercepts 3 and 1, crossings −1,7 and −5/2,3/2, and the distinction between strict inside and non-strict outside solutions.
Both original V graphs and level 4 are retained, including vertices 3 and −1/2, y-intercepts 3 and 1, crossings −1,7 and −5/2,3/2, and the distinction between strict inside and non-strict outside solutions.
Both original V graphs and level 4 are retained, including vertices 3 and −1/2, y-intercepts 3 and 1, crossings −1,7 and −5/2,3/2, and the distinction between strict inside and non-strict outside solutions.

Common mistakes

A modulus less than a positive level gives an inside interval; greater than or equal gives two outside intervals with included endpoints.

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