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Worked solution

Compare absolute-value graphs in two ways

Find the solution sets of the following inequalities.

(a) x2<∣x−2∣x^2<|x-2|,

(b) ∣x+13x−5∣>1\left|\dfrac{x+1}{3x-5}\right|>1.

CJC · Review Problems · Question 10

Answer

(a) (−2,1)(-2,1); (b) (1,5/3)∪(5/3,3)(1,5/3)\cup(5/3,3).

(a) Method 1: one graph

Sketch y=x2−∣x−2∣y=x^2-|x-2|. The graph crosses the axis at x=−2,1x=-2,1 and has the source-labelled minimum (−12,−94)(-\frac12,-\frac94). It lies below the axis exactly for −2<x<1-2<x<1.

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(a) Method 2: two graphs

Sketch y=x2y=x^2 and y=∣x−2∣y=|x-2| on the same axes. They intersect at (−2,4)(-2,4) and (1,1)(1,1); the absolute-value graph has vertex (2,0)(2,0). The parabola lies below the absolute-value graph between the intersections, so the set is {x∈R:−2<x<1}\{x\in\mathbb R:-2<x<1\}.

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(b) Method 1: one graph

Sketch y=∣x+13x−5∣−1y=\left|\dfrac{x+1}{3x-5}\right|-1. Its domain excludes x=53x=\frac53, its vertical asymptote is x=53x=\frac53, and its horizontal asymptote is y=−23y=-\frac23. It has minimum (−1,−1)(-1,-1) and zeros 1,31,3. It is positive between 1 and 3, with the pole removed: {x∈R:1<x<3, x≠53}\{x\in\mathbb R:1<x<3,\ x\ne\frac53\}.

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(b) Method 2: two graphs

Compare y=∣x+13x−5∣y=\left|\dfrac{x+1}{3x-5}\right| with y=1y=1. The ratio has minimum (−1,0)(-1,0), horizontal asymptote y=13y=\frac13 and vertical asymptote x=53x=\frac53. The intersections with y=1y=1 occur at x=1,3x=1,3; the ratio is larger on (1,53)∪(53,3)(1,\frac53)\cup(\frac53,3).

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Compare your sketch

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Four panels preserve both methods for each part. For (a), the difference has zeros −2 and 1 and minimum (−1/2,−9/4); the two graphs intersect at (−2,4) and (1,1), with the V vertex at (2,0). For (b), the difference has zeros 1 and 3, minimum (−1,−1), horizontal asymptote −2/3 and vertical asymptote 5/3. The unshifted ratio has minimum (−1,0), horizontal asymptote 1/3, the same pole and crossings with y=1 at 1 and 3.
Four panels preserve both methods for each part. For (a), the difference has zeros −2 and 1 and minimum (−1/2,−9/4); the two graphs intersect at (−2,4) and (1,1), with the V vertex at (2,0). For (b), the difference has zeros 1 and 3, minimum (−1,−1), horizontal asymptote −2/3 and vertical asymptote 5/3. The unshifted ratio has minimum (−1,0), horizontal asymptote 1/3, the same pole and crossings with y=1 at 1 and 3.
Four panels preserve both methods for each part. For (a), the difference has zeros −2 and 1 and minimum (−1/2,−9/4); the two graphs intersect at (−2,4) and (1,1), with the V vertex at (2,0). For (b), the difference has zeros 1 and 3, minimum (−1,−1), horizontal asymptote −2/3 and vertical asymptote 5/3. The unshifted ratio has minimum (−1,0), horizontal asymptote 1/3, the same pole and crossings with y=1 at 1 and 3.

Common mistakes

An interval bounded by two intersections can still contain a forbidden denominator input.

Solution notes

The diagrams are independently reconstructed mathematical equivalents. The plotted geometry is illustrative; use the labelled exact or approximate values and the stated domain.

The first printed graph for (b) appears to label its left zero as 2; the printed final set and second graph give 1. Solving |x+1|=|3x−5| gives 1 and 3, so the reconstruction labels 1. No marks are printed for Q10.

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