Porisma

Original question & working

Absolute values and the correct inverse branch · source reference

[2011/AJC/Promo/8] · Question pp. 3 · Working pp. 12, 13

Open the interactive lesson

The working and teaching notes are included in the interactive lesson.

View source review notes
Full original question, working & notes

Complete question

For real x≠−1/2x\ne-1/2, f(x)=∣(1−2x)/(2x+1)∣f(x)=|(1-2x)/(2x+1)|; for all real x, g(x)=tan⁡−1(2x)g(x)=\tan^{-1}(2x). (i) Show f is not one-one. Find the least k such that restricting the domain to [k,∞)[k,\infty) makes it one-one. [2] On this restricted domain, (ii) define the inverse of f. [3] (iii) Find the exact range of gf−1gf^{-1}. [2]

[2011/AJC/Promo/8]

Printed working

Part (i)

The horizontal line y=3y=3 cuts the graph of ff at more than one point (two points), so ff is not one-one. Least k=1/2k=1/2.

The graph labels vertical asymptote x=−1/2x=-1/2, horizontal asymptote y=1y=1, cusp (1/2,0)(1/2,0) and the counter-example line y=3y=3. Its two intersections are at x=−1x=-1 and x=−1/4x=-1/4, both giving3.

Part (ii)

Let y=∣(1−2x)/(2x+1)∣y=|(1-2x)/(2x+1)|, x≥1/2x\ge1/2:

y=−1−2x2x+1=2x−12x+1=1−22x+122x+1=1−yx=12(21−y−1)=−12+11−y.\begin{aligned} y&=-\frac{1-2x}{2x+1}\\ &=\frac{2x-1}{2x+1}=1-\frac2{2x+1}\\ \frac2{2x+1}&=1-y\\ x&=\frac12\left(\frac2{1-y}-1\right)=-\frac12+\frac1{1-y}. \end{aligned}

Therefore f−1(x)=−1/2+1/(1−x)f^{-1}(x)=-1/2+1/(1-x), Df−1=Rf=[0,1)D_{f^{-1}}=R_f=[0,1).

Thus f−1:x↦−1/2+1/(1−x)f^{-1}:x\mapsto-1/2+1/(1-x), 0≤x<10\le x<1.

Part (iii)

Rf−1=Df=[1/2,∞)→gRgf−1=[π/4,π/2).R_{f^{-1}}=D_f=[1/2,\infty)\xrightarrow{g}R_{gf^{-1}}=[\pi/4,\pi/2).

Thus Rgf−1=[π/4,π/2)R_{gf^{-1}}=[\pi/4,\pi/2). The source arctangent graph labels both horizontal asymptotes y=±π/2y=\pm\pi/2, point (1/2,π/4)(1/2,\pi/4), restricted domain ray Rf−1R_{f^{-1}} beginning at 1/21/2 and range with closed lower and open upper endpoints.

Teacher margin notes

For the graphical method, draw horizontal line y=3y=3 on the same diagram as the graph of ff, showing it cuts that graph at two points.

Note: (1) To show ff is not1–1, only give an example of one particular horizontal line cutting its graph more than once. (2) Either write “graph of f”, “graph of y=f(x)”, or “the curve y=f(x)”.

The source says for x≥1/2x\ge1/2, 1−2x<01-2x<0, 2x+1>02x+1>0, so ∣1−2x∣=−(1−2x)|1-2x|=-(1-2x) and ∣2x+1∣=2x+1|2x+1|=2x+1, hence ∣(1−2x)/(2x+1)∣=∣1−2x∣/∣2x+1∣=−(1−2x)/(2x+1)|(1-2x)/(2x+1)|=|1-2x|/|2x+1|=-(1-2x)/(2x+1). At the included endpoint x=1/2x=1/2, the first strict sign should be ≤0\le0; the displayed absolute-value identity and inverse remain valid there.

Source diagrams

Complete source absolute rational graph, both asymptotes, cusp1/2 and y=3 counter-example. Source arctangent curve y=arctan(2x), ±π/2 limits, point(1/2,π/4) and domain/range endpoint rays.

Provenance

Question page references: 3. Solution page references: 12, 13. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.