Absolute values and the correct inverse branch

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. marks On this restricted domain, (ii) define the inverse of f. marks

(iii) Find the exact range of gf−1gf^{-1}. marks

[2011/AJC/Promo/8]

Total marks: 7

Solution

(i)

One-one restrictions

For real x≠−12x\ne-\frac12, f(x)=∣1−2x2x+1∣f(x)=\left|\dfrac{1-2x}{2x+1}\right|. Show that ff is not one-one, and find the least kk such that restricting the domain to [k,∞)[k,\infty) makes ff one-one.

Explore this idea

Hint

  1. Find two inputs with the same output.

  2. Look for the turning point: where does the graph change direction?

Answer · (i)

f(−1)=f(−1/4)=3f(-1)=f(-1/4)=3; least k=1/2k=1/2.

Explore the idea

Open interactive exploration

The horizontal line test on |(1 − 2x)/(2x + 1)|

At y = 3 the line meets the graph twice: at x = −1 and x = −¼. The graph rises monotonically only for x ≥ ½.

Height of the horizontal linek = 3
Try an example
Test boundary heights

Buttons use full-precision boundary values; numbers shown here are rounded.

Number of distinct intersections
2

Two intersections, one on each side of the vertical asymptote.

All listed intersections are within the original drawing window. A touching point counts once; hollow endpoints are excluded.

Coordinates are rounded to 6 significant figures. Inspect a row to highlight that intersection and bring it into view.
Pointxy
-13
-0.253
  • F
  • Y = k
  • Intersections
  • Intersection selected in the coordinate table
Key idea

Because the graph folds at the cusp (12,0)(\frac12,0), only the branch x≥12x\ge\frac12 is one-one.

Ideas to try
Which values of kk give exactly one intersection? Which kk would you get if you used only x≥12x\ge\frac12?