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

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?
(ii)

Inverse of a restricted absolute value

On the restricted domain [k,∞)[k,\infty), define f−1f^{-1}.

Explore this idea

Hint

  1. For x≥12x\ge\frac12, 1−2x≤01-2x\le0 so ∣1−2x∣=2x−1|1-2x|=2x-1.

  2. Write y=2x−12x+1y=\frac{2x-1}{2x+1} and solve for xx.

  3. Domain of f−1f^{-1} is the range of the restricted ff.

Answer · (ii)

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

Explore the idea

The rising branch and its inverse

f(x) = (2x − 1)/(2x + 1) on x ≥ ½ rises from 0 towards 1. Its inverse is a mirror image with a vertical asymptote x = 1.

Point on f: input xx = 2.00
Reset the controls

Reflection swaps the coordinates. The segment PQ is perpendicular to y = x, and its blue midpoint M lies on that line.

P = (x, f(x))
(2, 0.6)
Q = (f(x), x)
(0.6, 2)
on the inverse
Domain of f
x≥12x\ge\frac12
Range of f = domain of f⁻¹
0≤y<10\le y<1
Midpoint M on y = x
(1.3, 1.3)
  • F restricted to x ≥ ½
  • F⁻¹
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

ff's horizontal asymptote y=1y=1 becomes f−1f^{-1}'s vertical asymptote x=1x=1, which is why its domain stops short of 1.

Ideas to try
Where is f−1f^{-1}'s vertical asymptote? Where is ff's horizontal asymptote?
(iii)

Range of a composite

g(x)=tan⁡−1(2x)g(x)=\tan^{-1}(2x) for all real xx. Find the exact range of gf−1gf^{-1}.

Explore this idea

Hint

  1. f−1f^{-1} outputs x≥12x\ge\frac12.

  2. Apply gg to that set and use the shape of tan⁡−1\tan^{-1}.

Answer · (iii)

Rgf−1=[π/4,π/2)R_{gf^{-1}}=[\pi/4,\pi/2).

Explore the idea

tan⁻¹(2x) on x ≥ ½

Only the part of the arctangent graph to the right of x = ½ is used.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • G on the part of its domain used
  • Asymptote π/2
Ideas to try
What would the range be if f−1f^{-1} output all real numbers?