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

If this part uses an earlier result, revisit (i) One-one restrictions or open all working together.

(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

Open interactive exploration

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?