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 (ii) Inverse of a restricted absolute value or open all working together.

(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

Open interactive exploration

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?