Quick Summary

f(x)=(x−1)/(x+1) is a self-inverse-flavoured Möbius map: computing f², f³, f⁴ by hand shows f⁴(x)=x — the identity — so f has period 4 under composition. Since 26 = 4×6+2, f²⁶ = f², and f²(x) = −1/x. So g(x) = −f²⁶(x) = 1/x. The region bounded by y=g(x), the line y=x−3/2, the x-axis, and x=4 is the area lying below BOTH curves simultaneously — the upper boundary switches from the line to g(x) at their intersection x=2. Integrating piece by piece gives exactly 1/8 + ln2.

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Why This Answer

The 26th iterate of f looks intimidating, but Möbius-type maps like (x−1)/(x+1) very often cycle back to the identity after a small number of compositions — testing f², f³, f⁴ by hand (rather than trying to guess a closed formula for fⁿ) reveals the period-4 structure almost immediately. Once that shortcut is spotted, the problem reduces entirely to standard area-between-curves work.

The area itself is the subtlest part: with g(x)=1/x decreasing and the line y=x−3/2 increasing, crossing once at x=2, the region "below both curves, above the x-axis, left of x=4" is bounded above by whichever curve is LOWER at each x — the line for x∈[1.5,2] (since g>line there) and g(x) for x∈[2,4] (since line>g there). Missing this switch and integrating the wrong curve on the wrong sub-interval is the single most likely place to lose the question.

Core Concept

Cycle-finding for iterated Möbius maps. For f(x)=(ax+b)/(cx+d), repeated composition frequently returns to the identity after a small period. Rather than deriving a general formula for fⁿ, compute f², f³, f⁴, … directly until a pattern (often the identity) appears, then reduce n modulo that period.

Area below two curves and above an axis. When a region is described as bounded by two curves, the axis, and a vertical line, sketch (or reason through) which curve is lower on each sub-interval — the "area below both" is a two-piece integral, switching integrands exactly at the curves' intersection point.

Step-by-Step Solution

1. Find the composition period. f(x)=(x−1)/(x+1). Compute f²(x)=f(f(x)) = −1/x, f³(x)=−(x+1)/(x−1), f⁴(x)=x. So f has period 4.

2. Reduce the index. 26 = 4×6+2, so f²⁶ = f² = −1/x. Then g(x) = −f²⁶(x) = 1/x.

3. Find where g meets the line y=x−3/2. Set 1/x = x−3/2 ⇒ x²−1.5x−1=0 ⇒ x=2 (taking the positive root relevant to x>1).

4. Find where the line meets y=0: x−3/2=0 ⇒ x=1.5.

5. Identify the lower envelope on each piece. For x∈(1.5,2), g(x)>line(x) (e.g. at x=1.75: g≈0.571, line=0.25) — so the LINE is the lower curve. For x∈(2,4), line(x)>g(x) (e.g. at x=3: line=1.5, g≈0.333) — so g is the lower curve.

6. Integrate the lower envelope from x=1.5 to x=4:

   Area = ∫₁.₅² (x - 3/2) dx + ∫₂⁴ (1/x) dx
   = [x²/2 - 1.5x]₁.₅² + [ln x]₂⁴
   = (2-3)-(1.125-2.25) + (ln4-ln2)
   = (-1)-(-1.125) + ln2
   = 0.125 + ln2 = 1/8 + ln2

Final Answer

The bounded area is exactly 1/8 + ln2. In the exam, when a function is composed with itself a large number of times, always test the first 3-4 iterates by hand before assuming you need a general formula — periodic behaviour is common and turns a scary-looking problem into simple arithmetic.

Key Facts to Remember

• f(x)=(x−1)/(x+1) has period 4 under composition: f⁴=identity.
• To find fⁿ for large n, reduce n modulo the period found by direct computation.
• For "area below both curves" regions, identify the lower envelope on each sub-interval by comparing the two functions, not by assuming one is always on top.
• ∫dx/x = ln|x| + C; ∫(x−c)dx = x²/2 − cx + C.

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Prerequisites

Before practising this question type, make sure the following are in place:

• Composing functions and recognising periodic behaviour under iteration.
• Finding intersections of two curves algebraically.
• Area between curves, including identifying the correct upper/lower envelope on each sub-interval.

Revise these from the Area Under Curves chapter, then attempt the related questions below.

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