Self-inverse functions alternate by parity

A function ff is said to be a self-inverse function if f(x)=f−1(x) f(x)=f^{-1}(x). The function ff is defined by f(x)=ax+bcx−2 f(x)=\frac{ax+b}{cx-2}, x∈Rx\in\mathbb R, x≠2/cx\ne2/c, where a,b,c,da, b, c, d are non-zero constants.

(i) By finding f−1(x) f^{-1}(x), show that a=2 a=2 for ff to be self-inverse. marks

For the rest use a=2,b=3,c=5 a=2, b=3, c=5. (ii) Find f2(x) f^2(x). mark (iii) Evaluate f71(4) f^{71}(4). marks

The function gg is defined by g(x)=2x2−3 g(x)=2x^2-3, x∈Rx\in\mathbb R. (iv) Explain why composite fgfg does not exist. marks (v) Find exact solutions of gf(x)=5 gf(x)=5. marks

[2020/CJC/Promo/10]

Total marks: 10

Solution

(i)

Self-inverse functions

f(x)=ax+bcx−2f(x)=\dfrac{ax+b}{cx-2} for x≠2cx\ne\frac2c. By finding f−1(x)f^{-1}(x), show that a=2a=2 for ff to be self-inverse.

Explore this idea

Hint

  1. Write y=ax+bcx−2y=\frac{ax+b}{cx-2} and make xx the subject.

  2. Compare f−1f^{-1} with ff.

Answer · (i)

f−1(x)=(2x+b)/(cx−a)f^{-1}(x)=(2x+b)/(cx-a); self-inverse requires a=2a=2.

Inverse requires 2a+bc≠0; its input excludes a/c.

Explore the idea

Open interactive exploration

A self-inverse graph is its own mirror image

With a = 2, b = 3, c = 5 the curve and its inverse lie on top of each other. One branch is shown (x > 2/5).

Point on f: input xx = 1.00
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))
(1, 1.66667)
Q = (f(x), x)
(1.66667, 1)
on the inverse
Domain of f
x>25x>\frac25 (right branch)
Range of f = domain of f⁻¹
y>25y>\frac25
Midpoint M on y = x
(1.33333, 1.33333)
  • F
  • F⁻¹, which is the same curve
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

A self-inverse graph is symmetric about y=xy=x: the reflection of any point on it is also on it.

Ideas to try
Move xx and compare the coordinates of PP and QQ: are they on the same curve?