Composition conditions and a hidden constant case

Functions f:x↦(x2−6x−6)/(x+1) f:x\mapsto(x^2-6x-6)/(x+1), x∈Rx\in\mathbb R, x≠−1 x\ne-1; g:x↦(ax+1)/(x+b) g:x\mapsto(ax+1)/(x+b), x∈Rx\in\mathbb R, x≠−bx\ne-b, where a,ba, b are constants.

(i) Sketch y=f(x) y=f(x), giving coordinates of turning points and equations of asymptotes. Write down the range of ff. marks

(ii) Find the value of aa and range of bb such that both fgfg and gfgf exist. marks

(iii) Find g−1(x) g^{-1}(x). Given g−1(x)=g(x) g^{-1}(x)=g(x) for all real xx, x≠−bx\ne-b, find bb in terms of aa. marks

[2019/TJC/Promo/9]

Total marks: 10

Solution

If this part uses an earlier result, revisit (ii) Existence of composites or open all working together.

(iii)

Self-inverse functions

Find g−1(x)g^{-1}(x). Given that g−1(x)=g(x)g^{-1}(x)=g(x) for all x≠−bx\ne-b, find bb in terms of aa.

Explore this idea

Hint

  1. Solve y=ax+1x+by=\frac{ax+1}{x+b} for xx.

  2. Compare with gg.

Answer · (iii)

g−1(x)=(1−bx)/(x−a)g^{-1}(x)=(1-bx)/(x-a), x≠ax\ne a; self-inverse requires b=−ab=-a.

Inverse exists only when ab≠1.

Explore the idea

Open interactive exploration

An example: a = −1, b = 1

g(x) = (1 − x)/(1 + x) satisfies b = −a. Its graph is symmetric about y = x, so it equals its own inverse. One branch is shown.

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, 0)
Q = (f(x), x)
(0, 1)
on the inverse
Domain of f
x>−1x>-1 (right branch)
Range of f = domain of f⁻¹
y>−1y>-1
Midpoint M on y = x
(0.5, 0.5)
  • G
  • G⁻¹, the same curve
  • Y = x
  • PQ is perpendicular to y = x; its blue midpoint M lies on that line
Key idea

b=−ab=-a makes the asymptotes x=−b=ax=-b=a and y=ay=a swap into each other under reflection in y=xy=x, so the curve maps onto itself.

Ideas to try
Move xx and compare PP and QQ: is QQ on the same curve?