Three compositions return the input

Define f(x)=(x−1)/xf(x)=(x-1)/x for real x≠0,1x\ne0,1.

(i) Define its inverse. marks

(ii) Show the composite f2f^2 exists and equals the inverse. marks

(iii) State f3(x)f^3(x). mark

(iv) Given f2014g(x)=1−e−xf^{2014}g(x)=1-e^{-x} for x>0x>0, find g(x). marks

[2014/JJC/Promo/8]

Total marks: 10

Solution

(i)

Inverse functions

Define f(x)=x−1xf(x)=\dfrac{x-1}{x} for real x≠0,1x\ne0,1. Define f−1f^{-1}.

Hint

  1. Write y=x−1xy=\frac{x-1}{x} and make xx the subject.

  2. State the domain of the inverse.

Answer · (i)

f−1(x)=1/(1−x)f^{-1}(x)=1/(1-x), x≠0,1x\ne0,1.

(ii)

Powers of a function

Show that the composite function f2f^2 exists and that f2=f−1f^2=f^{-1}.

Hint

  1. f2f^2 exists if Rf⊆DfR_f\subseteq D_f.

  2. Compute f(f(x))f(f(x)) and simplify.

Answer · (ii)

f2f^2 exists and f2(x)=1/(1−x)=f−1(x)f^2(x)=1/(1-x)=f^{-1}(x).

(iii)

Cycles of functions

State f3(x)f^3(x).

Explore this idea

Hint

  1. f3=f f2=f f−1f^3=f\,f^2=f\,f^{-1}, which is the identity.

Answer · (iii)

f3(x)=xf^3(x)=x.

x≠0,1x\ne0,1.

Explore the idea

f, then f², then f³ = identity

Each step applies f once more. After three applications the graph is the line y = x.

Teal: f(x) = (x − 1)/x. No comparison overlay.

Start: f(x) = (x − 1)/x

Vertical asymptote x=0x=0, horizontal asymptote y=1y=1.

  • Graph after the selected number of applications
Ideas to try
Follow the asymptotes: where do they sit after each application?
(iv)

Reducing large powers

Given that f2014g(x)=1−e−xf^{2014}g(x)=1-e^{-x} for x>0x>0, find g(x)g(x).

Hint

  1. Because f3f^3 is the identity, reduce 20142014 modulo 3.

  2. 2014=3×671+12014=3\times671+1, so f2014=ff^{2014}=f.

  3. Then undo ff using the inverse.

Answer · (iv)

g(x)=exg(x)=e^x.

x>0x>0.