Three compositions return the input · source reference

Complete question

Define $f(x)=(x-1)/x$ for real $x\ne0,1$. (i) Define its inverse. [3] (ii) Show the composite $f^2$ exists and equals the inverse. [3] (iii) State $f^3(x)$. [1] (iv) Given $f^{2014}g(x)=1-e^{-x}$ for $x>0$, find g(x). [3]

[2014/JJC/Promo/8]

Printed working

Part (i)

Let $y=(x-1)/x$, $x\in\mathbb R$, $x\ne0,1$. Then

$$xy=x-1\Rightarrow1=x-xy\Rightarrow1=x(1-y)\Rightarrow\frac1{1-y}=x.$$

Thus $f^{-1}(x)=1/(1-x)$ and $D_{f^{-1}}=R_f=\mathbb R\setminus\{0,1\}$.

Alternatively write the domain as $(-\infty,0)\cup(0,1)\cup(1,\infty)$.

Define $f^{-1}:x\mapsto1/(1-x)$, $x\in\mathbb R$, $x\ne0,1$.

The source sketch of $y=f(x)$ labels vertical asymptote $x=0$, horizontal asymptote $y=1$ and an open point $(1,0)$.

Part (ii)

Since $R_f=\mathbb R\setminus\{0,1\}\subseteq\mathbb R\setminus\{0,1\}=D_f$, $f^2$ exists (shown).

$$f^2(x)=f(f(x))=\frac{\frac{x-1}{x}-1}{\frac{x-1}{x}} =-\frac1{x-1}=\frac1{1-x}=f^{-1}(x).\quad\text{(Shown.)}$$

Part (iii)

$f^2(x)=f^{-1}(x)$ gives $f(f^2(x))=f(f^{-1}(x))$, so $f^3(x)=x$ on the original punctured domain.

Part (iv)

$f^{2014}(g(x))=1-e^{-x}$. Note $2014=671\times3+1$:

$$f\underbrace{f^3\cdots f^3}_{671\text{ blocks}}(g(x))=1-e^{-x}.$$

Replace the innermost $f^3(g(x))$ by $g(x)$, leaving670 such blocks, and use the earlier result repeatedly. This reduces to $f(g(x))=1-e^{-x}$, hence

$$g(x)=f^{-1}(1-e^{-x})=\frac1{1-(1-e^{-x})}=e^x.$$

Teacher margin notes

From (iii), $f^3(x)=x\Rightarrow f^3(g(x))=g(x)$. Example: $f^{10}(g(x))=f\underbrace{f^3f^3f^3}_{3\text{ blocks}}(g(x))=f\underbrace{f^3f^3}_{2\text{ blocks}}(g(x))=f\underbrace{f^3}_{1\text{ block}}(g(x))=f(g(x))$.

Source diagrams

Provenance

Question page references: 2. Solution page references: 11. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.