Solutions

Let

$$f(x)=\frac{1}{1+x}.$$

i)

$$\begin{align*} f(f(x)) &= f\left(\frac{1}{1+x}\right) \\ &= \frac{1}{1+\frac{1}{1+x}}. \end{align*}$$

This function is defined for $x$ such that

$$\begin{align*} 1+\frac{1}{1+x} \ne 0 &\implies -1 \ne \frac{1}{1+x} \\ &\implies x \ne -1, \end{align*}$$

and

$$\begin{align*} -1-x \ne 1 &\implies x \ne -2. \end{align*}$$

ii)

$$f\left(\frac{1}{x}\right)=\frac{1}{1+\frac{1}{x}}.$$

iii)

$$f(cx)=\frac{1}{1+cx}.$$

iv)

$$f(x+y)=\frac{1}{1+x+y}.$$

v)

$$f(x)+f(y)=\frac{1}{1+x}+\frac{1}{1+y}.$$

vi)

$$\begin{align*} f(cx)=f(x) &\implies \frac{1}{1+cx}=\frac{1}{1+x} \\ &\implies 1+x=1+cx \\ &\implies x=cx \\ &\implies x(c-1)=0. \end{align*}$$

This means that either $x=0$ or $c=1$. Now notice that if $x=0$, then $c$ can take on any value and satisfy the equality. Therefore, for any value of $c\in\mathbb{R}$, we can set $x=0$ so that

$$f(cx)=f(x)=f(0).$$

vii) We have that for $c=1$,

$$f(cx)=f(1\cdot x)=f(x).$$

So the equality holds for any $x$ in the domain of $f(x)$. Now, we prove by contradiction that it is the only value of $c$ for which the equality holds for two different numbers $x$.

Suppose that $c\ne 1$. Then $c-1\ne 0$ and, by the equality in vi),

$$\begin{align*} x(c-1)=0 &\implies x=0. \end{align*}$$

Hence, there are no two different numbers $x$ that satisfy $f(cx)=f(x)$ for $c$ other than $c=1$.