Proof in Spivak's Calculus I'm working through Spivak's Calculus, and am trying to show that the following is true:
$$
f(x) = \frac{1}{1 + x} \to f( c \cdot x ) = f(x) \text{ for all } c \in \mathbb{R}
$$
The book just says:
$f( c \cdot 0 ) = f( 0 )$ for all $c$. QED
which I find kind of lame. 
Can anyone fill me in here? I feel like Spivak has left some of the explaining out. 
What am I missing?
 A: You have misunderstood the question.
Spivak's Calculus, third edition, page 48, problem 1 from chapter 3:


*

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


...
(vi) For which numbers $c$ is there a number $x$ such that $f (cx) = f (x)$.
Hint: There are a lot more than you might think at first glance.
Answer:
(vi) $f(cx)=f(x)$ means that $\frac{1}{1+cx}=\frac{1}{1+x} \ \Rightarrow \ cx=x.$
If $x=0$, then $c$ can be any real number. Thus we conclude that for any $c$, there is at least one $x$ such that $f(cx)=f(x)$, namely, $x=0$.
A: You didn't take into account everything that the problem said.  The problem says:
"For which numbers $c$ is there a number $x$ such that $f(cx) = f(x)$." (It actually has a full stop rather than a question mark.  A typo.)
Where you found "for all $c=\mathbb R$" is not clear. (I'm guessing you meant "for all $c\in\mathbb R$".)
It doesn't say "for all $x$" but rather "is there a number $x$".  That's exactly the kind of thing that you need to focus on on in a problem like this.
Some simple algebra reduces the question to this: For which numbers $c$ is there a number $x$ such that $cx=x$?  If $c$ is any number at all, then this equality holds when $x=0$.  So the answer is: for every number $c$ there is a number $x$ such that $f(cx) = f(x)$.
