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\begin{document}
\vspace*{.3in}
\noindent{\bf Rudin, page 239/16:}
From the definition of derivative, we have
$$
f'(0) = \lim_{h\to 0} \frac{h+ 2h^2\sin(1/h) - 0}{h}
= 1 + \lim_{h\to 0} 2h\sin(1/h) = 1
$$
since $|2h\sin(1/h)|\leq 2|h|\to 0$ as $h\to 0$.  In particular, $f'(0)$
is invertible (i.e. non-zero).

Moreover, for $t\neq 0$, we have
$$
f'(t) = 1 + 4t\sin(1/t) - 2\cos(1/t).
$$
Hence $|f'(t)| \leq 1 + 4|t| + 2 < 7$ for $t\in(-1,1)$.  So $f'$ is bounded
on $(-1,1)$.

Now suppose that $f$ is actually injective on some neighborhood
$I = (-\epsilon,\epsilon)$ of $0$.  Then because $f$ is continuous, it
follows that $f$ is actually strictly monotone---say for the moment
that $f$ is strictly increasing.  Then at any point
$x\in I$, we have
$$
f'(x) = \lim_{h\to 0^+} \frac{f(x+h) - f(x)}{h} \geq 0
$$
because $h>0$ implies that $x+h > x$ which implies in turn that
$f(x+h) > f(x)$.

But in fact $f'$ is not non-negative on $I$:  at any point $x = 1/2n\pi$
we have $f'(x) = 1 + 4\cdot (1/2n\pi)\cdot 0 - 2\cdot 1 = -1$.  It follows
that $f$ cannot be strictly increasing on $I$.  

So it must be that $f$ is strictly \emph{decreasing} on $I$.  As before
we conclude that $f'(x) \leq 0$ for every $x\in I$.  This contradicts the
fact that $f'(0) = 1$, though.  So $f$ is not strictly decreasing,
therefore not monotone, and therefore not injective on $I$.  Too bad.
\qed
\end{document}