Proposition. Let I be an interval, and let $f: I \to \mathbb{R}$ be a strictly monotone and continuous on I. Let $J := f(I)$ and let $g:J \to \mathbb{R}$ be the inverse function of f. Prove that if f is differentiable at $c \in I$ and $f'(c) = 0$, then g is not differentiable at $d:=f(c)$.

Proof: Suppose g is differentiable at $d:=f(c)$. Therefore, $\displaystyle \lim_{y\to d} \frac{g(y)-g(d)}{y-d}$ exists.

Since g is differentiable, $g$ is also continuous at $d$, so $\displaystyle\lim_{y\to d} \frac{g(y)-g(d)}{y-d} = \frac{g(y)-g(d)}{y-d}.$ Since for every $y$ in $J$, $y=f(x)$ for some $x$ in $I$, we can rewrite the limit as $\displaystyle\frac{g(f(x))-g(f(c))}{f(x)-f(c)} = \frac{x-c}{f(x)-f(c)}$.

Since $f$ differentiable and hence continuous at $c$ and $f'(c) = 0$, we have $\displaystyle\lim_{x\to c} \frac{f(x)-f(c)}{x-c} =\frac{f(x)-f(c)}{x-c}=0.$ This implies that $\displaystyle\frac{x-c}{f(x)-f(c)} = \frac{1}{\frac{f(x)-f(c)}{x-c}}=\frac{1}{0}$ which is undefined.

This gives a contradiction.

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    $\begingroup$ That last equality is fishy... I'm sure someone will comment on this; that is, the contradiction is in the fact that the limit does not exist, not that $1/0$ is undefined. $\endgroup$ – DanZimm Jan 19 '15 at 17:29
  • $\begingroup$ @DanZimm I also know that it is fishy but I don't know how to resolve this problem. Can you give me some hints? $\endgroup$ – user10024395 Jan 19 '15 at 17:33

Note that $\iota(x):=g\bigl(f(x)\bigr)\equiv x$. If $g$ were differentiable at $d=f(c)$ then $$1=\iota'(c)=g'\bigl(f(c)\bigr)\cdot f'(c)\ ,$$ which contradicts the assumption $f'(c)=0$.


The second line of your proof is arguable

Since g is differentiable, $g$ is also continuous at $d$, so $\displaystyle\lim_{y\to d} \frac{g(y)-g(d)}{y-d} = \frac{g(y)-g(d)}{y-d}.$

g is continuous at d but there you are using continuity of $\frac{g(y)-g(d)}{y-d}$.

Another way to approach the problem is to chain's rule, I think this is more direct.


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