2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $f'(c)=0$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.