If $f$ is continuous at $c$ and $f ′(c) = 0$, then there exists an $h > 0$ such that $f$ is differentiable in the interval $(c – h, c + h)$.

Question

My book states the following: if $f$ is continuous at $c$ and $f ′(c) = 0$, then there exists an $h > 0$ such that $f$ is differentiable in the interval $(c – h, c + h)$. But I don't understand this. It is not as if $f'$ is given to be continuous, rather $f$ is given continuous and differentiable at $x=c$. So how can we possibly comment about the existence of $f'$ in the neighborhood of $c$ too? Note that f is defined on an open interval I and c belongs to I.

Answer 1

Let $\;g(x):\left]-\infty,+\infty\right[\to\mathbb{R}\;$ be the Darboux function defined as follows

$g(x)=\sum\limits_{n=1}^{\infty}\dfrac1{n!}\sin\bigl((n+1)!x\bigr)\quad\forall x\in\left]-\infty,+\infty\right[\;.$

$g(x)\;$ is continuous everywhere on $\;\left]-\infty,+\infty\right[\;$ and nowhere differentiable.

Let $\;c\;$ be a real number that is $\;c\in\mathbb{R}\;.$

Let $\;f(x):\left]-\infty,+\infty\right[\to\mathbb{R}\;$ be the function defined as

$f(x)=(x-c)^2g(x)\quad\forall x\in\left]-\infty,+\infty\right[\;,\;$

then $\;f(x)\;$ is continuous everywhere on $\;\left]-\infty,+\infty\right[\;,\;$ differentiable only at $\;c\;$ and $\;f’(c)=0\;,\;$ indeed

$\begin{align} f’(c)&=\lim\limits_{x\to c}\dfrac{f(x)-f(c)}{x-c}=\lim\limits_{x\to c}\dfrac{(x-c)^2g(x)}{x-c}=\\ &=\lim\limits_{x\to c}(x-c)g(x)=0\;. \end{align}$

Since $\;f(x)\;$ is differentiable only at the point $\;c\;,\;$ there does not exist any $\;h>0\;$ such that $\;f(x)\;$ is differentiable on the interval $\;\left]c-h,c+h\right[\;.$

Hence, it is wrong what your book states about continuous functions differentiable at a point $\;c\;$ such that $\;f’(c)=0\;.$

It is wrong even if the function is continuous everywhere, not only at the point $\;c\;,\;$ that is why my counterexample is different from those proposed by other users.

Answer 2

Just start with the function $Q$ that’s the characteristic function of $\Bbb Q$, i.e., $Q(x)=1$ if $x$ is rational, zero otherwise.

Then take $f(x)=x^2Q(x)$, with $c=0$.