Is there a continuous and differential function which have an unbounded derivative on a bounded point?

Question

Assume $f$ is continuous and differentiable on $[a,b]$, can $f$ has a point where its derivative goes to infinity within that interval? In other words, if $f$ is continuous and differentiable on a finite interval, can $f'$ has a singularity point on that interval?

Answer 1

by definition if a function $f$ is continuous on a compact interval $[a,b]$ and differentiable on the open interval $(a,b)$ then in particular the following limits exists and are equal: $$\lim_{y\to x^+}\frac{f(y)-f(x)}{y-x}=\lim_{y\to x^-}\frac{f(y)-f(x)}{y-x}:=f'(x)$$ for each $x\in (a,b)$. This automatically excludes the possibility that there is such an $x\in (a,b)$ such that $f'(x)=\pm\infty$ otherwise it means the limits above don't exist.

Answer 2

For piecewise continuity, we can certainly have that there is a singularity point. But for strict continuity then this would not be true since if we let $f$ be continuous on the finite interval $x\in [a,b]$, with $a<b$. Then we know $c_1\leq f \leq c_2$ , so we can write:

$$\frac{|f(x)-f(y)|}{x-y}\leq\frac{c_2-c_1}{x-y}$$ By the M.V.T, there exists a $\hat{c}$ in $[x,y]$ such that $f'(\hat{c})=\frac{c_2-c_1}{x-y}$ which is finite.

Answer 3

In fact, it is possible for a function to be differentiable but to have an unbounded derivative. An answer to this question gives the example $$f(x)=\cases{ x^2\sin(1/x^2),&$x\ne0$ \cr 0,&$x=0$}$$ for $-1\le x\le 1.$ It is easy to confirm this: $$f'(x)=\cases{2x\sin \frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2},&$x\ne0$ \\0,&$x=0$}$$ where $f'(0)$ is computed from the definition of derivative.

Note that $\lim_{x\to 0-}f'(x)=+\infty, \lim_{x\to 0+}f'(x)=-\infty.$ If we define $$g(x)=\cases{-f(x),&$x\ge0$ \\f(x), &$x<0$}$$ then $lim_{x\to 0 } g(x)=+\infty.$