# Thread: Constructing functions on closed intervals whose derivatives are unbounded

1. ## Constructing functions on closed intervals whose derivatives are unbounded

Hello folks,

I'm looking for a way to construct a function F on a closed interval [a, b] whose derivative is unbounded. Why properties does F need to possess?

Is there a discontinuity requirement?

2. ## Re: Constructing functions on closed intervals whose derivatives are unbounded

It is already discontinuous at $a$ and $b$. So, make it unbounded at the endpoints. For example: $f(x) = \begin{cases}(x-a)\sin\left( \dfrac{1}{x-a} \right) & x > a \\ 0 & x = a\end{cases}$. Note that this function is continuous, but its derivative at a is not defined. In fact, its derivative approaches all real numbers as x approaches a. But, this function is differentiable on $(a,b)$.

3. ## Re: Constructing functions on closed intervals whose derivatives are unbounded

Hey jumbo1985.

You will have a discontinuity if you allow unbounded derivatives.

You can think about integrating this to show that continuity is violated [Fundamental theorem of calculus doesn't work over interval].

4. ## Re: Constructing functions on closed intervals whose derivatives are unbounded

Originally Posted by chiro
Hey jumbo1985.

You will have a discontinuity if you allow unbounded derivatives.

You can think about integrating this to show that continuity is violated [Fundamental theorem of calculus doesn't work over interval].
This is false. The function I gave above is continuous (no discontinuities) but it's derivative is unbounded at x=a.

You can extend the function to the entire real line and it will still be continuous at x=a and its derivative still unbounded and discontinuous at x=b. Continuity does not imply differentiability.