# Thread: function differentiable, but derivative not bounded

1. ## function differentiable, but derivative not bounded

Question: Give an example of a function f that is differentiable on [0,1] but its derivative is not bounded on [0,1].

Ok, I know that the derivative f' cannot be continuous, because then it would be bounded on [0,1]. I also know that it cannot be increasing with jump discontinuities, because the derivative has to have the intermediate value property. I also do not think (but am not 100 percent sure) that it can have any infinite discontinuities, because I think that would make f unbounded, which it cannot be on [0,1] if its differentiable on [0,1].

Any ideas?

2. No, that's impossible. Saying that the function is differentiable on [0, 1] means that it has a finite derivative at each point on [0, 1]. Since [0, 1] is a closed and bounded set, that derivative must be bounded.

3. Halls, what about $\displaystyle f(x) = x^2 sin(1/x^2)$ when x isn't 0, and f(0) = 0.

f is differentiable at all non-zero points, and note that $\displaystyle f(0 +h) - f(0) = h^2 sin(1/h^2)$. Thus $\displaystyle \frac{ f(0+h) - f(0) } {h} = h sin(1/h^2) \rightarrow 0$. Thus f is also differentiable at 0 and f(0) = 0.

Note that on (0, 1], $\displaystyle f'(x) = 2x sin(1/x^2) - \frac{2cos(1/x^2)}{x}$, which goes to -inf as x goes to 0 through points of (0, 1], thus f' is not bounded on [0,1]