A function bounded and differentiable, but have an unbounded derivative?

danielkyulee
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Can a function f: (a,b) in R be bounded and diffferentiablle, but have an unbounded derivative. I believe it can, but can not think of any examples where this is true. Anyone have any ideas?
 
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Sure, so long as you properly choose your domain.

So let's think about this, we want a function whose derivative approaches infinity, but such that the function itself is bounded. Probably the easiest way to do this is to find a function whose derivative goes to zero, then rotate about the axis y=x.

You don't need anything complicated. A monomial will do fine.
 
Try to think of a function with a graph that looks a bit like the graph of a periodic function, but is such that the distance between the peaks get shorter and shorter instead of staying the same.
 
Fredrik said:
Try to think of a function with a graph that looks a bit like the graph of a periodic function, but is such that the distance between the peaks get shorter and shorter instead of staying the same.

Very nice. Perhaps somewhat more complicated, but nice.
 
Sqrt(x), 0 < x < 1.
 
Ray, here in the homework forum it's important to answer with hints, not complete solutions.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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