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#### teddybear9

##### New member

- Feb 11, 2012

- 3

Thanks!

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- Thread starter
- #1

- Feb 11, 2012

- 3

Thanks!

In what sense do you mean that the square root function is not differentiable?

Thanks!

Some functions to consider are:

|x -1|

|cos(x)|

x^{2/3}

|cos(x)|

x

- Feb 15, 2012

- 1,967

the function SammyS no doubt intended was:

$f(x) = \sqrt[3]{x^2}$, as $f(x) = x^{\frac{2}{3}}$ is undefined for real x < 0.

- Jan 31, 2012

- 54

Is there continuous nowhere differentiable functions?

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- #5

- Feb 11, 2012

- 3

I don't think so :S Is there?

Is there continuous nowhere differentiable functions?

- Jan 31, 2012

- 54

I don't think so :S Is there?

Yes, there are "many", here is one famous example:

http://en.wikipedia.org/wiki/Weierstrass_function

- Feb 15, 2012

- 1,967

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- #8

- Feb 11, 2012

- 3

oh haha thanks guys!

$\displaystyle g(x)=\left(\,\sqrt[3]{x}\,\right)$ is defined for all x, when considered as a real function.

the function SammyS no doubt intended was:

$f(x) = \sqrt[3]{x^2}$, as $f(x) = x^{\frac{2}{3}}$ is undefined for real x < 0.

So $\large{ f(x)=x^{\frac{2}{3}}}$ is also defined for all real x

- Feb 15, 2012

- 1,967

when i learned calculus the first time, $\sqrt[3]{x}$ was a perfectly good function, defined on all of $\mathbb{R}$. things change once you consider the reals in terms of the complex field.