# Continuity and Derivatives

#### teddybear9

##### New member
Can anyone give me an example of a continuous function that is NOT differentiable(other than the square root function)? I have to prove that not all continuous functions are differentiable.

Thanks!

#### SammyS

##### New member
Can anyone give me an example of a continuous function that is NOT differentiable(other than the square root function)? I have to prove that not all continuous functions are differentiable.

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

Some functions to consider are:
|x -1|

|cos(x)|

x2/3

• teddybear9 and Jameson

#### Deveno

##### Well-known member
MHB Math Scholar
the "standard" example is f(x) = |x|.

the function SammyS no doubt intended was:

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

• teddybear9

#### Also sprach Zarathustra

##### Member
Now, the natural question that you should ask yourself is:

Is there continuous nowhere differentiable functions?

#### teddybear9

##### New member
Now, the natural question that you should ask yourself is:

Is there continuous nowhere differentiable functions?
I don't think so :S Is there?

#### Also sprach Zarathustra

##### Member
• teddybear9

#### Deveno

##### Well-known member
MHB Math Scholar
in fact, "most" functions aren't differentiable, in the sense that the "bad" ones vastly out-number the "good" ones. this may come as a surprise to you. your teachers have been sheltering you from this terrible truth, and only letting you play with the "nice" functions.

• teddybear9

#### teddybear9

##### New member
oh haha thanks guys! #### SammyS

##### New member
the "standard" example is f(x) = |x|.

the function SammyS no doubt intended was:

$f(x) = \sqrt{x^2}$, as $f(x) = x^{\frac{2}{3}}$ is undefined for real x < 0.
$\displaystyle g(x)=\left(\,\sqrt{x}\,\right)$ is defined for all x, when considered as a real function.

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

• teddybear9

#### Deveno

##### Well-known member
MHB Math Scholar
i understand what you're saying, every real number has a cube root. for reasons that are probably too far off-topic for this thread, some authors do not define $a^q$, when a < 0, and q is rational. it can be done for odd denominators, if one is careful. but some treat it as "complex-valued" (mathematica does this, i believe). there's a certain ambiguity in what the symbols mean.

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