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Continuity and Derivatives

teddybear9

New member
Feb 11, 2012
3
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
Feb 15, 2012
3
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
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
the "standard" example is f(x) = |x|.

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

Is there continuous nowhere differentiable functions?
 

teddybear9

New member
Feb 11, 2012
3

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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

New member
Feb 11, 2012
3
oh haha thanks guys! :)
 

SammyS

New member
Feb 15, 2012
3
the "standard" example is f(x) = |x|.

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.
$\displaystyle g(x)=\left(\,\sqrt[3]{x}\,\right)$ 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
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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, $\sqrt[3]{x}$ was a perfectly good function, defined on all of $\mathbb{R}$. things change once you consider the reals in terms of the complex field.