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Posted

how do you show/prove that a function is differentiable at EVERY real x??

 

lol i just cant figure out the method? i thought of induction but yeah....guess not :(

 

 

anyways yep that it

:)

Sarah

Posted

every function at every point except at infinity has a gradient therefore every function is differentiable as long as you can prove it does go to infinity - is this statement correct?

Posted

It would depend on te function.

 

 

sqrt(x) on the positive reals (ie including zero) is not differentiable. On the strictly positives it is differentiabel at all points.

 

You simply show that for any given point of the domain the function is differentiable. Unless you have specific example in mind there is nothing more that anyone can say to you.

 

alext87's post is absolutely wrong.

Posted

Like matt grime said, it depends on the function. You can check a few things though. If you want it to be differentiable at every real x then it has to be continous at every real x. Also, make sure there are no sharp points, like in the absolute value function at 0.

Posted

ok ok the question is how do you use the defintion of dervative (f'(x) = (f(x+h - f(x))/h) to calucalte the dervative of |x| or |x|^3 ??

Posted

or you could use the fact that |x|^3 is just x^3 if x is positive, and -x^3 if x is negative, and thus clearly differentiable at all points but 0. so it suffices to check the derivative at 0, from the right you get the derivative being as for x^3, and the left as -x^3, both of which give 0 at the origin, so it is differentiable.

Guest eKo.RavanDi
Posted

first of all the function should be continous at all of the points in its domain. because there is no derivation in the end-points. then you should check for the sharp points. each point is NOT a sharp point if the left and right derivatives are equal. f'+(a)=f'-(a). the left and right derivatives are the left and right limits of f(x)-f(a)/x-a when x gets near to a.

sorry for bad english, ;)

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