Differentiability Of Functions

by user677491   Last Updated July 12, 2019 06:20 AM

Why the condition of differentiability is always given on an open interval not on a closed interval, even if we consider the function $$f(x)=|x|$$ defined for all $$x>0$$, this function is derivable at all the points $$[0, ∞)$$

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The derivative, by definition, is a two-sided limit, so the boundary would cause technical difficulties mostly.

That's why you might often see early in an Analysis course the assumptions of continuity on the boundary and differentiability on the interior.

Jay
July 12, 2019 05:44 AM

Differentiability is a pointwise definition. Saying a function is "differentiable" on an interval $$I$$ means it is differentiable, pointwise, for every point in that interval. Boundary points of a closed interval don't work very well with this definition.

For example $$f(x)=\sqrt{x}$$ is not differentiable at $$x=0$$ because a tangent line does not exist there. However for all $$x\in(0, \infty)$$ it does.

David Peterson
July 12, 2019 05:56 AM

Well, the example taking the interval $$[0,\infty)$$ makes not much sense. Open intervals (neighborhoods) are key.

$$\lim_{h\rightarrow 0+}\frac{|x+h|-|x|}{h}$$ goes to $$1$$ if $$x=0$$.

$$\lim_{h\rightarrow 0-}\frac{|x+h|-|x|}{h}$$ goes to $$-1$$ if $$x=0$$.

So at $$x=0$$ the function $$x\mapsto|x|$$ is not differentiable.

Wuestenfux
July 12, 2019 05:58 AM

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