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 $[0, ∞)$

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.

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.

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.

Updated November 30, 2017 17:20 PM

Updated July 22, 2017 01:20 AM

Updated November 19, 2018 19:20 PM

Updated January 11, 2019 21:20 PM

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