Hey everyone,(adsbygoogle = window.adsbygoogle || []).push({});

I was just curious about the nature of the cube root function [itex]f(x)=x^{1/3}[/itex]. I know that its derivative is obviously [itex]\frac{1}{3}x^{-2/3}[/itex] which has a discontinuity at [itex]x=0[/itex]. However, in the non-mathematical sense, the graph of [itex]y=f(x)[/itex] looks smooth - I don't see any angles or cusps like [itex]y=|x|[/itex] or [itex]y=x^{2/3}[/itex].

Now, granted, the discontinuity in [itex]f'(x)[/itex] is a vertical asymptote where [itex]\delta(M)[/itex] can always be chosen such that:

[tex]|x|<\delta \Rightarrow f'(x)>M[/tex]

So unlike the asymptotes of [itex]\frac{1}{x}[/itex] for example, or the derivative of [itex]y=x^{2/3}[/itex], the function both sides of the discontinuity "goes to positive infinity" so the tangent line to [itex]x^{1/3}[/itex] approaches the same slope on either side of [itex]x=0[/itex].

My question is: Is this a good enough explanation as to why an apparently smooth function has a discontinuous derivative? If not, why doesn't this violate the definition of a (mathematically) smooth function? (This may border on a philosophical question, so my apologies in advance if so)

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Continuity of d/dx [x^(1/3)]

**Physics Forums | Science Articles, Homework Help, Discussion**