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In Pugh's Real Mathematical Analysis there is an exercise, marked with three stars (which denotes that the author doesn't know the answer), whether there exist a nonsmooth function $f : \mathbb{R} \to \mathbb{R}$ such that $f^2$ and $f^3$ are both smooth.

My question is not strictly about this exercise, but rather about cases when we weaken the hypotheses when only one of $f^2$ and $f^3$ are smooth.
The fact that the exercise comes with this hypotheses suggest we should be able to find those functions. For the case when $f^2$ need to be smooth we have a function $f(x) = x$ if $x$ is rational and $-x$ if $x$ is irrational, but what about the case when $f^3$ needs to be smooth?

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    Does the function have to be onto? Otherwise something simple like $f(1)=-1$, $f(x\neq 1)=0$ would do the trick2012-10-02

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How about: $y=\sin^{1/3} x,\;\;x>0$

(if by $f^3$ you mean $f(x)^3$ of course)

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    @Victor not according to his second sentence?2012-10-02
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For a resolution of the exercise, see

http://www.math.ucla.edu/~tao/whatsnew.html

The Feb 16,2007 comment contains a proof that if $f^2$ and $f^3$ are smooth, $f$ is smooth.

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The idea is that it's easy to have a cusp that is "cured" by either squaring or cubing the function (or raising it to any other particular power), but more difficult to think of a case where both operations work. The simplest parametrized family of examples where $f$ is not smooth, but $f^{1+a}$ is, is probably $f(x; a)=\lvert x \rvert^{2/(1+a)}$. Choosing $a=1$ or $a=2$ gives the desired examples where $f^2$ and $f^3$ are smooth.