Is there example of two real functions $f$ and $g$ such that: $g$ has a local minimum at $x=0$ ($g$ is not necessarily differentiable at $x=0$), $f\circ g$ is differentiable at $x=0$ but $(f\circ g)'(0)\neq 0$, and $f$ is differentiable at $g(0)$?
Composition of non-differentiable functions
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real-analysis
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0$g$ would have to be non-differentiable at $0$. Otherwise $(f\circ g)^{\prime}(0) = f^{\prime}(g(0))\cdot g^{\prime}(0) = 0$ since $g$ has a local minimum at $0$. – 2012-10-02
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0$g$ can be even discontinuos and the local min point $x=0$ can be an end-point of the domain. – 2012-10-02
1 Answers
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Sure -- let $f(x)=x(x+1)$ and $$ g(x) = \begin{cases} -1 & x=0 \\ x & x \ne 0 \end{cases} $$
It gets more difficult if you want $g$ to be continuous.
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0Thank you Mr. Makholm! Of course we restrict g to the interval $(-2\pi,\infty)$. – 2012-10-02
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0(Sorry, just rewrote it to a polynomial instead of a sine). You don't need to restrict $g$, since it was only specified that it had a _local_ minimum. But if you want a global minimum, just take the $x\ne 0$ branch of $g$ to be $|x+\frac12|-\frac12$ instead. – 2012-10-02
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0I liked the other example better and I am going to use it. – 2012-10-02
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0What to you need it for, anyway? – 2012-10-02
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0I am writing a paper about the derivative notion.I wanted to invalidate the chain rule under certain assumptions. Thank you again! – 2012-10-02