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Consider the function $f:[-1,1] \to \mathbb{R}$ defined piecewise by \begin{align*}f(x) = \begin{cases}0: -1 \leq x < 0 \\ g(x): 0\leq x \leq 1 \end{cases}\end{align*} where $g:[0,1] \to \mathbb{R}$ is some function. Suppose $g$ is a non-zero function given by $$g(x) = a_0 + a_1x + a_2x^2 + a_3x^3.$$ Find non-negative real numbers $a_0, a_1, a_2, a_3$ (not all of which are 0) such that $f$ is continuous but not differentiable.

I've been able to find some examples of general continuous, but not differentiable functions, but I have not been able to find one that fits this specific example AND to supply non-negative real numbers that makes it true. My problem is I can't preserve continuity at $f(0)$ but make $f(x)$ non-differentiable given the restriction of the $a_i$'s.

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  • $\begingroup$ Do you know under what circumstances would $f$ would be continuous but not differentiable? Especially at $0$? $\endgroup$
    – player3236
    Sep 8, 2020 at 16:37
  • $\begingroup$ "I can't preserve continuity and $f(0)$ but make $f(x)$ non-differentiable" - well, it's easier to make something continuous than it is to make it differentiable. If you just think about continuity and forget about differentiability you'll probably have an example. Hint: why not try a linear function? $\endgroup$
    – Jair Taylor
    Sep 8, 2020 at 16:40
  • $\begingroup$ $f$ won't be differentiable if the limits approaching 0 (from the right and left) are not the same? For example I know that if $g(x) = |x|$, the function would be continuous but not differentiable at $f(0)$. However, I wouldn't be able to supply explicit $a_i$ values which the problem seemed to want. $\endgroup$
    – Evan Kim
    Sep 8, 2020 at 16:41

1 Answer 1

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What about $g(x) = x+x^2$? $g(0)=0$ and $g^\prime(0) = 1 \neq 0$ so $f$ is continuous at $0$ but not differentiable at that point.

By the way any $g(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ with $a_0= 0, \ a_1 \neq 0$ and $a_0,a_1,a_2,a_3$ non-negative would work.

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    $\begingroup$ $g(x)=x$ is simpler $\endgroup$
    – Mars
    Sep 8, 2020 at 16:40
  • $\begingroup$ I can see why that would not be differentiable, but how do I satisfy "finding non-negative real numbers $a_0, a_1, a_2, a_3$" part? $\endgroup$
    – Evan Kim
    Sep 8, 2020 at 16:43
  • $\begingroup$ $0, 1, 1, 0$ are all non-negative. $\endgroup$
    – mathcounterexamples.net
    Sep 8, 2020 at 16:48
  • $\begingroup$ Sorry, I don't find this answer very pedagogical. How have you arrived to this function? $\endgroup$
    – Miguel
    Sep 8, 2020 at 16:52
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    $\begingroup$ @Miguel Whatever $g(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ with $a_0= 0, \ a_1 \neq 0$ would work. The one I picked up is my preferred one as I like the prime number $2$ . $\endgroup$
    – mathcounterexamples.net
    Sep 8, 2020 at 16:56

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