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My high-school calculus teacher has asserted that a function $f(x)$ can only fail to be differentiable at a point $x=a$ if one of the following is true:

  • The function is discontinuous at $x=a$: $\lim_{x\to a}f(x) \ne f(a)$

  • The function has a cusp or vertical tangent at $x=a$: $\lim_{x\to a}\left|{{f(x)-f(a)}\over{x-a}}\right| = \infty$

  • The function has a corner at $x=a$: $\lim_{x\to a^+} {{f(x)-f(a)}\over{x-a}} \ne \lim_{x\to a^-} {{f(x)-f(a)}\over{x-a}}$

While in most cases that is probably correct, I find it somewhat hard to swallow that it is that way for all functions. Specifically, the function

$$f(x)=\begin{cases}x\sin \ln x^2, & x\ne0 \\ 0, & x=0\end{cases}$$

is most definitely not differentiable at $x=0$, but it also doesn't appear to satisfy any of the properties listed above.

The derivative $\frac{\mathrm{d} }{\mathrm{d} x}f(x)$ of the function for $x\ne0$ appears to be $\sin{{\ln x^2}}+2\cos{{\ln x^2}}$, which doesn't show any signs of increasing without bounds as $x\to0$ or suddenly changing at $x=0$, and $f$ is most definitely continuous at that point.

So, the Question is: What's up with$f$? Does it actually fall into one of the cases above, or are they only good as a rough guide for some sorts of functions?

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I think your derivative is incorrect. –  Daryl Feb 25 '13 at 21:50
    
Actually, I think you misinterpreted the expression. This is what you probably meant. –  AJMansfield Feb 25 '13 at 22:08
    
I added a more elaborate example to my answer. There may or may not be a way to accomplish the same thing with a single formula. I have errands to run and will fiddle with this later. Be nice to your teacher. –  Will Jagy Feb 25 '13 at 22:17
    
Not by the way, it appears your derivative is correct and you have my second example in a single formula. Good. I would have put in $|x|$ instead of $x^2.$ –  Will Jagy Feb 25 '13 at 22:23
    
@WillJagy That is actually what I did at first when I was playing around with it, but then I figured out that I can put $x^2$ in with nearly no extra complication to the derivative, and that way I don't have to use a function that does have a corner at $x=0$. –  AJMansfield Feb 25 '13 at 22:26

1 Answer 1

up vote 3 down vote accepted

Your teacher gave a rough guide. The standard example with oscillation, the next item on a fuller list, is $$ f(x) = x \sin \left( \frac{1}{x} \right) $$ which is continuous at $x=0,$ with the proviso that $f(0) = 0.$

I think you will find that your teacher was aware of this and did not want to muddy the waters.

I do not know what you mean by the word cusp.

EEDDIIIITTTTT: There may or may not be a way to build this next one with a single formula: take any function you like for $1 \leq x \leq 2,$ a polynomial should be possible, such that the graph $y = f(x)$ is tangent to the line $y=x$ at both $x=1,2$ and is tangent to $y = -x$ at $x = 3/2.$ Next, put in a half size version for $1/2 \leq x \leq 1,$ with the result that we have a differentiable function on the larger domain owing to the tangency at $x=1.$ Do the same for $1/4 \leq x \leq 1/2.$ And so on forever. Then make $f(0) = 0,$ and $f(-x) = - f(x).$ This results in a function of bounded derivative and $|f(x)| \leq |x|$ for $x \neq 0,$ but no derivative at the origin.

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I actually looked at that one first, but the derivative actually diverges to infinity. –  AJMansfield Feb 25 '13 at 21:54
    
Plus, the function, as you have it written, is not actually continuous at $x=0$. It needs the zero case defined. –  AJMansfield Feb 25 '13 at 21:55
    
@AJMansfield, in that case I do not know what you mean by cusp. Which is alright. –  Will Jagy Feb 25 '13 at 21:56
    
@AJMansfield It is standard to define $f(c)$ as $\lim\limits{x\to c} f(x)$ where the limit exists and the equation defining $f$ is not valid at $c$, so this is usually omitted. Also, note that failing to do so does not lead the function to be discontinuous at $c$, but rather undefined, so it is not a function. Furthermore the slope of this function does not diverge to infinity at $0$, it oscillates between $-1$ and $1$. –  Alex Becker Feb 25 '13 at 21:58
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@AJMansfield The value of the derivative at other points is irrelevant. The difference quotient at $0$ is $$\frac{x\sin(1/x)-0}{x-0}=\sin(1/x)$$ which oscillates between $-1$ and $1$, rather than diverging to infinity. –  Alex Becker Feb 25 '13 at 22:02

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