Wikipedia:Reference desk/Archives/Mathematics/2015 March 11

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March 11

Derivative limit

Does the derivative of $x^{2}-|x^{2}-1|+2||x|-1|+2|x|$ exist at $x=\pm 1$? 70.190.182.236 (talk) 04:39, 11 March 2015 (UTC)

This looks rather like a homework question. Show what you've tried and we can help you where you get stuck. —Quondum 05:23, 11 March 2015 (UTC)

It's not homework- I was just hoping for an answer. 70.190.182.236 (talk) 19:36, 11 March 2015 (UTC)

Have you tried drawing a graph of the function? This should give you an idea of what's happening at the points of interest. (Later hint, after drawing the graph myself: think very carefully about how derivatives are defined...) RomanSpa (talk) 12:19, 11 March 2015 (UTC)

Plugging it into the limit definition of the derivative, no, it looks like it does not exist. (limit of (3 - abs(1 - (1 + h)^2) + 2*abs(1 + h) + 2*abs(1 - abs(1 + h)) + (1 + h)^2)/h as h -> 0)70.190.182.236 (talk) 19:45, 11 March 2015 (UTC)

Take a look at the graph, and see what you think. —Quondum 20:39, 11 March 2015 (UTC)

Cut the crap- looking at the graph I would expect the derivative to exist- but the limit says it does not. 70.190.182.236 (talk) 21:28, 11 March 2015 (UTC)

It exists. For the positive branch of the x = +1 side, we have:

$\lim _{h\to 0^{+}}{\left((1+h)^{2}-|(1+h)^{2}-1|+2||1+h|-1|+2|1+h|\right)-3 \over h}$

$=\lim _{h\to 0^{+}}{\left(1+2h+h^{2}-(2h+h^{2})+2h+2+2h\right)-3 \over h}$

$=\lim _{h\to 0^{+}}{\left(3+4h\right)-3 \over h}=4$

I'll leave showing the negative branch and the x = -1 case to the reader. Dragons flight (talk) 03:22, 12 March 2015 (UTC)

I improved the LaTeX notation a bit [1]. Please check the point, where the limit is calculated: did you really mean x approaching zero from above? --CiaPan (talk) 06:04, 12 March 2015 (UTC)

Not x. $h=x-1$ is approaching 0 from above, as in $f'(x_{0})=\lim _{x\to x_{0}}{\frac {f(x)-f(x_{0})}{x-x_{0}}}=\lim _{h\to 0}{\frac {f(x_{0}+h)-f(x_{0})}{h}}$. -- Meni Rosenfeld (talk) 08:37, 12 March 2015 (UTC)

The function is clearly even (x only appears as an absolute value or a square), so -1 doesn't need to be checked separately. -- Meni Rosenfeld (talk) 08:37, 12 March 2015 (UTC)

The problem terms are those involving $||x|-1|$. Complete the square appropriately. If $||x|-1|$ appears in a linear term, then the function is not differentiable. If it appears only as a quadratic term, then the function is differentiable. Sławomir Biały (talk) 13:14, 11 March 2015 (UTC)

Sławomir Biały, that would lead me to draw the wrong conclusion. $|x^{2}-1|$ is also a problem term, and it interacts in a nontrivial way at the points of interest. I think RomanSpa's approach leads to a more intuitive understanding. —Quondum 18:07, 11 March 2015 (UTC)

Yes, but $|x^{2}-1|=||x|-1|(|x|+1)$. Hence my vague "complete the square appropriately" ;-) Sławomir Biały (talk) 18:26, 11 March 2015 (UTC)

Sorry, I missed that. Besides I would have missed what it meant. Back to remedial class for me —Quondum 18:51, 11 March 2015 (UTC)