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

= 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 |x^2+-+1|+%2B+2||x|+-+1|+%2B+2|x| 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:
 * $$\lim_{h\to 0^+} { \left((1+h)^2 - |(1+h)^2 - 1| + 2||1+h| - 1| + 2|1+h|\right) - 3 \over h}$$
 * $$= \lim_{h\to 0^+} { \left(1+2h+h^2 - (2h+h^2) + 2h + 2+2h\right) - 3 \over h}$$
 * $$= \lim_{h\to 0^+} { \left(3 + 4 h \right) - 3 \over h} = 4$$
 * 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 . 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 $$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$$. -- 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)


 * , 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)