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Explain why $|x^2-x|$ is not differentiable at $x=1$.....

lets go....

we need to show that $\lim_{a\to0}$ of $\frac{f(1+a)-f(1)}{a}$ doesn't exist...which is to say

$\lim_{a\to0}$ of $\frac{(|(1+a)^2 - (1+a)|- |1^2-1|)}{a}$ doesn't exist which is like

$\lim_{a\to0}$ of $\frac{(|(a)^2 + a|)}{a}$ so we take off absolute value (Is this allowed?)

and get $\lim_{a\to0}$ of $\frac{((a)^2 + a)}{a}$ which is $1$

Not quite what expected...How do we do this?

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    $\begingroup$ Consider the left hand limit and the right hand limit. If $x < 1$ and close to $1$ in as much as $0 < x < 1$, then $|x^2 - x| =|x(x-1)| = -x(x-1)$. If $x > 1$, $|x^2 - x| = +x(x-1)$. $\endgroup$ – Simon S Dec 3 '14 at 16:36
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    $\begingroup$ The point is exactly you cannot take off absolute value (indeed if $a\to 0^-$ then absolute value will give you -1 as (left)-limit) $\endgroup$ – Milly Dec 3 '14 at 16:36
  • $\begingroup$ Crap I didn't see that... $\endgroup$ – Bak1139 Dec 3 '14 at 16:42
  • $\begingroup$ I wouldn't say we "take off" the absolute value; we evaluate it. For the right-hand limit, the argument of the absolute value is positive (so yes, the | | goes away, in a sense). For the left-hand limit, the argument is negative, so we negate it when evaluating the absolute value. $\endgroup$ – The Chaz 2.0 Dec 3 '14 at 16:48
  • $\begingroup$ Sketch $x^2-x$ and hence $|x^2-x|$ to understand why it isn't differentiable. $\endgroup$ – JP McCarthy Dec 9 '14 at 15:39
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From the right hand side, $$\lim_{h\to0 } \frac{|(1+h)^2 -(1+h)|}{h} = 1$$

From the left hand side, $$\lim_{h\to0 } \frac{|(1-h)^2 -(1-h)|}{-h} = -1$$

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Note that $\mid x^2-x \mid= \left\{ \begin{array}{ccc} x-x^2&\mbox{if}&0\le x \le 1\\ x^2-x & &\mbox{otherwise} \end{array} \right.$

Now what happens with the left and right sided derivatives at $1$?

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we have $\frac{f(1+h)-f(1)}{h}=\frac{|(1+h)^2-(1+h)|-|f(1)|}{h}=\frac{|h|}{h}|1+h|=\pm |1+h|$ thus the limit $h$ tends to $0$ doesn't exist.

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