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 December 27th, 2014, 07:37 AM #1 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 Differentiation Show that $\displaystyle y=\left |x-2 \right |$ is not differentiable at $\displaystyle x=2$ How to show?
 December 27th, 2014, 08:33 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,031 Thanks: 1620 derivative at a point $\displaystyle x = a$ is defined as $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$ for $\displaystyle f(x) = |x-2|$, show that the limit ... $\displaystyle \lim_{x \to 2} \frac{f(x)-f(2)}{x-2}$ ... does not exist.
December 27th, 2014, 01:34 PM   #3
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Quote:
 Originally Posted by skeeter derivative at a point $\displaystyle x = a$ is defined as $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$ for $\displaystyle f(x) = |x-2|$, show that the limit ... $\displaystyle \lim_{x \to 2} \frac{f(x)-f(2)}{x-2}$ ... does not exist.
The limits for x > 2 and x < 2 are different.

 December 27th, 2014, 06:01 PM #4 Senior Member     Joined: Oct 2014 From: EU Posts: 224 Thanks: 26 Math Focus: Calculus You can consider the function as a piecewise function: $\displaystyle f(x) = \left \{ \begin{matrix} f_a(x) = x - 2 &\text{if}\,x \ge 2 \\ f_b(x) = -x+2 &\text{if}\,x < 2 \end{matrix}\right.$ Now, you can see that $\displaystyle f(x)$ is continous at $\displaystyle x=2$ evaluating the limits for $\displaystyle f_a(x)$ and $\displaystyle f_b(x)$ when $\displaystyle x$ tends to 2, but each derivative is different: 1 and -1, so $\displaystyle f(x)$ is not differentiable when $\displaystyle x = 2$. Last edited by szz; December 27th, 2014 at 06:21 PM.

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