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December 27th, 2014, 07:37 AM   #1
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Differentiation

Show that $\displaystyle y=\left |x-2 \right |$ is not differentiable at $\displaystyle x=2$

How to show?
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December 27th, 2014, 08:33 AM   #2
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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.
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December 27th, 2014, 01:34 PM   #3
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Quote:
Originally Posted by skeeter View Post
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.
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December 27th, 2014, 06:01 PM   #4
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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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