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How would I solve the following problem?

Where would the function $|2x-1|$ not be differentiable?

I am thinking it would not be differentiable at $x=1/2$ because there it would be zero.

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I presume abs is the absolute value. Then yes, there is only one point where it is not differentiable and it is $1/2$. –  1015 Jan 31 '13 at 20:27
    
@Fernando Martinez That's correct. It might help to think about the composition of the fucntion $abs$ with the function $2x-1$, i.e., $abs\circ f$, where $f(x)=2x-1$. –  Git Gud Jan 31 '13 at 20:28
    
yes it is absolute value, my question is that my question aks me to explain why it can not be differentiable at x=1/2 is it because there y=0, I am not sure. –  Fernando Martinez Jan 31 '13 at 20:29
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If you sketch the graph, you will see it has a 'kink' at that point. It is not because it is zero there. –  copper.hat Jan 31 '13 at 20:31
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2 Answers 2

up vote 3 down vote accepted

Given $f(x) = |2x - 1|$,

The function is not differentiable at $x = 1/2$; you can check the explanation below, and view the graph of the function, to see why.

When $x > 1/2$, $f(x) = 2x - 1$. When $x\lt 1/2$, $f(x) = 1 - 2x$. If you graph these lines, you'll seen that they form a "upward V" where the graph abruptly changes direction at $x = 1/2$, at the point $(1/2, 0)$.

By non-differentiable, I mean $$\lim_{x \downarrow \large\frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \downarrow \large\frac{1}{2}}\frac{2x-1}{x-\frac{1}{2}} = +2,$$ while $$\lim_{x \uparrow \large\frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \uparrow \large \frac{1}{2}}\frac{1-2x}{x-\frac{1}{2}} = -2$$ Hence, $lim_{x \to \frac{1}{2}} \dfrac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} $ does not exist, and it follows by defintion that $f(x)$ is therefore not differentiable at $x = 1/2$


Graph of $\;f(x) = \left|2x - 1\right|$:

enter image description here

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the function has no discontinuity in $x=1/2$ –  Emanuele Paolini Jan 31 '13 at 20:59
    
So the function is not differentiable when $x = 1/2$, and it just happens to be the case that that is where $f(x) = y = 0$. It is not differentiable by definition at $x = 1/2$, not because $f(x) = 0$. –  amWhy Jan 31 '13 at 20:59
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being continuous is not enough. A differentiable function (at a point or on an interval) must be continuous there, but being non-differentiable at a point doesn't necessarily mean it's not continuous: The absolute value function is continuous, but fails to be differentiable at x = 1/2 since the tangent slopes do not approach the same value from the left as they do from the right of $x = 1/2$. –  amWhy Jan 31 '13 at 21:30
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Because $x \gt 1/2 \implies 2x - 1 > 0 \implies |2x - 1| = 2x - 1$. Likewise $x \lt 1/2 \implies 2x - 1 < 0 \implies 1 - 2x > 0 \implies |2x - 1| = 1 - 2x$ –  amWhy Jan 31 '13 at 21:35
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Yes, exactly! You've got it. –  amWhy Jan 31 '13 at 21:41
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Let $f(x) = |2x-1|$. Then, if $x<\frac{1}{2}$, $f(x) = 1-2x$, if $x\geq \frac{1}{2}$, then $f(x) = 2x-1$.

Hence $\lim_{x \downarrow \frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \downarrow \frac{1}{2}}\frac{2x-1}{x-\frac{1}{2}} = +2$, but $\lim_{x \uparrow \frac{1}{2}}\frac{f(x) - f(\frac{1}{2})}{x-\frac{1}{2}} = \lim_{x \uparrow \frac{1}{2}}\frac{1-2x}{x-\frac{1}{2}} = -2$.

So, the limit $x \to \frac{1}{2}$ does not exist.

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