Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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
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

share|cite|improve this answer
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
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
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
Yes, exactly! You've got it. – amWhy Jan 31 '13 at 21:41

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.