Differentiation question? 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.
 A: 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 $\;f(x) = \left|2x - 1\right|$:

A: 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.
