Is it possible to take a continuous function and add an absolute value somewhere (e.g. [math]8x^2 + 7x[/math] to [math]8x^2 + 7|x|[/math]) so that it ends up being discontinuous?
Absolute values are still continuous, so it's the sum of two continuous functions.
>>8462974
continuous: [math]\frac{d}{dx}(x)[/math]
Not continues: [math]\frac{d}{dx}(|x|)[/math]
>>8462978
This.
In fact, the Weierstrass function is continuous everywhere but fails to be differentiable at every point where it is defined.
>>8463028
rofl no. d/dx (|x|) simply does not exist.
>>8463260
yeah it does
[eqn] \frac{\mathrm{d} |x|}{\mathrm{d} x} = \begin{cases}
\:\:\; 1 & x>0\\
-1 & x<0
\end{cases}
[/eqn]
Jesus you fuckboys. A function g(x) = f(|x|) is continuous as a composition of continuous functions if g itself is continuous, but not everywhere differentiable. That does not mean its nowhere differentiable. Iconically |x| is differentiable everywhere except for the origin. You can see this by taking the derivative as x→0 using the limit definition and recalling a (real) limit exists if and only if the limits for x↑0 and x↓0 are equal.