Is $|f(x) + C_1|$ the same thing as $|f(x)| + C_2$, where the $C$'s are just arbitrary constants? In other words, can one take out the constant from an absolute value sign?
The context of this is that I'm trying to take the absolute value of an integral whose antiderivative evaluates to $f(x) + C$.
Thanks in advance!
Here is a graphical example (not a substitute for rigorous analysis) that may help someone having difficulty following a more rigorous explanation.
Below is a blue plot with $C_{1} = \frac12$, red plot ($C<0$), orange plot ($C_2=0$), and green plot ($C_2>0$) for some chosen values of $C_{2}$.
As can be seen, for any value of $C_{2}$, the shape remains the same, only shifting in position on the $y$-axis. However, the blue plot fundamentally has a different shape that cannot be achieved with any shift on $y$-axis of the other plots.
Set $f(x)=x,~$ $g(x)=\lvert x+C_1\rvert,~$ $h(x)=\lvert x\rvert+C_2$. Assume $g(x)=h(x)$ everywhere.
Then $g(-C_1)=0$ and $h(-C_1)=\lvert C_1 \rvert+C_2.$ Thus, $\lvert C_1 \rvert=-C_2,$ so $C_2 \leq 0$.
$g(0)=\lvert C_1 \rvert$ and $h(0)=C_2$, so $\lvert C_1 \rvert=C_2$ and $C_2 \geq 0.$ Therefore, $C_2=0$. It also means $C_1=0$ contradicting to $C_1$ being arbitrary integration constant.
