so I understand that in using the three-point midpoint formula, $$f^{(1)}(x)= \frac{y_{+1}-y_{-1}}{2h} + \frac{e_{+1}-e_{-1}}{2h} - \frac{h^2}{6} f^{(3)}(\xi^*), \ \ \ \ \ \xi^* \in [a,b]$$
round-off error: $E_{round} = \frac{e_{+1}-e_{-1}}{2h}$
truncation error: $E_{trunc} = -\frac{h^2}{6} f^{(3)}(\xi^*)$
in order to find the optimal error of a point $f(x)$, when considering the round-off & truncation error, we must impose upper bounds on both $E_{round}$ & $E_{trunc}$. I get that this is because there's a 'sweet spot' at which the two combined errors $E_{total}$ is optimal.
So to derive an expression for optimal of $E_{total}$, in this slide from here, we let $\varepsilon$ & $M$ be some numbers that represent the limits that allow for the optimal of $E_{round}$ & $E_{trunc}$ to be found.
I understand why the final expression for the truncation error is $E_{trunc} = \frac{h^2}{6}M$, but I don't understand why the round-off error is $E_{round} = \frac{\varepsilon}{h}$, specifically how the person in the slide jumped from $$ \frac{|e(x_0 + h) - e(x_0-h)|}{2h} $$ to the following $$ \frac{|e(x_0 + h)| + |e(x_0-h)|}{2h} $$ why can we add the errors $e(x_0 \pm h)$?
The solution is probably simple but here's some more context from Burden's 'Numerical Analysis' just in case. Help is much appreciated
