In the simplest case, I have
$$f(x) = a x^2 + b x^3, $$
where $a$ and $b$ have some error $\Delta a$ and $\Delta b$ respectively. I know that in the absence of error, the equation
$$\frac{df}{dx} = 0$$
has solutions at $x = 0$ and $x = - \frac{2 a}{3b}$. How do I find the changes to the minima in the presence of the errors $\Delta a$ and $\Delta b$?
EDIT: And now, in general, I have an expansion
$$f(x_1, x_2, ..., x_N) = \sum\limits_{i} \alpha_{i} x_i + \sum\limits_{i} \sum\limits_j \beta_{ij} x_i x_j + \sum\limits_{i} \sum\limits_j \sum\limits_k \gamma_{ijk} x_i x_j x_k + ...$$
up to sixth order, where the coefficients $\alpha_i, \beta_{ij}, \gamma_{ijk}, ...$ in principle could have their own errors $\Delta \alpha_i, \Delta \beta_{ij}, \Delta \gamma_{ijk}, ...$.
Does there exist some general expression (of course depending on the functional form of $f$) that the solutions to
$$\frac{df}{dx_k} = 0, \,\,\mathrm{for}\,\,k = 1, 2, ... N$$
can be found with the influence of the errors?
$x=0$ is still $x=0$ after a change in $a$ or $b$.
$$ - \frac{2(a+\Delta a)}{3(b + \Delta b} = - \frac{2a}{3b} + \frac{2(a \Delta b - b \Delta a)}{3 b (b + \Delta b)}$$ If $\Delta b$ is small compared to $b$, the term on the right is approximately $$ \dfrac{2 a}{3b^2} \Delta b - \dfrac{2}{3b} \Delta a$$