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An implicit equation specifies a relation between variables without isolating a designated output as an explicit function of the inputs. For example, $x^2+y^2=1$ specifies a circle. On $-1<x<1$, it determines the two real branches $y=\pm \sqrt{1-x^2}$, each an implicit function when one branch is selected. The implicit function theorem gives local conditions for a unique differentiable branch; implicit differentiation computes its derivative without first solving explicitly.

An implicit function is a function specified through a relation between its input and output, rather than an explicit output formula. Locally, an equation $F(x,y)=0$ may specify $y=g(x)$ if one output $y$ is selected for each input $x$ in a suitable domain. Different branches can define different functions; the relation itself need not give a unique output globally. For example, $x^2+y^2=1$ gives the two branches $y=\pm \sqrt{1-x^2}$ on $-1<x<1$.

The implicit function theorem gives sufficient conditions for a unique local differentiable branch, while implicit differentiation computes its derivative by the chain rule. If $F$ is continuously differentiable and $F_y\ne 0$, that derivative is $g^{\prime }(x)=-F_x/F_y$ along the branch.

The tight axis-aligned bounding box is $Q=[0,2]\times [0,2]$. Bisect vertically at $x=1$, giving a right child $R=[1,2]\times [0,2]$. For a scalar implicit function representing the same constructive solid geometry solid, useWith the convention $f_i\le 0$ inside, set intersection corresponds to the maximum, and set union to the minimum. These operations encode membership; $\Phi$ is not a signed distance function.

Since every $f_i$ is an affine function, its extreme values on $R$ occur at corners. Exact real-arithmetic interval arithmetic gives:

Each leaf range is exact, but combining independent intervals forgets that the expressions share $x,y$. The final enclosure need not be the exact range of $\Phi$. For example, equality $\Phi =-1$ would require both $f_6\le -1$, hence $y\ge 1$, and $\min (f_2,f_3)\le -1$, hence $y\le 0$ on $R$, which is impossible. The enclosure is still sufficient for a reliable three-way decision:Here the result is unresolved. This is a real mixed box, rather than only a pessimistic test: $(5/4,3/4)$ is strictly inside, whereas $(3/2,3/2)$ is outside. A zero endpoint alone can also mean contact with the boundary, so equality must not be treated as proof of strict interior or strict exterior. In floating-point arithmetic, certified bounds require outward rounding.

Continue splitting unresolved boxes, normally alternating axes or splitting the longest side. Propagate the interval arithmetic results through the constructive solid geometry tree, discard exterior boxes and stop refining strict interior boxes. In sufficiently small boundary boxes, locate crossings of the active implicit equations with box edges and connect them into boundary line segments. Retain corners and resolve ambiguous multiple crossings by further refinement; here every genuine boundary piece is straight, so its supporting implicit equation allows exact intersection rather than a curved approximation. A maximum boundary-box diagonal $\epsilon$ gives spatial localization at that scale, provided no boundary components are discarded; corner-sign sampling alone provides no such guarantee.

The tight box saves empty-space work, but it is awkward for extracting facets. Entire boundary line segments, notably $EO$, $OA$ and $DE$, lie on the initial box boundary, and $CD$ lies on the first splitting straight line. A sign-change-only method may have no exterior sample on the other side and lose such pieces; exact zeros also persist through refinement. A slightly padded initial axis-aligned bounding box, together with an explicit rule for boundary-coincident pieces, makes the faceting more robust. The tight box is therefore economical for enclosure but not the best unqualified choice for a simple faceting implementation.

Put $q={\phi }^{\prime }(y-\beta z)$. Apply implicit differentiation with respect to $x$ and $y$, using the chain rule:ConsequentlyAt a point where the assumed differentiable function $z(x,y)$ exists, the first equation forces $\alpha -\beta q\ne 0$. Combining the equations therefore givesThe assertion is understood on such differentiable branches of the implicit function; arbitrary parameter choices need not produce one everywhere.