Domain and Evaluation Error-Checking Workshop

Why Error-Checking Matters

A multivariable function assigns an output to an ordered input. A correct evaluation therefore depends on two things: using the coordinates in their stated order and checking that the input is allowed by the function’s domain.

Many errors occur before the arithmetic begins. A coordinate can be substituted into the wrong variable, a restriction can be forgotten, or a boundary point can be treated as included when it is excluded. Even a neatly simplified expression is not a function value if the original formula is undefined at the input.

Core check: Before evaluating $f(a,b)$, confirm that $(a,b)$ belongs to the domain of $f$. Then substitute $a$ for the first input variable and $b$ for the second, in that order.

A Reliable Checking Procedure

The domain of a function $f$ of two variables is the set of all ordered pairs $(x,y)$ for which its defining rule gives a valid output. When an expression contains several operations with restrictions, an input must satisfy every applicable restriction at once.

Definition: If a formula requires conditions $C_1(x,y), C_2(x,y), \ldots, C_n(x,y)$ to be defined, its domain is the set of ordered pairs satisfying all of them: $$ \{(x,y): C_1(x,y)\text{ and }C_2(x,y)\text{ and }\cdots\text{ and }C_n(x,y)\}. $$

For common real-valued expressions, an even root requires a nonnegative radicand, a logarithm requires a positive argument, and a denominator must be nonzero. When a root occurs in a denominator, its radicand must be strictly positive: it must be defined, and the resulting denominator cannot equal zero.

These conditions are combined by intersection, not by choosing whichever one seems most restrictive. For example, if one part of a formula requires $x\geq 0$ and another requires $y>0$, the domain consists only of pairs satisfying both inequalities. Fulfilling one does not compensate for failing the other.

Write the input in order.
For $f(x,y)$, the input $(a,b)$ means $x=a$ and $y=b$. Do not rearrange the coordinates.

List every restriction.
Check roots, logarithms, denominators, and any stated branch or contextual conditions.

Test the proposed input.
Substitute its coordinates into each restriction. If even one fails, the function has no value there.

Evaluate and verify.
Once the input is admissible, substitute in the original formula, simplify each operation, and check that the final expression is defined.

Worked Examples: Catching Common Errors

Worked Example: A Swapped Input and an Excluded Line

Consider $$ f(x,y)=\frac{\sqrt{9-x^2-y^2}}{x-y}. $$ Find the domain and evaluate $f(1,2)$. Also check whether the boundary point $(2,2)$ is allowed.

The square root requires $$ 9-x^2-y^2\geq 0, $$ which is equivalent to $x^2+y^2\leq 9$. The denominator also requires $x-y\neq 0$, or $x\neq y$. Thus both restrictions must hold: $$ D_f=\{(x,y)\in\mathbb R^2:x^2+y^2\leq 9,\ x\neq y\}. $$

At $(1,2)$, the disk condition gives $1^2+2^2=5\leq9$, and the denominator is $1-2=-1\neq0$, so the input is in the domain. Substitution in the stated order gives $$ f(1,2)=\frac{\sqrt{9-1^2-2^2}}{1-2} =\frac{\sqrt{9-1-4}}{-1} =\frac{\sqrt4}{-1} =\frac{2}{-1}=-2. $$

The calculation checks directly: the numerator is $2$, the denominator is $-1$, and their quotient is $-2$. Reversing the inputs would instead produce $f(2,1)=2$, because the radicand remains $4$ while the denominator becomes $2-1=1$. That is a different ordered input and a different function value.

The point $(2,2)$ lies in the disk since $2^2+2^2=8\leq9$, but its denominator is $2-2=0$. It is therefore excluded. A point can satisfy the root condition and still fail the domain test.

Worked Example: A Missing Strict Logarithm Restriction

Consider $$ g(x,y)=\ln(y-x^2)+\sqrt{4-x-y}. $$ Find its domain, then evaluate $g(1,2)$ and decide whether $g(1,1)$ exists.

The logarithm requires its argument to be strictly positive: $$ y-x^2>0,\qquad\text{so}\qquad y>x^2. $$ The square root requires $$ 4-x-y\geq0,\qquad\text{so}\qquad x+y\leq4. $$ Both conditions are needed, giving $$ D_g=\{(x,y)\in\mathbb R^2:y>x^2,\ x+y\leq4\}. $$

For $(1,2)$, the logarithm argument is $2-1^2=1>0$, and the radicand is $4-1-2=1\geq0$. Thus the point is in the domain. Its value is $$ g(1,2)=\ln(2-1^2)+\sqrt{4-1-2} =\ln(1)+\sqrt1 =0+1=1. $$ Here $\ln(1)=0$ and $\sqrt1=1$, so the result is defined.

For $(1,1)$, the square-root radicand is $4-1-1=2$, which is allowed. But the logarithm argument is $1-1^2=0$, and $\ln(0)$ is undefined. Therefore $g(1,1)$ does not exist.

A common wrong domain calculation is to replace $y>x^2$ with $y\geq x^2$, on the grounds that the square-root condition permits equality at a boundary. That would incorrectly include $(1,1)$. The correction is to preserve the strict inequality from the logarithm: equality makes its argument zero, which is not positive.

Worked Example: A Boundary That Must Be Excluded

Let $$ h(x,y)=\frac{x^2+2y}{\sqrt{5-x-y}}. $$ Find its domain and evaluate $h(1,2)$. Check the tempting boundary input $(2,3)$.

The square root in the denominator must be defined and nonzero. Therefore its radicand must satisfy the strict condition $$ 5-x-y>0, $$ or $x+y<5$. Hence $$ D_h=\{(x,y)\in\mathbb R^2:x+y<5\}. $$

For $(1,2)$, $1+2=3<5$, so the input is allowed. Substituting in order, $$ h(1,2)=\frac{1^2+2(2)}{\sqrt{5-1-2}} =\frac{1+4}{\sqrt2} =\frac{5}{\sqrt2}. $$ The numerator is $5$, and the denominator is $\sqrt2$, which is positive and nonzero. Thus the value is defined. If desired, rationalizing gives $5\sqrt2/2$, an equivalent value.

At $(2,3)$, the radicand is $5-2-3=0$. The square root itself would be defined, but it appears in the denominator, making the denominator zero. So $h(2,3)$ is undefined. The boundary $x+y=5$ is excluded, not included.

Why the Checks Work

Each restriction comes from a specific operation in the formula. If an operation is undefined at an input, then the entire expression does not assign a real output there. Conversely, if every operation in the formula is defined at that input, the operations can be carried out in their stated order to produce its value.

This gives a simple justification for the procedure: testing every restriction is necessary because any failed operation prevents evaluation; testing them all is sufficient when those are the formula’s only restrictions because every component can then be evaluated and combined. In particular, simplification after substitution cannot repair an operation that was undefined in the original formula. The domain belongs to the given function, not merely to a simplified-looking result.

Common Error Patterns

Error	Why it fails	Correction
Swapping the coordinates	An ordered pair $(a,b)$ assigns $a$ to the first variable and $b$ to the second.	Write $x=a,\ y=b$ before substituting.
Checking only one restriction	A formula with multiple restricted operations requires all of them to be valid.	Make a separate check for each root, logarithm, and denominator.
Including every boundary	Different operations impose different boundary rules; a denominator cannot be zero.	Translate the exact condition, including whether it is strict.
Calculating before checking the domain	The proposed input may make an operation undefined.	Test admissibility first; report undefined if any condition fails.

Why Careful Evaluation Matters

A domain check is not an optional formality. It determines whether a function value exists at all. This matters in formula-based models, tables, and later calculations: an invalid input cannot be made valid by careful arithmetic or by rounding.

When the proposed value is valid, showing the intermediate substitutions makes the result auditable. It reveals whether the coordinates were placed correctly and whether each operation used the intended inputs. When the value is invalid, naming the failed restriction gives a precise reason rather than an unsupported statement that the expression “does not work.”

Takeaway: Preserve input order, check every domain condition with its correct strict or non-strict inequality, and only then evaluate the formula.

Check Your Understanding

For each question, state the relevant domain check before evaluating or deciding that a value is undefined.

For $p(x,y)=\sqrt{x+y}$, is $(-2,2)$ in the domain? What is $p(-2,2)$?
For $q(x,y)=\ln(x-y)$, is $(3,1)$ allowed? What changes for $(1,3)$?
For $r(x,y)=1/(x+2y)$, find the restriction on $x,y$. Is $(2,-1)$ in the domain?
For $s(x,y)=\sqrt{4-x^2-y^2}/(y-x)$, list both restrictions and decide whether $(0,2)$ is allowed.
Explain why a zero radicand is permitted in a numerator root but not in a denominator root.

Error	Why it fails	Correction
Swapping the coordinates	An ordered pair \((a,b)\) assigns \(a\) to the first variable and \(b\) to the second.	Write \(x=a,\ y=b\) before substituting.
Checking only one restriction	A formula with multiple restricted operations requires all of them to be valid.	Make a separate check for each root, logarithm, and denominator.
Including every boundary	Different operations impose different boundary rules; a denominator cannot be zero.	Translate the exact condition, including whether it is strict.
Calculating before checking the domain	The proposed input may make an operation undefined.	Test admissibility first; report undefined if any condition fails.