Polynomial and Absolute-Value Domains

Introduction

A formula in several variables can accept some inputs and exclude others. To determine its natural domain, ask which real inputs allow every operation in the formula to produce a real output.

For the formulas considered here, the answer is especially simple. A polynomial in two or three real variables is defined for every real input, and taking the absolute value of a real-valued polynomial does not remove any inputs. The key is to distinguish absolute value from operations that require a sign condition: the absolute value itself accepts both positive and negative real numbers.

Key takeaway: A polynomial formula, or a formula formed from polynomials using absolute value, has natural domain \(\mathbb{R}^2\) when it has two real inputs and \(\mathbb{R}^3\) when it has three real inputs.

Polynomials in Several Variables

A polynomial in two variables \(x\) and \(y\) is a finite sum of terms made from real coefficients and nonnegative integer powers of \(x\) and \(y\). For example, \(3x^2y-5xy^3+7\) is a polynomial. A polynomial in three variables \(x\), \(y\), and \(z\) is defined in the same way, using nonnegative integer powers of all three variables.

In these formulas, the allowed operations are addition, subtraction, multiplication, multiplication by real constants, and nonnegative integer powers. Each operation is defined for real inputs. In particular, the zeroth power of a real number is defined as \(a^0=1\), including when \(a=0\), as is standard for polynomial evaluation.

Definition: The natural domain of a real-valued formula is the set of all real input tuples for which the formula can be evaluated as a real number, using its stated operations. For a formula in two variables, the ambient input space is \(\mathbb{R}^2\); for a formula in three variables, it is \(\mathbb{R}^3\).

The natural domain is not always the same as a declared domain. A problem may specify that a function is to be considered only on a particular subset of its possible inputs. But if no smaller domain is declared, a polynomial in two variables has natural domain \(\mathbb{R}^2\), and a polynomial in three variables has natural domain \(\mathbb{R}^3\).

Why Every Polynomial Input Is Allowed

Take any real input tuple, not just a convenient example. Substituting its coordinates into a polynomial replaces the variables with real numbers. A nonnegative integer power of a real number is real; the product of real numbers is real; and the sum or difference of real numbers is real. Since a polynomial contains only finitely many such operations, evaluating it gives a real number.

This argument establishes the domain claim for every input, not merely for inputs that have been tested by calculation.

Result: If \(P\) is a polynomial in \(x\) and \(y\), its natural domain is \(\mathbb{R}^2\). If \(P\) is a polynomial in \(x\), \(y\), and \(z\), its natural domain is \(\mathbb{R}^3\).

Absolute Value Adds No Input Restriction

For a real number \(a\), its absolute value is defined by

|a|= \begin{cases} a, & a\ge 0,\\ -a, & a<0. \end{cases}

These two cases cover every real number. If the input is nonnegative, its absolute value is the input itself. If the input is negative, its absolute value is the opposite number, which is positive. Therefore \(|a|\) is defined for every real \(a\); there is no requirement that \(a\) already be nonnegative.

If \(P(x,y)\) is a polynomial, then for every \((x,y)\in\mathbb{R}^2\), the value \(P(x,y)\) is real. Consequently \(|P(x,y)|\) is defined. The same reasoning applies to a polynomial in three variables, and to sums of polynomial terms and their absolute values.

Result: Placing a polynomial inside absolute-value bars does not restrict its input domain. A formula formed from polynomials by addition, subtraction, multiplication, and absolute value is defined at every real input tuple of the appropriate dimension.

Worked Examples

Worked Example: A Polynomial in Two Variables

Determine the natural domain of \(f(x,y)=2x^2-3xy+y^2+4\), and evaluate it at \((1,-2)\).

The formula uses only real constants, nonnegative integer powers, multiplication, addition, and subtraction. By the polynomial result, every real pair is allowed. Thus the natural domain is \(\mathbb{R}^2\).

For the requested value, substitute \(x=1\) and \(y=-2\), showing each term:

\begin{aligned} f(1,-2) &=2(1)^2-3(1)(-2)+(-2)^2+4\\ &=2(1)+6+4+4\\ &=2+6+4+4\\ &=16. \end{aligned}

As a check, the four terms evaluate to \(2\), \(6\), \(4\), and \(4\); their sum is \(16\), as calculated. The point \((1,-2)\) belongs to \(\mathbb{R}^2\), so it is included in the domain.

Worked Example: Absolute Values of Polynomials in Two Variables

Determine the natural domain of \(g(x,y)=|x^2-4y|+|3x+y-2|\), and evaluate it at \((-1,2)\).

Both expressions inside the absolute-value bars are polynomials, so each has a real value for every real pair \((x,y)\). Absolute value is defined for positive, zero, and negative real inputs. Therefore \(g\) has natural domain \(\mathbb{R}^2\).

At \((-1,2)\), first evaluate each polynomial inside its absolute-value bars:

\begin{aligned} x^2-4y &=(-1)^2-4(2)\\ &=1-8\\ &=-7, \end{aligned} \qquad \begin{aligned} 3x+y-2 &=3(-1)+2-2\\ &=-3+2-2\\ &=-3. \end{aligned}

Now take the absolute values and add:

\begin{aligned} g(-1,2) &=|-7|+|-3|\\ &=7+3\\ &=10. \end{aligned}

Checking the result directly, the two absolute values are \(7\) and \(3\), whose sum is \(10\). In particular, the negative values inside the bars cause no domain problem.

Worked Example: A Formula in Three Variables

Determine the natural domain of \(h(x,y,z)=|x-y+2z|+(x+z)^2-2yz+3\), and evaluate it at \((1,-2,2)\).

The expression \(x-y+2z\) is a polynomial, so it gives a real number for every real triple. Its absolute value is therefore defined for every triple. The remaining terms are also polynomials. Hence the natural domain is \(\mathbb{R}^3\).

At \((1,-2,2)\), evaluate the absolute-value input and the other terms separately:

\begin{aligned} x-y+2z &=1-(-2)+2(2)\\ &=1+2+4\\ &=7, \end{aligned} \qquad \begin{aligned} (x+z)^2 &=(1+2)^2\\ &=3^2\\ &=9, \end{aligned} \qquad \begin{aligned} -2yz &=-2(-2)(2)\\ &=4(2)\\ &=8. \end{aligned}

Thus the function value is

\begin{aligned} h(1,-2,2) &=|7|+9+8+3\\ &=7+9+8+3\\ &=27. \end{aligned}

As a check, substituting the computed term values into the original expression gives \(7+9+8+3=27\). The input is a real triple and is among the allowed inputs in \(\mathbb{R}^3\).

A General Justification

The examples illustrate a general fact. The following argument explains why it applies to every polynomial and not merely to the particular formulas above.

Choose any input.
Let \((a,b)\) be any pair of real numbers, or let \((a,b,c)\) be any triple of real numbers. No sign condition is imposed on these coordinates.

Evaluate the polynomial operations.
Every nonnegative integer power of each chosen coordinate is real. Multiplying real numbers and multiplying by real coefficients gives real numbers, and finite sums and differences of real numbers are real. Thus a polynomial evaluates to a real number at the chosen input.

Apply absolute value if it appears.
The polynomial's value is real, and the definition of absolute value assigns a real value to every real input. Applying absolute value therefore introduces no further restriction.

Conclude the domain.
Because the input was arbitrary, every pair or triple of real numbers is permitted. The natural domain is the full ambient space: \(\mathbb{R}^2\) or \(\mathbb{R}^3\), respectively.

The argument depends on the operations used. It establishes the full-domain result for polynomial operations and absolute value; it does not say that every possible formula in several variables accepts every real input.

Common Mistake: Requiring a Nonnegative Interior

A frequent error is to assume that an expression inside absolute-value bars must itself be nonnegative. That condition is not part of the definition of absolute value. For example, one might incorrectly reason that \(g(x,y)=|x^2-4y|\) requires \(x^2-4y\ge 0\), and then exclude points where \(x^2-4y<0\).

Test the point \((0,1)\). The proposed condition fails there, since

$$ x^2-4y=0^2-4(1)=-4<0. $$

But the formula itself is defined:

$$ g(0,1)=|-4|=4. $$

Thus \((0,1)\) is a valid input. The corrected conclusion is that the natural domain of \(g\) is \(\mathbb{R}^2\). The inequality \(x^2-4y\ge 0\) would describe where the inside is nonnegative, not where the absolute-value formula is defined.

Why This Distinction Matters

Finding a natural domain means checking whether each operation in a formula can be carried out at an input. Polynomial operations accept real inputs, and absolute value accepts every real number. Looking at whether a polynomial happens to be positive or negative can help determine its value, but it does not determine whether its absolute value exists.

This distinction is useful whenever a formula combines several operations. First identify what each operation requires; do not add a condition merely because an expression can take negative values. For the formulas in this tutorial, there is no input to exclude from \(\mathbb{R}^2\) or \(\mathbb{R}^3\).

Check Your Understanding

For each question, focus on whether the formula is defined, not on whether the expressions inside absolute-value bars are positive.

What is the natural domain of \(p(x,y)=5x^3-2xy+8\)?
Why is \(|x-3y|\) defined even at points where \(x-3y<0\)?
Determine the natural domain of \(q(x,y,z)=|xy-z^2|+x^2+y^2+z^2\).
Evaluate \(r(x,y)=|x^2-y|\) at \((1,3)\), showing the value inside the absolute-value bars before taking its absolute value.
Explain why checking a few points is not, by itself, a complete justification that a polynomial's natural domain is all of \(\mathbb{R}^2\) or all of \(\mathbb{R}^3\).