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Domains and Function Representations · Tutorial 217 of 999

Boundary Inclusion and Exclusion in Domains

Use the defining conditions for roots, logarithms, and division to decide whether a domain boundary is included or excluded.

Beginner 10 min read

What You'll Learn

  • Identify candidate boundaries by setting a restriction equal to its threshold.
  • Decide why an even-root boundary is included when its radicand is zero.
  • Explain why a logarithm’s zero-argument boundary is excluded.
  • Exclude points that make a denominator zero.
  • Check boundary points by substituting them into the original formula.
  • Describe when a domain includes, excludes, or only partly includes its boundary.

Why Boundary Points Need a Check

A domain condition often describes a region using an inequality. The equality case is where a boundary may occur, but it does not always tell you whether a point on that boundary is allowed. That depends on the operation in the formula.

For an even root, a radicand of zero is allowed. For a logarithm, an argument of zero is not allowed. For a denominator, a value of zero is not allowed. These differences determine whether an edge or curve is part of the domain.

The practical rule is simple: first find the candidate boundary by replacing the inequality with equality; then substitute a point on that boundary into the original formula. The equality condition alone is not a universal instruction to include or exclude the point.

Key takeaway: An even-root restriction uses a non-strict inequality, so its zero-radicand boundary is included. Logarithmic arguments must be strictly positive, and denominators must be nonzero, so their zero-value boundaries are excluded.

What Does Boundary Inclusion Mean?

Suppose a formula is defined on one side of a line or curve. The boundary is the edge separating that region from points outside it. A boundary point is included if it belongs to the domain, and excluded if it does not.

For example, the inequality \(x\geq 0\) describes the right half-plane together with the line \(x=0\). The inequality \(x>0\) describes the right half-plane without that line. The equality \(x=0\) identifies the candidate boundary; the inequality symbol says whether it is included.

Domain restrictions come from the formula’s operations. For the restrictions considered here, the conditions are:

OperationRequired conditionEquality case
Even root, such as a square rootRadicand is greater than or equal to zeroAllowed: the radicand may equal zero
LogarithmArgument is strictly greater than zeroNot allowed: the argument cannot equal zero
DenominatorDenominator is not equal to zeroNot allowed: the denominator cannot equal zero

These conditions apply to real-valued formulas. A zero radicand gives a real value for an even root, but a logarithm of zero and division by zero are undefined. For a formula with more than one restriction, a point belongs to the domain only if it satisfies every restriction.

Even Roots: Equality Is Allowed

The real square root \(\sqrt{u}\) is defined when \(u\geq 0\). In particular, \(\sqrt{0}=0\), so the equality case is valid. If the radicand is a function of several variables, the candidate boundary is where that radicand equals zero.

For example, a condition \(9-x^2-y^2\geq 0\) can be written as \(x^2+y^2\leq 9\). Its boundary is the circle \(x^2+y^2=9\), and the non-strict inequality includes that circle.

Worked Example: Checking the Boundary of a Square-Root Domain

Find the domain of \(f(x,y)=\sqrt{9-x^2-y^2}\), and decide whether the point \((0,3)\) belongs to it.

The radicand must be nonnegative:

$$ 9-x^2-y^2\geq 0. $$

Rearranging gives:

$$ x^2+y^2\leq 9. $$

Thus the domain is the closed disk of radius \(3\), including its circular boundary. At \((0,3)\), the radicand is:

$$ 9-0^2-3^2=9-0-9=0. $$

The formula evaluates to \(\sqrt{0}=0\), so \((0,3)\) is included. As a check on the inequality, \(0^2+3^2=9\), which satisfies \(x^2+y^2\leq 9\) with equality. A point such as \((0,3.1)\) is outside because \(0^2+3.1^2=9.61>9\), making the radicand negative.

Logarithms: Equality Is Excluded

The real logarithm \(\ln(u)\) is defined only when \(u>0\). It is not defined at \(u=0\). Therefore, when the argument approaches zero along a boundary, the boundary points where it equals zero do not belong to the domain.

The distinction between \(u\geq 0\) and \(u>0\) is essential. Replacing the strict inequality with a non-strict one would incorrectly include points where the logarithm has no real value.

Worked Example: A Logarithm with a Line Boundary

Find the domain of \(g(x,y)=\ln(4-x-y)\), and check the point \((1,3)\).

The argument of the logarithm must be strictly positive:

$$ 4-x-y>0. $$

Solving for the sum of the variables gives:

$$ x+y<4. $$

The domain is the open half-plane on the side of the line \(x+y=4\) where \(x+y<4\). The line itself is the candidate boundary, and the strict inequality excludes it.

At \((1,3)\), the logarithm’s argument is:

$$ 4-1-3=0. $$

Thus \(g(1,3)=\ln(0)\), which is undefined, so \((1,3)\) is excluded. For comparison, at \((1,2)\) the argument is \(4-1-2=1>0\), and \(g(1,2)=\ln(1)=0\), so that point is included. Substituting both points confirms the strict inequality describes the domain correctly.

Denominators: Zero Is Excluded

A quotient is defined only where its denominator is nonzero. Thus, if a denominator is \(D(x,y)\), the condition is \(D(x,y)\neq 0\). The equation \(D(x,y)=0\) identifies points to remove from the domain.

Unlike a root condition, this is not a region that includes one side of an equality curve. The denominator’s zero set is excluded wherever it occurs. Points on either side can still be in the domain if the denominator is nonzero there.

Worked Example: Removing a Circular Boundary

Find the domain of \(h(x,y)=\dfrac{1}{x^2+y^2-1}\), and determine whether the point \((1,0)\) is allowed.

The denominator must not equal zero:

$$ x^2+y^2-1\neq 0. $$

Equivalently, the domain condition is:

$$ x^2+y^2\neq 1. $$

The excluded set is the circle of radius \(1\). At \((1,0)\), the denominator is:

$$ 1^2+0^2-1=1+0-1=0. $$

Therefore \(h(1,0)\) would require division by zero, so \((1,0)\) is excluded. As a check, at \((0,0)\) the denominator is \(0+0-1=-1\), which is nonzero, so the point is included and \(h(0,0)=1/(-1)=-1\). At \((2,0)\) the denominator is \(4+0-1=3\), also nonzero, so that point is included and \(h(2,0)=1/3\). The domain consists of all points except those on the circle.

Why the Equality Test Works

The boundary test follows directly from the definitions of the real operations involved. It is not a separate convention about geometric edges.

1
Translate the operation into its condition.
An even root requires a nonnegative radicand; a logarithm requires a positive argument; a denominator requires a nonzero value.
2
Identify the equality case.
Set the radicand or argument equal to zero, or set the denominator equal to zero, to find candidate boundary points.
3
Apply the original condition at equality.
The root condition permits zero, while the logarithm and denominator conditions do not.

To justify the rule fully, let \(u\) be the value produced by a radicand, logarithm argument, or denominator at a point. The real square root exists exactly when \(u\geq 0\); therefore, if the boundary value is \(u=0\), the expression has value \(\sqrt{0}=0\). The real logarithm exists exactly when \(u>0\), so \(u=0\) fails its defining condition. A quotient with denominator \(u\) is defined exactly when \(u\neq 0\), so \(u=0\) also fails that condition. Hence the same equality value is included for the root and excluded for the other two operations.

When several restrictions occur in one formula, the domain is their intersection: a point must pass every test. An allowed equality for one operation cannot compensate for a failed condition elsewhere. For instance, a point where a square-root radicand is zero is still excluded if another part of the formula has a logarithm argument equal to zero or a denominator equal to zero.

Common Mistake: Treating Every Boundary the Same Way

A frequent error is to assume that equality always means inclusion because a boundary looks like an edge of the region. The reverse error is to assume that every equality boundary must be removed. Both mistakes ignore the operation that created the restriction.

For example, the following calculation is wrong for \(q(x,y)=\sqrt{x+y}\):

$$ x+y\geq 0 \quad\Longrightarrow\quad x+y>0. $$

The second line improperly removes the equality case. At \((1,-1)\), the radicand is \(1+(-1)=0\), and \(q(1,-1)=\sqrt{0}=0\). The correction is to keep \(x+y\geq 0\), including its boundary line.

For a logarithm, however, changing \(>\) to \(\geq\) creates the opposite error. If \(r(x,y)=\ln(x+y)\), the condition \(x+y\geq 0\) would include points on \(x+y=0\), where the argument is zero and the logarithm is undefined. The correct condition is \(x+y>0\).

A Reliable Boundary-Checking Routine

When reading a formula, use a short sequence of checks rather than relying on the shape of a sketch:

  1. Write the domain condition for each operation in the formula.
  2. Use equality to locate the candidate boundary for each inequality restriction.
  3. Check the equality case using the operation’s condition: nonnegative for an even-root radicand, positive for a logarithm argument, and nonzero for a denominator.
  4. Substitute a specific candidate boundary point into the original formula.
  5. If there are multiple restrictions, confirm that the point satisfies all of them.

A sketch can help show which side of a line or curve satisfies an inequality, but it cannot replace the equality check. A boundary may be included, excluded, or subject to different conditions at different parts of a domain when several restrictions are combined.

Boundary rule: Before deciding whether an edge belongs to a domain, ask which operation imposed the restriction. Include the equality case for an even root; exclude it for a logarithm or a denominator that becomes zero.

Check Your Understanding

For each question, identify the relevant equality case and decide whether the boundary points are included.

  1. What inequality describes the domain of \(f(x,y)=\sqrt{16-x^2-y^2}\), and is the circle \(x^2+y^2=16\) included?
  2. For \(g(x,y)=\ln(7-x+y)\), what line is the candidate boundary, and does a point on that line belong to the domain?
  3. For \(h(x,y)=1/(x-y)\), which set must be removed from the domain?
  4. Why is replacing the condition \(u>0\) for a logarithm with \(u\geq 0\) incorrect?
  5. If a point makes an even-root radicand zero but also makes a denominator zero, does it belong to the domain? Explain using the restrictions.