What puts the red cross at x = 1 on the middle line?

The denominator \ln x is 0 at x = 1.

The numerator 1 has no value at x = 1.

Combine domain restrictions

Builds on finding where a single expression is defined. 6 minutes.

A formula with two restrictions

The expression \(h(x) = \dfrac{\sqrt{x + 3}}{x - 1}\) has two parts, and each part restricts the input.

The square root needs \(x + 3 \ge 0\), so \(x \ge -3\). The upper teal bar under the graph shows these allowed inputs.

The denominator cannot be 0, so \(x \ne 1\). The lower teal bar shows the allowed inputs, and the red ring marks the input that the denominator excludes.

Drag the input along the axis. The orange dot appears only when the input lies on both teal bars.

The domain keeps the inputs that satisfy both restrictions. Press Show the domain. The teal bar on the axis starts at \(x = -3\) and has a gap at \(x = 1\).

The domain is the overlap

Each part of a formula allows a set of inputs, shown as a teal bar. The red cross marks an input that the second part excludes, such as a zero of a denominator.

Drag either edge. The bottom bar holds only the inputs that lie on both bars above it, so an input outside any one bar is outside the domain.

The domain is the overlap of the allowed sets.

Notation

Split the formula into its parts. Look for each root, each logarithm and each denominator.
Write the restriction of each part. Each restriction gives a set of allowed inputs, and the graph shows each one as a teal bar.
\[\begin{aligned} \sqrt{u}&: \; u \ge 0 \\ \ln u&: \; u > 0 \\ \frac{1}{u}&: \; u \ne 0 \end{aligned}\]
The domain is the intersection of the allowed sets, which holds the inputs that satisfy every restriction.
In a composition \(f(g(x))\), the input \(x\) must first be in the domain of \(g\). Then the output \(g(x)\) must be an input that \(f\) accepts.
\[g(x) \text{ must lie in the domain of } f\]
Each part is continuous where it is defined, so the combined function is continuous at every input of its domain.

When it fails

Press Cancel the common factor. Because \(x^2 - 4 = (x - 2)(x + 2)\), the expression \(\dfrac{x^2 - 4}{x - 2}\) simplifies to \(x + 2\), the orange dashed line, which has no restriction.

The original denominator is still 0 at \(x = 2\), so the original graph has a red ring there. Read the domain from the expression as it was given, before you cancel.

Example 1: A sum

Find the domain of \(f(x) = \sqrt{8 - x} + \ln(x + 1)\).

The root needs a radicand that is not negative. The upper teal bar shows the inputs it allows.
\[8 - x \ge 0 \Rightarrow x \le 8\]
The logarithm needs a positive input. The lower teal bar shows the inputs it allows.
\[x + 1 > 0 \Rightarrow x > -1\]
A sum has a value only where both parts do. Keep the inputs on both teal bars.
\[-1 < x \le 8\]
Test the edges. At \(x = 8\), the root is \(\sqrt{0} = 0\), so 8 belongs to the domain. At \(x = -1\), the logarithm receives 0, so \(-1\) does not.

Example 2: A composition

Find the domain of \(g(x) = \ln(\sqrt{x} - 2)\).

The inner function is \(\sqrt{x}\). It needs \(x \ge 0\), as the teal bar under the graph shows.
\[x \ge 0\]
A tempting wrong move: The root is the only part that restricts \(x\), so the domain is \(x \ge 0\).
\[x \ge 0\]
At \(x = 1\), the root gives 1, so the logarithm receives \(1 - 2 = -1\). The value \(\ln(-1)\) does not exist, and the curve has no point at \(x = 1\).
The outer function is the logarithm, so its input \(\sqrt{x} - 2\) must be positive. Then \(\sqrt{x} > 2\), and because both sides are positive, squaring gives \(x > 4\).
\[\sqrt{x} - 2 > 0 \Rightarrow x > 4\]
Combine the restrictions. Every input with \(x > 4\) also has \(x \ge 0\), so the domain is \(x > 4\).

Practice

Find the domain of \(p(x) = \dfrac{\ln x}{\sqrt{6 - x}}\).

Find the inputs that each part allows. The root stands in a denominator, so it cannot be 0. Drag the input to test each edge.

Solution

The logarithm needs a positive input, so \(x > 0\).
\[x > 0\]
The root needs \(6 - x \ge 0\). It also stands in a denominator, so it cannot be 0. Therefore \(6 - x > 0\), and \(x < 6\).
\[6 - x > 0 \Rightarrow x < 6\]
Keep the inputs on both teal bars. Both edges are left out, because at \(x = 6\) the denominator is \(\sqrt{0} = 0\) and at \(x = 0\) the logarithm receives 0.
\[0 < x < 6\]

Questions

Use the number lines for \(r(x) = \dfrac{1}{\ln x}\) for the first two questions. The top line comes from the logarithm, and the middle line comes from the denominator.

Which input is in the domain of \(r\)?
- \(x = 0.5\)
- \(x = 1\)
- \(x = 0\)
- \(x = -2\)
Answer
\(x = 0.5\)
What puts the red cross at \(x = 1\) on the middle line?
- The denominator \(\ln x\) is 0 at \(x = 1\).
- The logarithm needs \(x > 1\).
- The numerator 1 has no value at \(x = 1\).
Answer
The denominator \(\ln x\) is 0 at \(x = 1\).
Find the domain of \(s(x) = \ln(x - 2) + \dfrac{1}{x - 5}\).
- \(x > 2\) and \(x \ne 5\)
- \(x > 2\)
- \(x \ne 2\) and \(x \ne 5\)
- \(x \ge 2\) and \(x \ne 5\)
Answer
\(x > 2\) and \(x \ne 5\)

Summary

You find the restriction of each part of a formula and keep the inputs that satisfy all of them.

This works because a combined formula has a value only where every part has one, so its domain is the overlap of the allowed sets.

Use it for a sum, product, quotient or composition, and read the domain before you simplify.

This is one of four skills in topic 1.12 of Which & Why: AP Calculus.