The right branch is written (0, \infty). Why is the end at 0 round?

The function has no value at x = 0, so the branch has no point there.

The function has a value at 0, but it is not continuous from the right.

Infinity always takes a round bracket, so 0 does too.

State the intervals on which w(x) = \dfrac{x - 1}{x^2 - 7x + 10} is continuous.

(-\infty, 2) \cup (2, 5) \cup (5, \infty)

Topic 1.12

State intervals of continuity

Builds on combining the domain restrictions of a formula. 6 minutes.

A pencil that cannot cross a break

The graph of \(f(x) = \dfrac{6}{x^2 - 2x - 3}\) comes in three pieces, because the denominator is 0 at \(x = -1\) and at \(x = 3\). A faint dashed line marks each of these inputs.

Press Trace the left piece. The orange dot moves along the curve like a pencil that never leaves the paper, and it cannot cross the dashed line at \(x = -1\). A teal bar on the axis shows the inputs it covers.

Trace the other two pieces the same way. The middle piece lies between \(-1\) and 3, and the right piece starts after 3.

Each stretch that the dot traces without a gap is an interval of continuity. The breaks \(-1\) and 3, where the function has no value, belong to no interval.

Breaks cut the axis into stretches

The numbers are gone. Two breaks, the red crosses where the function has no value, cut the axis into three stretches, shown as teal bars.

Drag a break. The bars change length, but there are still three, because a break is never inside a bar.

The bars are the intervals of continuity. Each bar has an open end at a break, because the function has no value there.

Notation

An interval is the set of inputs between two numbers. A round bracket leaves an end out, and a square bracket includes it. The middle bar shows both kinds of end in turn.
\[(a, b) \qquad [a, b] \qquad [a, b)\]
Infinity is not a number, so it always takes a round bracket.
\[(a, \infty) \qquad (-\infty, b]\]
A function is continuous on an interval when it is continuous at every input in it, so the interval never includes a break.
To state the intervals, list the breaks from left to right. The intervals are the stretches between them, and the symbol \(\cup\) joins separate intervals.
\[(-\infty, b_1) \cup (b_1, b_2) \cup (b_2, \infty)\]

When it fails

Press Draw one interval across both breaks. The red bar crosses both breaks and has a hole at each one.

The function has no value at a break, so an interval that contains a break is not an interval of continuity. Write one interval for each stretch.

Example 1: A denominator with two zeros

State the intervals on which \(g(x) = \dfrac{x - 1}{x^2 - 16}\) is continuous.

The function has no value where the denominator is 0, so solve for these inputs. The red rings mark these breaks.
\[x^2 - 16 = 0 \Rightarrow x = -4 \text{ or } x = 4\]
The breaks \(-4\) and 4 cut the axis into three stretches. The numerator is 0 at \(x = 1\), but that gives the value \(g(1) = 0\), so 1 is not a break.
Write each stretch with round brackets at the breaks and at infinity.
\[(-\infty, -4) \cup (-4, 4) \cup (4, \infty)\]

Example 2: A root in a denominator

State the interval on which \(k(x) = \dfrac{x + 5}{\sqrt{x - 2}}\) is continuous.

The root needs \(x - 2 \ge 0\). It stands in a denominator, so it cannot be 0. Therefore \(x - 2 > 0\), and \(x > 2\).
A tempting wrong move: The root is defined at \(x = 2\), so the interval starts with a square bracket.
\[[2, \infty)\]
At \(x = 2\), the denominator is \(\sqrt{0} = 0\), so the fraction has no value. The curve rises beside the dashed line and never reaches it, so the red dot marks a point that is not on the graph.
The function is continuous at every input of its domain, \(x > 2\). Use a round bracket at 2, because 2 is not in the domain.
\[(2, \infty)\]

Practice

State the intervals on which \(p(x) = \dfrac{x}{(x + 2)(x - 1)(x - 5)}\) is continuous.

Find the inputs where the denominator is 0. Each one is a dashed line that cuts the axis, so count the stretches.

Solution

The denominator is 0 when one of its factors is 0, so the breaks are \(-2\), 1 and 5.
\[x = -2, \quad x = 1, \quad x = 5\]
Three breaks cut the axis into four stretches. Write each one with round brackets.
\[\begin{aligned} &(-\infty, -2) \cup (-2, 1) \\ &\cup (1, 5) \cup (5, \infty) \end{aligned}\]

Questions

Use the graph of \(f(x) = \ln(x^2)\) for the first two questions. The faint dashed line is at \(x = 0\).

On which intervals is \(f\) continuous?
- \((-\infty, 0) \cup (0, \infty)\)
- \((-\infty, \infty)\)
- \((0, \infty)\)
- \([0, \infty)\)
Answer
\((-\infty, 0) \cup (0, \infty)\)
The right branch is written \((0, \infty)\). Why is the end at 0 round?
- The function has no value at \(x = 0\), so the branch has no point there.
- The function has a value at 0, but it is not continuous from the right.
- Infinity always takes a round bracket, so 0 does too.
Answer
The function has no value at \(x = 0\), so the branch has no point there.
State the intervals on which \(w(x) = \dfrac{x - 1}{x^2 - 7x + 10}\) is continuous.
- \((-\infty, 2) \cup (2, 5) \cup (5, \infty)\)
- \((-\infty, 1) \cup (1, \infty)\)
- \((2, 5)\)
- \((-\infty, 2] \cup [5, \infty)\)
Answer
\((-\infty, 2) \cup (2, 5) \cup (5, \infty)\)

Summary

You find the breaks, the inputs where the function has no value, and write one interval for each stretch between them.

A function is continuous on an interval only if it is continuous at every input in it, so no interval contains a break.

You give these intervals when a question asks where a function is continuous.

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