AP Calculus AB/BC · Unit 1 · Topic 1.12
Confirming Continuity over an Interval
Use function families, domains, operations, and endpoint one-sided limits to find continuity intervals.
1. Topic Focus
Build the language of limits, connect numerical, graphical, and algebraic representations, and use continuity theorems with verified hypotheses.
This topic: Use function families, domains, operations, and endpoint one-sided limits to find continuity intervals.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
A rational function is continuous on intervals separated by denominator zeros.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
Continuity over an interval is a point-by-point claim
A function is continuous on an interval when it satisfies the appropriate continuity condition at every point belonging to that interval. One bad point is enough to make the claim false.
The endpoint condition depends on whether the interval includes that endpoint. This is why the interval notation must be read before any calculations begin.
Open, closed, and half-open intervals
| Interval | Interior requirement | Left endpoint | Right endpoint |
|---|---|---|---|
| \((a,b)\) | Two-sided continuity at every \(x\in(a,b)\) | Not included | Not included |
| \([a,b]\) | Two-sided continuity at every \(x\in(a,b)\) | \(\lim\limits_{x\to a^+}f(x)=f(a)\) | \(\lim\limits_{x\to b^-}f(x)=f(b)\) |
| \([a,b)\) | Two-sided continuity at every \(x\in(a,b)\) | Right-continuous at \(a\) | Not included |
| \((a,b]\) | Two-sided continuity at every \(x\in(a,b)\) | Not included | Left-continuous at \(b\) |
Infinity is never an included endpoint, so interval notation always uses a parenthesis beside \(\pm\infty\).
Basic functions are continuous on their domains
| Function family | Continuity domain | Restrictions to inspect |
|---|---|---|
| Polynomial, exponential, absolute value | All real numbers | None beyond the real input |
| Rational \(p(x)/q(x)\) | Where \(q(x)\ne0\) | Zeros of the original denominator |
| Even root \(\sqrt[n]{u(x)}\) | Where \(u(x)\ge0\) | Included zero boundaries and excluded negative regions |
| Logarithm \(\log(u(x))\) | Where \(u(x)>0\) | Zero and negative arguments are excluded |
| \(\sin x\), \(\cos x\) | All real numbers | None |
| \(\tan x\), \(\sec x\) | Where \(\cos x\ne0\) | Odd multiples of \(\pi/2\) |
| \(\cot x\), \(\csc x\) | Where \(\sin x\ne0\) | Integer multiples of \(\pi\) |
The phrase “continuous on its domain” does not mean “continuous on every real interval.” A rational function can be continuous at every point where it is defined while its domain is split by vertical asymptotes or holes.
Operations that preserve continuity
If \(f\) and \(g\) are continuous on an interval, then their sums, differences, constant multiples, and products are continuous there. A quotient is continuous wherever its denominator is nonzero:
For a composition \(f(g(x))\), require the inner function to be continuous and its outputs to stay in a portion of the outer function's continuity domain. Domain restrictions from every layer must be combined.
A workflow for maximal continuity intervals
- Find the real domain of the original expression.
- Factor denominators and identify every excluded input.
- Solve inequalities imposed by roots, logarithms, and other compositions.
- Place all excluded or boundary points on a number line.
- Split the domain into connected intervals.
- Include a finite boundary only when the function is defined there and has the required one-sided continuity.
- Write the maximal intervals; do not join intervals across a missing point.
Original restrictions survive algebraic cancellation
For
the simplified formula is continuous at 1, but the original function is not defined there. Thus \(r\) is continuous on \((-\infty,1)\) and \((1,\infty)\), not across all real numbers. Equivalent nearby formulas preserve limits but do not automatically restore excluded domain points.
Piecewise functions require junction checks
Each piece may be continuous on its own interval while the complete function fails at a boundary. At every included junction \(x=c\), verify
After checking every junction and every endpoint, combine adjacent intervals only when continuity actually holds across their shared boundary.
Confirming a stated closed interval
To prove continuity on \([a,b]\), organize the justification into three parts:
- Show continuity at every interior point of \((a,b)\), often by citing a continuous function family and its domain.
- Show right continuity at \(a\).
- Show left continuity at \(b\).
A discontinuity outside \([a,b]\) is irrelevant. A single discontinuity inside it invalidates the whole closed-interval claim.
Why interval continuity matters
Later existence theorems require continuity over a complete interval, not merely at selected points. Before invoking a theorem, verify that the function is continuous everywhere on the exact interval named in its hypotheses.
Common interval errors
- Using brackets at \(\pm\infty\).
- Including a logarithmic boundary where the argument equals zero.
- Excluding an even-root boundary where the radicand equals zero and the appropriate one-sided limit matches.
- Forgetting denominator zeros after factors cancel.
- Checking the pieces of a piecewise function but not their junctions.
- Requiring a two-sided limit at an included endpoint of a stated domain interval.
- Claiming continuity on \([a,b]\) after checking only the endpoint values.
6. Detailed Worked Example and Error Check
Example 1: Combine a root restriction with a denominator restriction.
The root requires \(x\ge1\), while the denominator excludes 4. The maximal continuity intervals are
The bracket at 1 is valid because \(f(1)=0\) and the right-hand limit equals 0. The point 4 must be excluded.
Example 2: Split a rational-function domain.
The denominator vanishes at \(-2\) and 0. Since a rational function is continuous wherever its denominator is nonzero,
Example 3: A logarithm has open domain boundaries.
The logarithm requires \(5-x^2>0\), so \(-\sqrt5<x<\sqrt5\). Therefore \(h\) is continuous on
The endpoints are excluded because the logarithm's argument would be zero.
Example 4: Verify a piecewise function over a closed interval. Let
Each polynomial piece is continuous on its portion. At the junction,
The function is right-continuous at \(-1\) and left-continuous at 3, so it is \(\boxed{\text{continuous on }[-1,3]}\).
Example 5: Simplification does not fill a hole.
Although \(q(x)=x+1\) for \(x\ne1\), the original expression is undefined at 1. Therefore \(q\) is not continuous on \([0,2]\). Its maximal continuity intervals remain \((-\infty,1)\) and \((1,\infty)\).
7. AP Reasoning Routine
Read one-sided behavior first, choose a matching limit procedure, and justify conclusions with definitions or theorem conditions.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Determine the requested continuity intervals and justify every included or excluded endpoint.
(a) State the maximal intervals of continuity of \(f(x)=x^4-3x+1\).
(b) Find the maximal intervals for \(g(x)=\frac{x+2}{x^2-x-6}\).
(c) Find the continuity interval of \(h(x)=\sqrt{9-x^2}\).
(d) Find the maximal intervals for \(p(x)=\frac{\ln(x+2)}{x-1}\).
(e) State the maximal intervals of continuity of \(\tan(2x)\) using an integer \(k\).
(f) Find \(m\) so that \(F(x)=mx+2\) for \(x<1\) and \(F(x)=x^2+3\) for \(x\ge1\) is continuous on all real numbers.
(g) Explain why \(r(x)=\sqrt{x}\) is continuous on \([0,4]\), explicitly checking the endpoints.
(h) Find the maximal continuity intervals of \(s(x)=\sqrt{\frac{x-1}{x+2}}\).
Check the solution
Part (a) is continuous on \((-\infty,\infty)\) because it is a polynomial. In part (b), \(x^2-x-6=(x-3)(x+2)\); the original denominator excludes \(-2\) and 3 even though one factor cancels, so the intervals are \((-\infty,-2)\), \((-2,3)\), and \((3,\infty)\). In part (c), \(9-x^2\ge0\) gives \([-3,3]\); the square-root function is right-continuous at \(-3\) and left-continuous at 3. In part (d), \(x>-2\) and \(x\ne1\), giving \((-2,1)\) and \((1,\infty)\). In part (e), \(\cos(2x)\ne0\), so the maximal intervals are \((-\pi/4+k\pi/2,\ \pi/4+k\pi/2)\) for integers \(k\). In part (f), the left limit is \(m+2\), while the right limit and \(F(1)\) are 4; therefore \(m=2\). In part (g), the function is continuous at every interior point, \(\lim\limits_{x\to0^+}\sqrt{x}=0=r(0)\), and \(\lim\limits_{x\to4^-}\sqrt{x}=2=r(4)\). In part (h), require \((x-1)/(x+2)\ge0\) with \(x\ne-2\); a sign chart gives \((-\infty,-2)\cup[1,\infty)\), and the square root is continuous on each of those maximal intervals.