AP Calculus AB Continuity and the Intermediate Value Theorem Lesson (Free PDF)

AP Calculus AB math lesson: Continuity and the Intermediate Value Theorem – Maths PDF

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This continuity lesson for AP Calculus AB covers continuity and the intermediate value theorem (Common Core AP CALC AB 1.10, AP CALC AB 1.11, AP CALC AB 1.12, AP CALC AB 1.16), with a free PDF to print.

A function is continuous at x = a when f(a) is defined, the limit as x approaches a exists, and the limit equals f(a). Discontinuities are removable (a hole), jump, or infinite. The Intermediate Value Theorem says a function continuous on [a, b] takes every value between f(a) and f(b).

AP Calculus AB · Common Core State Standards: AP CALC AB 1.10, AP CALC AB 1.11, AP CALC AB 1.12, AP CALC AB 1.16

Key facts in this continuity lesson

Key facts

Continuity needs three things: f(a) exists, the limit exists, and they are equal.
Removable discontinuities are holes; jump discontinuities have different one-sided limits; infinite discontinuities are vertical asymptotes.
To make a piecewise function continuous, set the one-sided limits equal at the boundary.
The IVT guarantees a value, but only if the function is continuous on the whole closed interval.

What does it mean to be continuous?

A function is continuous at a point if you can draw its graph through that point without lifting your pencil. In calculus, this idea is made precise with three conditions.

Continuity at x = a

A function \(f\) is continuous at \(x=a\) if (1) \(f(a)\) is defined, (2) \(\lim_{x\to a} f(x)\) exists, and (3) \(\lim_{x\to a} f(x)=f(a)\).

A function is continuous at x = a only when all three conditions hold.

What kinds of discontinuity are there?

Type	What happens	Example
Removable (hole)	The limit exists but \(f(a)\) is missing or different.	\(\frac{x^{2}-4}{x-2}\) at \(x=2\)
Jump	The one-sided limits exist but are different.	A step function such as postage by weight
Infinite	The outputs grow without bound (vertical asymptote).	\(\frac{1}{x-3}\) at \(x=3\)

To classify a discontinuity, simplify first. If a factor cancels, the discontinuity is removable. If a factor stays in the denominator while the numerator is not zero, it is infinite. If the formula changes at a boundary and the two sides disagree, it is a jump.

Removable: the limit is 3 but f(2) = 1.

Jump: the one-sided limits are 3 and 5.

How do you make a piecewise function continuous?

Set the left-hand and right-hand limits equal at the boundary, then solve for the unknown. For example, if \(f(x)=x^{2}\) for \(x < 2\) and \(f(x)=kx-2\) for \(x\ge 2\), continuity at \(2\) requires \(4=2k-2\), so \(k=3\).

What is the Intermediate Value Theorem?

Intermediate Value Theorem (IVT)

If \(f\) is continuous on the closed interval \([a,b]\) and \(k\) is any number between \(f(a)\) and \(f(b)\), then there is at least one \(c\) in \([a,b]\) with \(f(c)=k\).

A continuous graph must cross every height between f(a) and f(b).

To use the theorem on an exam, write three things: \(f\) is continuous on the interval, the values \(f(a)\) and \(f(b)\), and a sentence saying that \(k\) lies between them.

Example. Show that \(f(x)=x^{3}+x-5\) has a zero in \([1,2]\). Since \(f\) is a polynomial, it is continuous. \(f(1)=-3 < 0\) and \(f(2)=5 > 0\), so by the IVT there is a \(c\) in \((1,2)\) with \(f(c)=0\).

Continuity lesson: questions and answers

What are the three conditions for continuity?

A function is continuous at x = a if f(a) is defined, the limit of f(x) as x approaches a exists, and that limit equals f(a). If any one fails, the function is discontinuous there.

What is a removable discontinuity?

A removable discontinuity is a hole where the limit exists but the function is undefined or has a different value. Redefining the single value at that point repairs the graph.

How can you tell a jump from an infinite discontinuity?

At a jump, both one-sided limits are finite but unequal. At an infinite discontinuity, the outputs grow without bound and the graph has a vertical asymptote.

What does the Intermediate Value Theorem guarantee?

If f is continuous on [a, b] and k is between f(a) and f(b), then f(c) = k for at least one c in [a, b]. It proves a value exists but does not tell you where.

Why must f be continuous for the IVT?

A function with a jump can skip over values. For f(x) = 1/x on [-1, 1], f goes from -1 to 1 but never equals 0 because of the asymptote at x = 0.

How to use this continuity lesson

Read the definitions first, then follow the worked method line by line and redo the example on paper without looking. When every step makes sense, move on to the worksheet, then check yourself with the quiz.

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