Continuity and the Intermediate Value Theorem

Continuity at a point

f is continuous at x = a if all three hold:

1. f(a) is defined.
2. lim (x→a) f(x) exists.
3. lim (x→a) f(x) = f(a).

On the AP exam, justify continuity by checking all three conditions explicitly.

Types of discontinuity

Type	What happens	Limit exists?
Removable (hole)	limit exists but differs from f(a), or f(a) undefined	yes
Jump	left and right limits are different	no
Infinite	vertical asymptote	no

A removable discontinuity can be fixed by redefining one point.

Continuous functions

Polynomials, eˣ, sin x and cos x are continuous everywhere. Rational functions, roots, logarithms and tan x are continuous on their domains.

Making a piecewise function continuous

Set the two pieces equal at the boundary.

Example. f(x) = kx + 1 for x < 2 and x² for x ≥ 2. For continuity at x = 2: 2k + 1 = 4, so k = 3/2.

The Intermediate Value Theorem (IVT)

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

Example. f(x) = x³ + x − 1 is continuous, f(0) = −1 and f(1) = 1. Since 0 is between −1 and 1, f has a root in (0, 1).

IVT pitfalls

• Continuity on the closed interval is required. A jump can skip values.
• The IVT says a value is reached; it does not say where, or how many times.
• In free response, name the theorem and state continuity explicitly.

Quick summary

• Continuity: defined, limit exists, limit equals value.
• Removable, jump and infinite discontinuities.
• Piecewise: match the pieces at the boundary.
• IVT: continuous on [a, b] guarantees every in-between value.