Continuity and the Intermediate Value Theorem

Understanding Continuity

In calculus, a function is considered continuous at a point $x = a$ if you can draw its graph at that point without lifting your pen. Mathematically, three conditions must be met for $f(x)$ to be continuous at $x = a$:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists (the left-hand and right-hand limits are equal).
3. $\lim_{x \to a} f(x) = f(a)$.

Types of Discontinuities

When a function fails to be continuous, it has a discontinuity. There are three main types:

• Removable Discontinuity (Hole): The limit exists at $x = a$, but the actual function value $f(a)$ is either undefined or doesn't match the limit.
• Jump Discontinuity: The left-hand limit and right-hand limit both exist but are not equal to each other. This often happens in piecewise functions.
• Infinite Discontinuity: The function shoots off to positive or negative infinity as it approaches $a$, creating a vertical asymptote.

Example: Making a Piecewise Function Continuous

Problem: Find the value of $k$ that makes the following function continuous at $x = 2$:

$f(x) = \begin{cases} x^2 + k & x < 2 \\ 3x - 1 & x \geq 2 \end{cases}$

Solution: For $f(x)$ to be continuous at $x = 2$, the left-hand limit must equal the right-hand limit (and the function value).

1. Find the right-hand limit and $f(2)$: $\lim_{x \to 2^+} (3x - 1) = 3(2) - 1 = 5$
2. Find the left-hand limit: $\lim_{x \to 2^-} (x^2 + k) = 2^2 + k = 4 + k$
3. Set them equal to each other: $4 + k = 5$ $k = 1$

The Intermediate Value Theorem (IVT)

The Intermediate Value Theorem (IVT) is a powerful rule that relies on continuity.

It states: If a function $f(x)$ is continuous on a closed interval $[a, b]$, and $N$ is any number strictly between $f(a)$ and $f(b)$, then there must exist at least one number $c$ in the interval $(a, b)$ such that $f(c) = N$.

In plain English: A continuous function cannot skip values. If you are driving and accelerate from 30 mph to 50 mph, you must have been traveling at exactly 40 mph at some point.

Example: Using IVT to Find a Root

Problem: Use the IVT to show that the equation $x^3 - x - 1 = 0$ has a solution on the interval $[1, 2]$.

Solution: Let $f(x) = x^3 - x - 1$.

1. Since $f(x)$ is a polynomial, it is continuous everywhere, including on the interval $[1, 2]$.
2. Evaluate the function at the endpoints:
   • $f(1) = 1^3 - 1 - 1 = -1$ (This is negative)
   • $f(2) = 2^3 - 2 - 1 = 5$ (This is positive)
3. Since $f(1) < 0$ and $f(2) > 0$, the number $0$ falls strictly between $f(1)$ and $f(2)$.

By the Intermediate Value Theorem, there must be at least one value $c$ in $(1, 2)$ where $f(c) = 0$. Therefore, the equation has a solution in this interval.