MA161 - Lesson 8: Definition of Continuity and Types of Discontinuities

Graph Analysis: The handwritten notes show a piecewise function graph with several key points marked: a hole at x=1, a function value at (6,5), vertical asymptote at x=3, and various limit behaviors at different points including x=-5 and x=-6.

Types of Discontinuities

Three Main Types

Graph Features: The accompanying graph illustrates all three types: a removable discontinuity (hole), a jump discontinuity where function values jump from one level to another, and an infinite discontinuity showing a vertical asymptote.

Definition of Continuity

Informal Definition:

Mathematical Definition: \(f(x)\) is continuous at \(x = a\) if:

  1. \(f(a)\) is defined
  2. \(\lim_{x \to a} f(x)\) exists and is finite
  3. \(\lim_{x \to a} f(x) = f(a)\)

Examples of Continuous Functions

Functions that are continuous everywhere:

Rational functions: Continuous everywhere except where the denominator equals zero.

Example: \(f(x) = \frac{2x - 15}{x + 5}\) is continuous everywhere except at \(x = -5\).

Analyzing Discontinuities

Example 1: Removable Discontinuity

Analyze \(f(x) = \frac{x^2 + 2x - 15}{x + 5}\) at \(x = -5\).

Solution:

First, factor the numerator: \(x^2 + 2x - 15 = (x + 5)(x - 3)\)

So when \(x \neq -5\): \(f(x) = \frac{(x + 5)(x - 3)}{x + 5} = x - 3\)

Find the limit:

\[\lim_{x \to -5} f(x) = \lim_{x \to -5} (x - 3) = -5 - 3 = -8\]

Since \(f(-5)\) is not defined but the limit exists, this is a removable discontinuity.

To remove it: Define \(g(x) = x - 3\) for all \(x\), or add the point \((-5, -8)\) to the graph.

Example 2: Jump Discontinuity

Analyze the piecewise function:

\[f(x) = \begin{cases} -x^2 + 16 & \text{if } x \leq 0 \\ x + 8 & \text{if } x > 0 \end{cases}\]

Solution:

\(f(0) = -0^2 + 16 = 16\)

Left limit: \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2 + 16) = 16\)

Right limit: \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 8) = 8\)

Since the left and right limits are different (16 ≠ 8), this is a jump discontinuity at \(x = 0\).

Left and Right Continuity

Left Continuous at \(x = a\):

  1. \(f(a)\) is defined
  2. \(\lim_{x \to a^-} f(x)\) exists and is finite
  3. \(\lim_{x \to a^-} f(x) = f(a)\)

Right Continuous at \(x = a\):

  1. \(f(a)\) is defined
  2. \(\lim_{x \to a^+} f(x)\) exists and is finite
  3. \(\lim_{x \to a^+} f(x) = f(a)\)
Example 3: Finding Value for Continuity

Find the value of \(c\) such that the function is continuous at \(x = 0\):

\[f(x) = \begin{cases} -x^2 + c & \text{if } x \leq 0 \\ cx + 8 & \text{if } x > 0 \end{cases}\]

Solution:

\(f(0) = -0^2 + c = c\)

Left limit: \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x^2 + c) = c\)

Right limit: \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (cx + 8) = 8\)

For continuity: \(c = 8\)

Intermediate Value Theorem (IVT)

Intermediate Value Theorem:

Suppose \(f(x)\) is continuous on \([a,b]\). Then \(f(x)\) takes all values between \(f(a)\) and \(f(b)\).

More precisely: if \(L\) is between \(f(a)\) and \(f(b)\), then we can find \(c\) between \(a\) and \(b\) such that \(f(c) = L\).

Example: Applying IVT

Consider \(f(x) = x^3 + 2x^2 + 5\) on \([0,1]\). Show there exists \(c\) such that \(f(c) = 7\).

Solution:

\(f(x)\) is continuous (polynomial function)

\(f(0) = 0^3 + 2(0)^2 + 5 = 5\)

\(f(1) = 1^3 + 2(1)^2 + 5 = 8\)

Since \(5 < 7 < 8\), and 7 is between \(f(0)\) and \(f(1)\), IVT guarantees there exists \(c\) between 0 and 1 such that \(f(c) = 7\).

Solving: \(c^3 + 2c^2 + 5 = 7\), so \(c^3 + 2c^2 - 2 = 0\)

Key Applications of IVT: