Continuity, Discontinuity, and Intermediate Value Theorem

Understanding Continuity and Discontinuity

• Continuity at a point: A function $f$ is continuous at a point $a$ if $\lim_{x \to a} f(x) = f(a)$.

• Continuity on an interval: A function $f$ is continuous on an interval $I$ if it is continuous at every point $a$ within that interval.

• Types of Discontinuity:

  • Jump discontinuity: The function "jumps" from one value to another at a specific point.

  • Infinite discontinuity: Occurs when there is a vertical asymptote, meaning the function's value approaches positive or negative infinity.

  • Oscillation discontinuity: The function oscillates rapidly around a point, not approaching a single value.

Properties of Continuous Functions

• Lines: Any linear function ($y = mx + b$) is continuous at every point across the entire real line ($(-\infty, \infty)$).

• Rational Functions: A rational function (a quotient of two polynomials, e.g., $f(x) = \frac{P(x)}{Q(x)}$) is continuous everywhere its denominator is not zero.

  • Example: For $f(x) = \frac{x^2 + 1}{x^2 - 1}$, the denominator is $x^2 - 1 = (x-1)(x+1)$. This function is undefined when $x = 1$ or $x = -1$. As $x$ approaches $1$ or $-1$, the numerator approaches a non-zero number, while the denominator approaches zero, leading to the quotient approaching infinity. Therefore, there are vertical asymptotes (infinite discontinuities) at $x = 1$ and $x = -1$. The function is continuous on the intervals: $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.

• Combinations of Continuous Functions: If $f(x)$ and $g(x)$ are continuous functions, then:

  • Their sum ($f(x) + g(x)$) is continuous.

  • Their product ($f(x) \cdot g(x)$) is continuous.

  • Their quotient ($\frac{f(x)}{g(x)}$) is continuous wherever $g(x) \ne 0$.

Ensuring Continuity in Piecewise Functions

To make a piecewise function continuous, the segments must meet at their boundaries. This means the limit from the left, the limit from the right, and the function value at the boundary point must all be equal.

• Example: Consider the piecewise function: $f(x) = \begin{cases} x^2 - 1 & \text{for } x \ge 2 \ a - 2x & \text{for } x < 2 \end{cases}$ To find the value of $a$ that makes $f(x)$ continuous:

  1. Calculate $f(2)$: Using the first piece of the function (for $x \ge 2$):
     $f(2) = 2^2 - 1 = 4 - 1 = 3$

  2. Calculate the right-hand limit as $x \to 2^+$: Using the first piece:
     $\lim_{x \to 2^+} (x^2 - 1) = 2^2 - 1 = 3$

  3. Calculate the left-hand limit as $x \to 2^-$: Using the second piece:
     $\lim_{x \to 2^-} (a - 2x) = a - 2(2) = a - 4$

  4. Set the limits and function value equal for continuity: For continuity, the left-hand limit, right-hand limit, and the function value must all be equal at $x=2$.
     $a - 4 = 3 \implies a = 7$
     Thus, setting $a = 7$ makes the function continuous at $x=2$.

Intermediate Value Theorem (IVT)

• Statement: If a function $f$ is continuous on a closed interval $[a, b]$, and Z is any number between $f(a)$ and $f(b)$ (meaning f(a) < Z < f(b) or f(b) < Z < f(a)), then there exists at least one number $c$ in the open interval $(a, b)$ such that $f(c) = Z$.

• Visual Interpretation: If you draw a continuous curve between two points ($a, f(a)$) and ($b, f(b)$), the curve must cross every horizontal line $y=Z$ that lies between $f(a)$ and $f(b)$. This means the function takes on all intermediate values between $f(a)$ and $f(b)$.

• Importance of Continuity: The IVT relies fundamentally on the function being continuous. If a function has a discontinuity (e.g., a jump), it might skip over values, and the theorem would not hold.

• Application - Finding Roots: The IVT can be used to show that an equation has at least one root (a value of $x$ for which $f(x) = 0$).

  • Problem: Show that the function $f(x) = x^5 - x^3 - 10x + 1$ has at least one root.

  • Solution Strategy: We need to find two points, $a$ and $b$, such that $f(a)$ and $f(b)$ have opposite signs. Since $f(x)$ is a polynomial, we know it is continuous on all real numbers ($(-\infty, \infty)$) and thus on any closed interval $[a, b]$.

    1. Choose a value for $x$, e.g., $x = 1$:
       $f(1) = (1)^5 - (1)^3 - 10(1) + 1 = 1 - 1 - 10 + 1 = -9$
       So, f(1) < 0.

    2. Choose another value for $x$, e.g., $x = 2$:
       $f(2) = (2)^5 - (2)^3 - 10(2) + 1 = 32 - 8 - 20 + 1 = 5$
       So, f(2) > 0.

    3. Apply IVT: Since $f(1) = -9$ (negative) and $f(2) = 5$ (positive), and $f(x)$ is continuous on the interval $[1, 2]$, by the Intermediate Value Theorem, there must exist at least one value $c$ in the interval $(1, 2)$ such that $f(c) = 0$. This proves that the function has at least one root.

Common Continuous Functions

• Polynomial functions: Continuous on $(-\infty, \infty)$.

• Exponential functions: (e.g., $a^x$, $e^x$) are continuous for all real numbers.

• Trigonometric functions: (e.g., $\sin(x)$, $\cos(x)$) are continuous for all real numbers (though inverse trig functions, tangents, secants, etc. have restricted domains of continuity).