SF1625 Envariabel - Continuity

Continuity

Key Concepts

  • Interior Point: A point within an interval that is not an endpoint.
  • Endpoint: The boundary values of an interval.
  • Continuity at a Point: A function ff is continuous at x=cx = c if ff is defined in a neighborhood of cc and limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c).
  • Right Continuity: A function ff is right-continuous at x=ax = a if ff is defined on some interval (a,c)(a, c) and limxa+f(x)=f(a)\lim_{x \to a^+} f(x) = f(a).
  • Left Continuity: A function ff is left-continuous at x=bx = b if ff is defined on some interval (c,b)(c, b) and limxbf(x)=f(b)\lim_{x \to b^-} f(x) = f(b).
  • Continuous Function: A function that is continuous at every point in its domain.
  • Discontinuous Function: A function that is not continuous at one or more points in its domain.

Intervals

  • Open Interval: I = {x : a < x < b} = (a, b) (also a,b=±a, b = \pm \infty)
  • Closed Interval: I=x:axb=[a,b]I = {x : a \leq x \leq b} = [a, b] (a,b±a, b \neq \pm \infty)
  • Half-Open Interval: I = {x : a \leq x < b} = [a, b)

Definition 4 (Continuity at a Point)

  • A function ff is continuous at a point x=cx = c if:
    • ff is defined in a neighborhood of cc.
    • limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c)

Right and Left Continuity

  • Right Continuity at x = a: ff is defined in some interval (a,c)(a, c) and limxa+f(x)=f(a)\lim_{x \to a^+} f(x) = f(a).
  • Left Continuity at x = b: ff is defined in some interval (c,b)(c, b) and limxbf(x)=f(b)\lim_{x \to b^-} f(x) = f(b).

Theorem 5

A function ff is continuous at x=cx = c if and only if it is both right- and left-continuous at x=cx = c.

  • A function ff is continuous at its left endpoint if it is right-continuous there.
  • A function ff is continuous at its right endpoint if it is left-continuous there.
  • A function ff is continuous on an interval if it is continuous at every point in the interval.

Definition 7

A function ff is a continuous function if it is continuous at every point in its domain.

Examples of Functions

  • Function 1:

    y(x)={11(x1)2amp;if xlt;1 (x1)2amp;if x1y(x) = \begin{cases} 1 - \sqrt{1 - (x - 1)^2} &amp; \text{if } x &lt; 1 \ (x - 1)^2 &amp; \text{if } x \geq 1 \end{cases}

  • Function 2:

    y(x)={1x2amp;if x0 0amp;if x=0y(x) = \begin{cases} 1 - x^2 &amp; \text{if } x \neq 0 \ 0 &amp; \text{if } x = 0 \end{cases}

  • Function 3:

    y(x)=1x2y(x) = \sqrt{1 - x^2}

Examples and Exercises

  1. Sketch the graphs of the three functions.
  2. Identify points of continuity and discontinuity for each function.
  3. Determine intervals on which the functions are continuous.
  4. Classify which functions are continuous functions.

Properties of Continuous Functions

  • Elementary functions are continuous, including:

    • Polynomials
    • Rational functions
    • Trigonometric functions and their inverses (arcus functions)
    • Power functions (xrx^r)
    • Exponential functions (axa^x) and logarithmic functions (loga(x)\log_a(x)
    • Absolute value function x=x2|x| = \sqrt{x^2}
  • Combinations and compositions of elementary functions using the four arithmetic operations are also continuous (Theorems 6 and 7).

    • Example: f(x)=x2+(sin(x))4exln(1+x2)f(x) = \frac{\sqrt{x^2 + (\sin(x))^4} - e^x}{\ln(1 + x^2)}

Example 5

Is g(x)=1xg(x) = \frac{1}{x} a continuous function?

Example 6

Is the Heaviside function a continuous function?

H(x)={0amp;if xlt;0 1amp;if x0H(x) = \begin{cases} 0 &amp; \text{if } x &lt; 0 \ 1 &amp; \text{if } x \geq 0 \end{cases}

Example 7

Let f(x)=xx2+1f(x) = \frac{x}{x^2 + 1}.

  • a) Determine the domain of f.
  • b) Identify points where f is continuous, right-continuous, and left-continuous.
  • c) Determine if f is odd, even, or neither.
  • d) Is f bounded?

Continuous Extensions

Example 8

Determine the constant kk such that the function f(x)f(x) is continuous:

f(x)={sin(x)amp;if xπ2 x+kamp;if xgt;π2f(x) = \begin{cases} \sin(x) &amp; \text{if } x \leq \frac{\pi}{2} \ x + k &amp; \text{if } x &gt; \frac{\pi}{2} \end{cases}

Example 9

The function g(x)=xcos(1x)g(x) = x \cos(\frac{1}{x}) is defined for all x0x \neq 0. Can its definition be extended to x=0x = 0 such that the resulting function is continuous on all of R\mathbb{R}? In other words, does there exist a continuous function G(x)G(x) on R\mathbb{R} such that G(x)=g(x)G(x) = g(x) for all x0x \neq 0? If such a G(x)G(x) exists, it is a continuous extension of g(x)g(x).

Properties of Continuous Functions

Theorem 8 – The Min-Max Theorem

If f(x)f(x) is continuous on the closed, finite interval [a,b][a, b], then there exist numbers pp and qq in [a,b][a, b] such that for every xx in the interval, f(p)f(x)f(q)f(p) \leq f(x) \leq f(q).

  • ff has its absolute minimum value at x=px = p, m=f(p)m = f(p).
  • ff has its absolute maximum value at x=qx = q, M=f(q)M = f(q).
Theorem 9 – The Intermediate-Value Theorem

If ff is a continuous function on the interval [a,b][a, b], and ss is a number between f(a)f(a) and f(b)f(b), then there exists a point x=cx = c, c[a,b]c \in [a, b] such that f(c)=sf(c) = s.

  • If ff is not continuous, nothing can be said about intermediate values.
Example 10

Does the Intermediate-Value Theorem apply to the Heaviside function on the interval [1,1][-1, 1]?

  • H(1)=0H(-1) = 0 and H(1)=1H(1) = 1, but there is no value in the interval that yields s=12s = \frac{1}{2}. This does not contradict the theorem because HH is not continuous.
Example 11

Show that the equation x3+x1=0x^3 + x - 1 = 0 has a solution in the interval [0,1][0, 1].

Bisection Method

If f(a) \cdot f(b) < 0 and f(x)f(x) is continuous on the interval axba \leq x \leq b, then a root can be numerically found using the bisection method with arbitrary accuracy.

Bisection Method Steps
  1. Take a new point in the middle of the interval: x=c=(a+b)2x = c = \frac{(a+b)}{2} and calculate the function value at that point, f(c)f(c).
  2. If f(c)f(c) has the same sign as f(a)f(a), then there is a root between cc and bb. Let cc be the new lower bound of the interval.
  3. If f(c)f(c) has the same sign as f(b)f(b), then there is a root between aa and cc. Let cc be the new upper bound of the interval.
  4. Repeat the process until the interval is sufficiently small.
Example 12

Determine the solution to the equation x3+x1=0x^3 + x - 1 = 0 with an error of at most 0.2 using the bisection method.

Existence Proofs

  • Why are existence proofs important?
  • It ensures that calculations are meaningful and lead to valid solutions.
Example 13

Solve the system of equations:

2x+2yamp;=4 2x+3yamp;=6 4x+5yamp;=7\begin{aligned} 2x + 2y &amp;= 4 \ 2x + 3y &amp;= 6 \ 4x + 5y &amp;= 7 \end{aligned}

  • Subtracting equation 1 from equation 2 gives y=2y = 2. Substituting into equation 1 gives x=0x = 0. However, these values do not satisfy equation 3, indicating the system has no solution.
Example 14

Find the largest positive integer.

  • Assume the largest integer is NN. Then N2N^2 is also an integer, and N2NN^2 \leq N.

    N2N0    N(N1)=0    N=0 or N=1N^2 - N \leq 0 \implies N(N - 1) = 0 \implies N = 0 \text{ or } N = 1

  • The only positive integer solution is N=1N = 1, which is incorrect.

  • This example illustrates that if a problem has no solution, calculations might produce an incorrect answer. Demonstrating existence of a solution before attempting to find it is important.