Math Problem Statement

Sketch a single function f that is defined everywhere except x = 2 and satisfies the following conditions: f(1) = 1, f(3) = 1, lim_{x -> 1-} f(x) = 1, lim_{x -> 1+} f(x) = 0, lim_{x -> 2-} f(x) = infinity, lim_{x -> 2+} f(x) = infinity, lim_{x -> 3-} f(x) = 1, lim_{x -> 3+} f(x) = 1, and f is continuous at all x except at x = 1, x = 2, and x = 3.

Solution

The question asks to sketch a single function ff that is defined everywhere except at x=2x = 2, and the following conditions must hold:

  1. f(1)=1f(1) = 1, f(3)=1f(3) = 1
  2. limx1f(x)=1\lim_{x \to 1^-} f(x) = 1 and limx1+f(x)=0\lim_{x \to 1^+} f(x) = 0
  3. limx2f(x)=\lim_{x \to 2^-} f(x) = \infty and limx2+f(x)=\lim_{x \to 2^+} f(x) = \infty
  4. limx3f(x)=1\lim_{x \to 3^-} f(x) = 1 and limx3+f(x)=1\lim_{x \to 3^+} f(x) = 1
  5. ff is continuous everywhere except at x=1x = 1, x=2x = 2, and x=3x = 3.

Explanation of the Conditions:

  1. At x=1x = 1:

    • The function approaches 1 from the left (limx1f(x)=1\lim_{x \to 1^-} f(x) = 1), but approaches 0 from the right (limx1+f(x)=0\lim_{x \to 1^+} f(x) = 0). This indicates a jump discontinuity at x=1x = 1.
    • The function value at x=1x = 1 is f(1)=1f(1) = 1, which means the point (1,1)(1, 1) is on the graph.
  2. At x=2x = 2:

    • The limits limx2f(x)=\lim_{x \to 2^-} f(x) = \infty and limx2+f(x)=\lim_{x \to 2^+} f(x) = \infty suggest that there is a vertical asymptote at x=2x = 2, as the function tends to infinity on both sides of x=2x = 2.
  3. At x=3x = 3:

    • Both the left-hand and right-hand limits as x3x \to 3 are equal to 1 (limx3f(x)=1\lim_{x \to 3^-} f(x) = 1 and limx3+f(x)=1\lim_{x \to 3^+} f(x) = 1), and the function value at x=3x = 3 is also 1 (f(3)=1f(3) = 1). This indicates the function is continuous at x=3x = 3.
  4. Elsewhere:

    • The function is continuous at all other points except x=1x = 1, x=2x = 2, and x=3x = 3, meaning there are no other discontinuities.

Sketch of the Function:

  • At x=1x = 1: There is a jump discontinuity where the function jumps from 1 to 0 as xx passes from left to right.
  • At x=2x = 2: There is a vertical asymptote because the function tends to infinity from both sides.
  • At x=3x = 3: The function approaches 1 from both sides and is continuous.

The Graph:

The graph confirms these conditions. The function drops at x=1x = 1, there is a vertical asymptote at x=2x = 2, and it continues smoothly at x=3x = 3, maintaining a value of 1 on either side of 3.


Would you like more details or have any questions?

Here are some related questions to explore further:

  1. How can we determine the type of discontinuity a function has?
  2. What is the difference between jump discontinuities and vertical asymptotes?
  3. How can limits be used to describe the behavior of functions near points of discontinuity?
  4. How would the behavior of the function change if the limits at x=2x = 2 were not infinity?
  5. How do we ensure that a function is continuous at a point?

Tip: Remember that a vertical asymptote occurs when the function value tends to infinity as xx approaches a particular point.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Discontinuities
Vertical Asymptotes
Continuity

Formulas

lim_{x -> a} f(x)
Discontinuity at a point: jump, infinite
Vertical asymptote: lim_{x -> a} f(x) = infinity

Theorems

Definition of continuity at a point
Properties of limits
Vertical asymptotes

Suitable Grade Level

Grades 11-12 or early college