Math Problem Statement

Sketch the graph of a function y = f(x) such that f(x) is continuous on [-2, 0) ∪ (0, 1) ∪ (1, ∞) and discontinuous at x = 0 and x = 1. f(x) has a horizontal asymptote at y = 3. f(x) has a vertical asymptote at x = 0. lim x→1− f(x) = 1, lim x→1+ f(x) = 2, and f(1) = 3.

Solution

The problem asks for a sketch of a function y=f(x)y = f(x) based on a set of given conditions:

  1. Continuity and Discontinuity:

    • f(x)f(x) is continuous on [2,0)(0,1)(1,)[-2, 0) \cup (0, 1) \cup (1, \infty), meaning there are no breaks in the graph on these intervals.
    • f(x)f(x) is discontinuous at x=0x = 0 and x=1x = 1, implying breaks or jumps at these points.
  2. Asymptotes:

    • Horizontal asymptote at y=3y = 3, meaning the graph approaches y=3y = 3 as xx \to \infty.
    • Vertical asymptote at x=0x = 0, meaning the function grows unbounded near x=0x = 0.
  3. Limits and Function Values:

    • limx1f(x)=1\lim_{x \to 1^-} f(x) = 1, implying the left-hand limit as xx approaches 1 is 1.
    • limx1+f(x)=2\lim_{x \to 1^+} f(x) = 2, implying the right-hand limit as xx approaches 1 is 2.
    • f(1)=3f(1) = 3, implying a point on the graph at x=1x = 1 and f(x)=3f(x) = 3.

Interpretation:

  • For x=0x = 0: The function has a vertical asymptote, which means the function may shoot up to infinity or down to negative infinity as it nears x=0x = 0, but the function doesn't exist at x=0x = 0.

  • For x=1x = 1: The function is discontinuous here. The left-hand limit approaches 1, the right-hand limit approaches 2, and the actual value of the function at x=1x = 1 is 3. This suggests a jump discontinuity at x=1x = 1, with a value marked at 3.

  • Horizontal asymptote at y=3y = 3: As xx \to \infty, the function approaches y=3y = 3, meaning the graph levels off at this value but never touches it.

Sketch Approach:

  • Start by drawing a vertical asymptote at x=0x = 0.
  • On the left of x=0x = 0, sketch a function continuous from x=2x = -2 to 0, approaching the vertical asymptote.
  • After x=0x = 0, sketch the function so that it approaches the vertical asymptote but remains continuous until x=1x = 1.
  • At x=1x = 1, there is a jump discontinuity: the graph jumps from 1 on the left to 2 on the right and marks a point at (1,3)(1, 3).
  • Finally, for x>1x > 1, the graph should approach the horizontal asymptote at y=3y = 3 as xx \to \infty.

Would you like more details on how to draw the graph clearly?

Additional Questions:

  1. What is the significance of horizontal asymptotes in function behavior?
  2. How would the graph behave as xx \to -\infty?
  3. How does the discontinuity at x=1x = 1 affect the graph's shape?
  4. Could the vertical asymptote behavior differ depending on the direction x0x \to 0?
  5. How would you plot the function between 2-2 and 00 on this graph?

Tip: Asymptotes indicate the direction a graph heads but doesn't cross or touch at infinity (horizontal) or a specific xx-value (vertical).

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Math Problem Analysis

Mathematical Concepts

Continuity and Discontinuity
Asymptotes
Limits
Graph Sketching

Formulas

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Theorems

Limit Theorems
Asymptote Rules

Suitable Grade Level

Grades 11-12 (Pre-Calculus/Calculus)