Math Problem Statement

Show that f(x) has a jump discontinuity at x = 5 by calculating the limits from the left and right at x = 5. The piecewise function f(x) is defined as follows: - For x < 5: f(x) = x^2 + 3x + 7 - For x = 5: f(x) = 7 - For x > 5: f(x) = -2x + 4

Solution

The image shows a mathematical problem involving the identification of a jump discontinuity in a piecewise function f(x)f(x) at x=5x = 5. Here's a step-by-step approach to address this.

Problem Breakdown:

  1. Definition of a Jump Discontinuity: A function f(x)f(x) has a jump discontinuity at x=ax = a if:

    • limxaf(x)\lim_{x \to a^-} f(x) exists,
    • limxa+f(x)\lim_{x \to a^+} f(x) exists, and
    • The left and right limits are not equal.
  2. Piecewise Function Definition:

    \begin{cases} x^2 + 3x + 7 & \text{if } x < 5, \\ 7 & \text{if } x = 5, \\ -2x + 4 & \text{if } x > 5. \end{cases}$$
  3. Task: We need to show that f(x)f(x) has a jump discontinuity at x=5x = 5 by calculating the left-hand limit limx5f(x)\lim_{x \to 5^-} f(x) and the right-hand limit limx5+f(x)\lim_{x \to 5^+} f(x).

Step-by-Step Solution:

1. Left-hand Limit (x5x \to 5^-):

For x<5x < 5, f(x)=x2+3x+7f(x) = x^2 + 3x + 7. limx5f(x)=limx5(x2+3x+7).\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (x^2 + 3x + 7). Substitute x=5x = 5 into the expression: f(5)=52+3(5)+7=25+15+7=47.f(5^-) = 5^2 + 3(5) + 7 = 25 + 15 + 7 = 47.

2. Right-hand Limit (x5+x \to 5^+):

For x>5x > 5, f(x)=2x+4f(x) = -2x + 4. limx5+f(x)=limx5+(2x+4).\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (-2x + 4). Substitute x=5x = 5 into the expression: f(5+)=2(5)+4=10+4=6.f(5^+) = -2(5) + 4 = -10 + 4 = -6.

3. Check for Jump Discontinuity:

Since limx5f(x)=47\lim_{x \to 5^-} f(x) = 47 and limx5+f(x)=6\lim_{x \to 5^+} f(x) = -6, and these two limits are not equal, there is a jump discontinuity at x=5x = 5.

Graphing f(x)f(x):

The graph would show different behaviors for x<5x < 5, x=5x = 5, and x>5x > 5:

  • For x<5x < 5, the graph follows the parabola x2+3x+7x^2 + 3x + 7.
  • At x=5x = 5, the function takes a constant value 77.
  • For x>5x > 5, the graph follows the line 2x+4-2x + 4.

Would you like more details or any specific explanations on this problem?

Here are 5 related questions you can explore:

  1. What are the conditions for a function to have a removable discontinuity?
  2. How do you classify other types of discontinuities (e.g., infinite discontinuity)?
  3. What is the difference between a jump discontinuity and a removable discontinuity?
  4. How can you extend this piecewise function to make it continuous at x=5x = 5?
  5. What is the significance of limits when analyzing the behavior of piecewise functions?

Tip: When evaluating limits for piecewise functions, always make sure to consider the specific domain for each piece of the function carefully.

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Math Problem Analysis

Mathematical Concepts

Limits
Discontinuity
Piecewise Functions

Formulas

lim x→a f(x)
Left-hand and Right-hand Limits

Theorems

Definition of Jump Discontinuity

Suitable Grade Level

Grades 11-12 (High School Calculus)