AP Calculus BC 1.13 Removing Discontinuities Study Notes - New Syllabus
AP Calculus BC 1.13 Removing Discontinuities Study Notes- New syllabus
AP Calculus BC 1.13 Removing Discontinuities Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Removing Discontinuities
Key Concepts:
- Removing Discontinuities
Removing Discontinuities
Removing Discontinuities
If the limit of a function exists at a point of discontinuity in its graph, then the discontinuity is called a removable discontinuity.
It can be removed by defining or redefining the value of the function at that point to equal the limit.
Key Idea:
If \( \lim_{x \to a} f(x) = L \) exists but \( f(a) \) is either not defined or not equal to \( L \), then we can make the function continuous by setting \( f(a) = L \).
Conditions for Removing Discontinuity:
- The limit \(\lim_{x \to a} f(x)\) must exist and be finite.
- The function may be undefined at \( a \), or \( f(a) \neq \lim_{x \to a} f(x)\).
- Redefine \( f(a) = \lim_{x \to a} f(x)\) to make the function continuous at \( x = a \).
Piecewise Functions: For a piecewise-defined function to be continuous at a boundary point, the following must be true:
\( \text{Left-hand limit} = \text{Right-hand limit} = \text{Value of the function at the point.} \)
Example:
The function \( f(x) = \dfrac{x^2 – 4}{x – 2} \) has a discontinuity at \( x = 2 \). Show how to remove it.
▶️ Answer/Explanation
Step 1: Factorize the numerator:
\( x^2 – 4 = (x – 2)(x + 2) \).
So \( f(x) = \dfrac{(x – 2)(x + 2)}{x – 2} = x + 2, \text{ for } x \neq 2\).
Step 2: Compute \(\lim_{x \to 2} f(x)\):
\(\lim_{x \to 2} (x + 2) = 4\).
Step 3: Define \( f(2) = 4 \) to make the function continuous at \( x = 2 \).
Answer: Redefine the function as: \( f(x) = \begin{cases} x + 2, & x \neq 2 \\ 4, & x = 2 \end{cases} \)
Example:
Determine the value of \( k \) that makes the following piecewise function continuous at \( x = 1 \): \( f(x) = \begin{cases} x^2 + k, & x < 1 \\ 3x – 2, & x \geq 1 \end{cases} \)
▶️ Answer/Explanation
Step 1: For continuity at \( x = 1 \):
\(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)\).
\(\lim_{x \to 1^-} (x^2 + k) = 1^2 + k = 1 + k\).
\(\lim_{x \to 1^+} (3x – 2) = 3(1) – 2 = 1\).
Set equal: \( 1 + k = 1 \Rightarrow k = 0\).
Answer: \( k = 0 \).
Example:
Remove the discontinuity of \( f(x) = \dfrac{\sin x}{x} \) at \( x = 0 \).
▶️ Answer/Explanation
Step 1: Compute the limit as \( x \to 0 \):
\(\lim_{x \to 0} \dfrac{\sin x}{x} = 1\).
Step 2: Define \( f(0) = 1 \).
Answer: New definition: \( f(x) = \begin{cases} \dfrac{\sin x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases} \)
Example:
If the function \( f \) is continuous for all real numbers and if \( f(x) = \dfrac{x^2 – 5x + 4}{x – 1} \text{ when } x \neq 1, \) then find \( f(1) \).
▶️ Answer/Explanation
Step 1: Factorize the numerator: \( x^2 – 5x + 4 = (x – 4)(x – 1). \)
Step 2: Simplify: \( f(x) = \dfrac{(x – 4)(x – 1)}{x – 1} = x – 4, \text{ for } x \neq 1. \)
Step 3: Compute limit as \( x \to 1 \): \( \lim_{x \to 1} f(x) = 1 – 4 = -3. \)
Answer: \( f(1) = -3 \).
Example:
If the function \( f \) is continuous for all real numbers and if \( f(x) = \dfrac{x^2 + 14x + 48}{x + 8} \text{ when } x \neq -8, \) then find \( f(-8) \).
▶️ Answer/Explanation
Step 1: Factorize numerator: \( x^2 + 14x + 48 = (x + 8)(x + 6). \)
Step 2: Simplify: \( f(x) = \dfrac{(x + 8)(x + 6)}{x + 8} = x + 6, \text{ for } x \neq -8. \)
Step 3: Compute limit as \( x \to -8 \): \( \lim_{x \to -8} f(x) = -8 + 6 = -2. \)
Answer: \( f(-8) = -2 \).
Example:
Let \( f \) be the function defined by: \( f(x) = \begin{cases} \dfrac{x^2 – 2x – 15}{x – 5}, & x \neq 5 \\ a, & x = 5 \end{cases} \) For what value of \( a \) is \( f \) continuous at \( x = 5 \)?
▶️ Answer/Explanation
Step 1: Factorize numerator: \( x^2 – 2x – 15 = (x – 5)(x + 3). \)
Step 2: Simplify: \( f(x) = x + 3, \text{ for } x \neq 5. \)
Step 3: Compute limit as \( x \to 5 \): \( \lim_{x \to 5} f(x) = 5 + 3 = 8. \)
Answer: \( a = 8 \).
Example:
Let \( f \) be the function defined by: \( f(x) = \begin{cases} \dfrac{x^2 – 16x + 63}{x – 7}, & x \neq 7 \\ b, & x = 7 \end{cases} \) For what value of \( b \) is \( f \) continuous at \( x = 7 \)?
▶️ Answer/Explanation
Step 1: Factorize numerator: \( x^2 – 16x + 63 = (x – 7)(x – 9). \)
Step 2: Simplify: \( f(x) = x – 9, \text{ for } x \neq 7. \)
Step 3: Compute limit as \( x \to 7 \): \( \lim_{x \to 7} f(x) = 7 – 9 = -2. \)
Answer: \( b = -2 \).