AP Calculus BC 1.12 Confirming Continuity over an Interval Study Notes - New Syllabus
AP Calculus BC 1.12 Confirming Continuity over an Interval Study Notes- New syllabus
AP Calculus BC 1.12 Confirming Continuity over an Interval Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
Confirming Continuity Over an Interval
Key Concepts:
- Confirming Continuity Over an Interval
Confirming Continuity Over an Interval
Confirming Continuity Over an Interval
A function is said to be continuous on an interval if it is continuous at every point in that interval.
Key Idea: If a function is continuous on an interval, there are no breaks, jumps, or holes in its graph throughout that interval.
Important Facts:
- Polynomial functions are continuous on all real numbers.
- Rational functions are continuous on their domain (except where the denominator is zero).
- Power, exponential, and logarithmic functions are continuous on their respective domains.
- Trigonometric functions are continuous wherever they are defined (e.g., \(\sin x\), \(\cos x\) continuous for all real numbers, but \(\tan x\) is undefined at odd multiples of \(\frac{\pi}{2}\)).
Procedure to Determine Continuity Over an Interval:
- Find the domain of the function.
- Check if the function is continuous at every point in that domain or in the given interval.
- Identify any points of discontinuity (holes, vertical asymptotes, jumps).
Example:
Determine the intervals where the function \( f(x) = \dfrac{x^2 – 4}{x^2 – x – 6} \) is continuous.
▶️ Answer/Explanation
Step 1: Find the domain.
The denominator is \( x^2 – x – 6 = (x – 3)(x + 2) \). So, the function is undefined at \( x = 3 \) and \( x = -2 \).
Step 2: The function is rational, so it is continuous everywhere except where the denominator is zero.
Step 3: Intervals of continuity:
\((-\infty, -2) \cup (-2, 3) \cup (3, \infty)\)
Answer: The function is continuous on \((-\infty, -2) \cup (-2, 3) \cup (3, \infty)\).
Example:
Determine where the function \( f(x) = \ln(x – 1) \) is continuous.
▶️ Answer/Explanation
Step 1: The function is logarithmic, so it is continuous on its domain.
\( x – 1 > 0 \Rightarrow x > 1 \).
Domain: \( (1, \infty) \).
Step 2: The function is continuous for all \( x > 1 \).
Answer: Continuous on \( (1, \infty) \).
Example:
Determine where \( f(x) = \tan x \) is continuous on \( [0, 2\pi] \).
▶️ Answer/Explanation
Step 1: The function \( \tan x = \dfrac{\sin x}{\cos x} \) is undefined where \( \cos x = 0 \).
\( \cos x = 0 \) at \( x = \dfrac{\pi}{2}, \dfrac{3\pi}{2} \).
Step 2: So the function is continuous on:
\([0, \dfrac{\pi}{2}) \cup (\dfrac{\pi}{2}, \dfrac{3\pi}{2}) \cup (\dfrac{3\pi}{2}, 2\pi]\).
Answer: Continuous on \( [0, \dfrac{\pi}{2}) \cup (\dfrac{\pi}{2}, \dfrac{3\pi}{2}) \cup (\dfrac{3\pi}{2}, 2\pi] \).
Example:
Let \( f \) be the function given below. On which of the following intervals is \( f \) continuous?
\( f(x) = \begin{cases} 3^x, & x \leq -1 \\ \frac{2x + 3}{x + 4}, & -1 < x \leq 0 \\ x^2 + 2x, & 0 < x < 4 \\ \tan(x), & x \geq 4 \end{cases} \)
(A) \((-5, 0)\)
(B) \((-0.5, 3)\)
(C) \((3, 5)\)
(D) \((5, \infty)\)
▶️ Answer/Explanation
Check each piece:
- \( 3^x \): Continuous for all real \( x \).
- \( \dfrac{2x + 3}{x + 4} \): Continuous except at \( x = -4 \) (not in interval).
- \( x^2 + 2x \): Continuous everywhere.
- \( \tan(x) \): Continuous except at \( \dfrac{\pi}{2} + n\pi \).
Check interval options:
(A) \((-5, 0)\): This covers pieces 1 & 2, boundary at -1:
LHL at -1 = \( 3^{-1} = \dfrac{1}{3}\), RHL at -1 = \( \dfrac{1}{3}\) → matches.
At 0, interval ends before polynomial starts (0 is endpoint), so no issue.
Continuous on (-5, 0).
(B) \((-0.5, 3)\): Includes 0 where function changes from rational to polynomial.
LHL at 0 = \( \dfrac{3}{4}\), RHL at 0 = 0 → discontinuous.
(C) \((3, 5)\): Includes 4 where jump occurs between polynomial and \( \tan x\).
(D) \((5, \infty)\): Includes discontinuities at multiples of \( \dfrac{\pi}{2}\).
Correct Answer: (A) \((-5, 0)\).
Example:
Which of the following functions are continuous on the interval \( 1 < x < 6 \)?
I. \( f(x) = \dfrac{x – 4}{x^2 – 16} \)
II. \( g(x) = \dfrac{x – 4}{x^2 + 16} \)
III. \( h(x) = \ln\left( \dfrac{1}{x} \right) \)
(A) I only
(B) I and II only
(C) I and III only
(D) II and III only
▶️ Answer/Explanation
Analyze each function in \( 1 < x < 6 \):
I: \( f(x) = \dfrac{x – 4}{x^2 – 16} = \dfrac{x – 4}{(x – 4)(x + 4)} \). Zero denominator at \( x = 4 \) (inside interval) → discontinuous.
II: \( g(x) = \dfrac{x – 4}{x^2 + 16} \). Denominator never zero → continuous everywhere.
III: \( h(x) = \ln\left( \dfrac{1}{x} \right) = -\ln(x) \). Domain: \( x > 0 \). For \( 1 < x < 6 \), continuous.
Answer: (D) II and III only.