#AP Calculus AB/BC: Continuity Over an Interval 🚀

Hey there, future calculus master! Let's make sure you're totally solid on continuity over an interval. This is a big topic, and we're going to break it down so it feels like a piece of cake. 🍰

This topic is super important because it connects limits and functions. Expect to see it in both multiple-choice and free-response questions.

# 📈 Continuity on an Interval

The official definition from the College Board is that a function is continuous on an interval if it's continuous at every single point within that interval. Think of it like a smooth, unbroken road—no potholes, no sudden cliffs. 🛣️

For example, $f(x) = ln(3x)$ is continuous on its domain, which is $(0, \infty)$.

# 🏁 Continuity for Piecewise Functions

Piecewise functions are where things get a bit more interesting! We need to check continuity for each piece and at the points where the function changes its definition. It's like making sure each road segment connects perfectly to the next.

#⛳ Checking Domain Restrictions

Watch out for these common domain restrictions:

• Square Roots: $\sqrt{z}$ requires $z \geq 0$. No imaginary numbers allowed! 🚫

  For example, $f(x) = \sqrt{3x+1}$ has a domain of $[-\frac{1}{3}, \infty)$.

  

  The graph of $f(x) = \sqrt{3x+1}$ showing its domain.

• Rational Functions: $\frac{1}{z}$ requires $z \neq 0$. Division by zero is a big no-no! 🙅‍♀️

  For example, $g(x) = \frac{1}{x+2}$ has a domain of $(-\infty, -2) \cup (-2, \infty)$.

  

  The graph of $g(x) = \frac{1}{x+2}$ showing its domain.

$$lim_{x\to a^{-}} f(x) = lim_{x\to a^{+}} f(x) = f(a)$$

# 🧮 Practice Problems

Let's get our hands dirty with some practice! 💪

#1) Continuity of a Piecewise Function

Given:

$$f(x) = \begin{cases} x+2, & x<3 \\ x^{2}- 2x + 2, & x\geq 3 \end{cases}$$

Is this function continuous on the interval $(-2,6)$?

Step 1: Check for Discontinuities in the Domains

Both $x+2$ and $x^2 - 2x + 2$ are polynomials, so they are continuous everywhere.

Step 2: Check Continuity at the Point the Function Changes Expressions

We need to check at $x = 3$:

• Left-hand limit: $lim_{x\to 3^{-}} (x + 2) = 5$
• Right-hand limit: $lim_{x\to 3^{+}} (x^2 - 2x + 2) = 5$
• Function value: $f(3) = 5$

Since all three values are equal, the function is continuous at $x=3$.


Graph of the piecewise function.

Step 3: Answering the Question

The function is continuous on the interval $(-2,6)$. You nailed it! 🎉

#2) Continuity of Rational Function

Is $f(x) = \frac{3x}{x+2}$ continuous on the interval $(-10,10)$?

Step 1: Check for Discontinuities in the Domain

The denominator cannot be zero, so $x \neq -2$. The domain is $(-\infty, -2) \cup (-2, \infty)$.

Since $x=-2$ is within the interval $(-10, 10)$, the function is not continuous on the interval.


Graph of the rational function.

# 🌟 Final Exam Focus

Alright, let's talk strategy for the big day! Here's what to keep in mind:

• High-Priority Topics: Continuity, limits, and piecewise functions are frequently tested. Make sure you understand the definitions and how to apply them.

• Common Question Types: Expect questions that ask you to identify discontinuities, determine if a function is continuous over an interval, and work with piecewise functions.

• Time Management: Practice these types of problems so you can solve them quickly and accurately.

• Common Pitfalls: Don't forget to check domain restrictions, especially for rational and radical functions. Also, make sure you're checking the left-hand and right-hand limits when dealing with piecewise functions.

Practice Question

#Practice Questions

Multiple Choice Questions

1. For what value of $k$ is the following function continuous at $x = 2$?

$$f(x) = \begin{cases} kx^2 - 1, & x < 2 \\ 3x + k, & x \geq 2 \end{cases}$$

(A) -1 (B) 1 (C) 2 (D) 3

2. Which of the following functions is continuous on the interval $(-2, 2)$?

   (A) $f(x) = \frac{1}{x}$ (B) $g(x) = \frac{1}{x+2}$ (C) $h(x) = \sqrt{x+2}$ (D) $j(x) = \frac{x}{x^2 + 1}$

Free Response Question

Consider the function $f$ defined by

$$f(x) = \begin{cases} ax^2 + bx + 1, & x \leq 1 \\ x^3 + 2x + 1, & x > 1 \end{cases}$$

(a) Find the values of $a$ and $b$ such that $f$ is continuous and differentiable at $x = 1$.

(b) With these values of $a$ and $b$, is $f$ continuous on the interval $[-2, 2]$? Justify your answer.

Answers and Scoring:

MCQ 1:

For continuity at $x=2$, we need $k(2)^2 - 1 = 3(2) + k$. Solving for $k$, we get $4k - 1 = 6 + k$, so $3k = 7$, and $k = \frac{7}{3}$. The correct answer is not listed, this is a common trick on the AP exam.

MCQ 2:

(A) has a discontinuity at $x=0$. (B) has a discontinuity at $x=-2$. (C) has a domain restriction at $x<-2$. (D) is continuous everywhere. So the correct answer is (D).

FRQ:

(a) For continuity at $x=1$, we need $a(1)^2 + b(1) + 1 = (1)^3 + 2(1) + 1$, which simplifies to $a + b + 1 = 4$, or $a + b = 3$. For differentiability, we need the derivatives to match at $x=1$. The derivative of $ax^2 + bx + 1$ is $2ax + b$, and the derivative of $x^3 + 2x + 1$ is $3x^2 + 2$. So, $2a(1) + b = 3(1)^2 + 2$, which simplifies to $2a + b = 5$. Solving the system of equations $a+b=3$ and $2a+b=5$, we get $a=2$ and $b=1$.

(b) With $a=2$ and $b=1$, the function is continuous at $x=1$. The polynomial pieces are continuous on their domains. Therefore, $f$ is continuous on the interval $[-2, 2]$.

Scoring Breakdown for FRQ:

(a) 2 points: 1 point for setting up the continuity equation, 1 point for setting up the differentiability equation, and 1 point for solving for $a$ and $b$.

(b) 1 point for correct conclusion with justification.

You've got this! Keep up the great work, and remember to stay calm and confident. You're going to do awesome! 💯