#AP Calculus AB/BC: Removing Discontinuities - Your Ultimate Guide

Hey there, future AP Calculus master! 👋 Let's tackle discontinuities together, making sure you're super confident for the exam. We'll focus on making those tricky functions smooth and continuous. Let's dive in!

#1.13 Removing Discontinuities

This section is all about understanding and fixing those pesky breaks in graphs. Remember, a continuous function is one you can draw without lifting your pencil!

#⭕ What are Discontinuities?

Discontinuities are points where a function isn't... well, continuous. Think of them as gaps, jumps, or holes in your graph. We've got three main types, but we're focusing on removable discontinuities today.

Image courtesy of LibreTexts Mathematics

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#📍 Removable Discontinuities

Removable discontinuities are like "holes" in a graph. The function isn't defined at that point, but the limit exists. We can "fill" this hole to make the graph continuous.

Image courtesy of LibreTexts Mathematics

In this graph, the limit as $x$ approaches $a$ exists, but the function is undefined at $x = a$.

#🌀 Filling the Gap

To remove a discontinuity, we redefine the function's value at that point to equal the limit of the function as x approaches that point.

Example:

$$f(x) = \frac{x - 1}{x^2 - 1} \quad \text{for} \quad x \neq 1$$

There's a hole at $x = 1$. To fix it, factor and cancel:

$$\frac{x-1}{x^2-1} = \frac{x-1}{(x-1)(x+1)} = \frac{1}{x+1}$$

Now, the function is defined at $x=1$ as $\frac{1}{2}$. The discontinuity is gone!

#✏️ Practice Filling the Gap

Consider the function $g(x)$:

$$g(x) = \frac{(x^2 - 4x + 3)}{(x - 1)} \text{ for } x \neq 1$$

$$g(x) = k \text{ for } x = 1$$

What value of $k$ makes $g(x)$ continuous at $x = 1$?

Solution:

1. Factor the numerator: $x^2 - 4x + 3 = (x - 1)(x - 3)$
2. Cancel the $(x - 1)$ terms: $\frac{(x - 1)(x - 3)}{(x - 1)} = x - 3$
3. Evaluate at $x = 1$: $g(1) = 1 - 3 = -2$
4. Set $k = -2$ to make the function continuous.

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#📈 Piecewise Functions

Piecewise functions have different rules for different parts of their domain. To ensure continuity, the pieces must "meet" at the boundaries.

#✔️ Ensuring Continuity

For a piecewise function $f(x)$ to be continuous at $x = a$:

• The left-hand limit as $x \to a$ must exist (let's call it $L$).
• The right-hand limit as $x \to a$ must exist (let's call it $M$).
• $L = M = f(a)$.

Example:

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

Image courtesy of mathcoachblog

At $x=2$, both pieces approach $1$, and $f(2) = 1$. Thus, the function is continuous.

#✏️ Practice Ensuring Continuity

Consider the function:

$$f(x) = \begin{cases} \frac{x^2 + 5x + 4}{a(x + 4)}, & x \neq -4 \\ a, & x = -4 \end{cases}$$

What must $a$ be for the function to be continuous?

Solution:

1. Factor and cancel: $\frac{x^2 + 5x + 4}{a(x + 4)} = \frac{(x + 1)(x + 4)}{a(x + 4)} = \frac{x + 1}{a}$
2. Evaluate at $x = -4$: $\frac{-4 + 1}{a} = \frac{-3}{a}$
3. Set equal to $a$: $\frac{-3}{a} = a$
4. Solve for $a$: $-3 = a^2 \implies a = \pm \sqrt{-3}$. Since we can't have imaginary value for a, there is no value of a that makes the function continuous.

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#📷 Visualizing Continuity

Graphs are super helpful for spotting discontinuities. Remember, a continuous graph is one you can draw without lifting your pencil!

#✏️ Practice Visualizing Continuity

Graph $f(x) = \frac{x^2 - 9}{x + 3}$ over the interval $[-5, 5]$. Is it continuous? Can you make it continuous?

Image courtesy of Emery

There's a discontinuity at $x = -3$. Let's remove it:

$$\frac{x^2 - 9}{x + 3} = \frac{(x - 3)(x + 3)}{x + 3} = x - 3$$

Now, $f(-3) = -6$, and the function is continuous.

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#⭐ Closing

📚 AP Calc is practice-driven! Attempt more problems, especially from past AP exams, to strengthen your understanding. Always check for continuity and practice "patching up" those functions. Great work! 👏

Practice Question

#Practice Questions

Multiple Choice Questions

1. Let $f$ be the function given by $f(x) = \frac{x^2 - 4}{x - 2}$. Which of the following statements is true? (A) $f$ is continuous at $x = 2$. (B) $f$ has a removable discontinuity at $x = 2$. (C) $f$ has a jump discontinuity at $x = 2$. (D) $f$ has an infinite discontinuity at $x = 2$.

2. The function $g$ is defined as $$g(x) = \begin{cases} cx^2 + 2x, & x \leq 1 \\ x^3 - cx, & x > 1 \end{cases}$$ where $c$ is a constant. For what value of $c$ is $g$ continuous at $x = 1$? (A) -1 (B) 0 (C) 1 (D) 2

Free Response Question

Let $f$ be the function defined by $$f(x) = \begin{cases} \frac{x^2 - 5x + 6}{x - 2}, & x \neq 2 \\ k, & x = 2 \end{cases}$$

(a) Find the limit of $f(x)$ as $x$ approaches 2. (b) Determine the value of $k$ that makes $f$ continuous at $x = 2$. (c) Sketch the graph of $f(x)$ for $x$ near 2, showing the discontinuity if it exists.

Solutions

Multiple Choice

1. (B): The function $f(x) = \frac{x^2 - 4}{x - 2}$ can be simplified to $f(x) = x + 2$ for $x \neq 2$. Thus, it has a removable discontinuity at $x = 2$.

2. (A): For $g$ to be continuous at $x = 1$, the two pieces must meet. So, $c(1)^2 + 2(1) = (1)^3 - c(1)$, which gives $c + 2 = 1 - c$. Solving for $c$, we get $2c = -1$, so $c = -\frac{1}{2}$.

Free Response

(a) To find the limit, we factor the numerator: $\frac{x^2 - 5x + 6}{x - 2} = \frac{(x - 2)(x - 3)}{x - 2} = x - 3$. The limit as $x$ approaches 2 is $2 - 3 = -1$. (2 points: 1 for factoring, 1 for limit)

(b) For $f$ to be continuous at $x = 2$, $k$ must equal the limit. Thus, $k = -1$. (1 point)

(c) The graph is a line $y = x - 3$ with a hole at $(2, -1)$. (2 points: 1 for the line, 1 for the hole)