AP Calculus AB Flashcards: Removing Discontinuities

What this deck covers

This deck focuses on Removing Discontinuities, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the simplified form of $f(x) = \frac{x^2 - 16}{x - 4}$?

Answer: $x + 4$, after factoring and canceling $(x-4)$. Factor $x^2-16=(x-4)(x+4)$ to cancel $(x-4)$ and get $x+4$.

Flashcard 2: How do you simplify $f(x) = \frac{x^2 - 1}{x - 1}$?

Answer: $x + 1$, after factoring and canceling $(x-1)$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and get $x+1$.

Flashcard 3: Does $f(x) = \frac{x^2 - 4x + 4}{x - 2}$ have a removable discontinuity?

Answer: Yes, at $x = 2$, as $f(x)$ simplifies to $x - 2$. Factor $(x-2)^2$ in numerator to cancel $(x-2)$ leaving $x-2$.

Flashcard 4: To remove a discontinuity, what must be true about $\lim_{x \to a} f(x)$?

Answer: It must exist and equal a finite value. Infinite limits cannot be used to remove discontinuities.

Flashcard 5: Determine if $f(x) = \frac{x^2 - 9}{x - 3}$ is continuous at $x = 3$.

Answer: No, it has a removable discontinuity at $x = 3$. The function is undefined at $x=3$ but has a finite limit there.

Flashcard 6: What does it mean if $\lim_{x \to a} f(x)$ does not exist?

Answer: The function has a non-removable discontinuity at $x = a$. Either one-sided limits differ or approach infinity.

Flashcard 7: What type of discontinuity does a piecewise function often have?

Answer: Jump discontinuity, if the pieces do not connect. Different function rules at boundaries often create jumps.

Flashcard 8: Identify the type of discontinuity in $g(x) = \frac{1}{x - 3}$ at $x = 3$.

Answer: Infinite discontinuity at $x = 3$. The denominator approaches zero causing infinite behavior.

Flashcard 9: Which discontinuity is not found in $y = x^3 - 3x + 2$?

Answer: No discontinuities; continuous for all real $x$. Polynomial functions are continuous everywhere with no discontinuities.

Flashcard 10: Identify the removable discontinuity in $f(x) = \frac{x^2 - 9}{x + 3}$.

Answer: At $x = -3$, as $f(x)$ can be redefined by simplifying. Factor $x^2-9=(x+3)(x-3)$ to cancel $(x+3)$ and get $x-3$.

Flashcard 11: Is $f(x) = x^2 - 4$ continuous for all real numbers?

Answer: Yes, it is a polynomial with no discontinuities. Polynomial functions have no breaks, holes, or asymptotes.

Flashcard 12: Which discontinuity type cannot be removed by redefining $f(x)$?

Answer: Jump and infinite discontinuities. These involve limits that don't exist or are infinite.

Flashcard 13: Can all discontinuities be removed?

Answer: No, only removable discontinuities can be redefined to be continuous. Only holes can be filled; jumps and infinite breaks cannot.

Flashcard 14: How can you remove the discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$?

Answer: Redefine $f(x)$ at $x = 2$ as $f(x) = x + 2$. Factor $x^2-4=(x-2)(x+2)$ to cancel $(x-2)$ and get $x+2$.

Flashcard 15: For $f(x) = x^2 - 4$, where is the removable discontinuity?

Answer: No removable discontinuity; $f(x)$ is continuous. Polynomial functions are continuous everywhere on their domains.

Flashcard 16: Identify the type of discontinuity in $f(x) = \frac{1}{x - 2}$ at $x = 2$.

Answer: Infinite discontinuity at $x = 2$. The denominator approaches zero causing the function to approach infinity.

Flashcard 17: What is a removable discontinuity?

Answer: A point where a function is not defined but can be redefined to make it continuous. This describes a hole that can be filled by redefining the function value.

Flashcard 18: Which type of discontinuity is present in $f(x) = \frac{1}{x}$ at $x = 0$?

Answer: Infinite discontinuity at $x = 0$. The denominator approaches zero while numerator stays constant.

Flashcard 19: What is the removable discontinuity in $f(x) = \frac{x^2 - 25}{x - 5}$?

Answer: At $x = 5$, as $f(x)$ can be redefined as $x + 5$. Factor $x^2-25=(x-5)(x+5)$ to cancel $(x-5)$ and get $x+5$.

Flashcard 20: What must be true for a function to be continuous at $x = a$?

Answer: $f(a) = \lim_{x \to a} f(x)$ and both must exist. All three conditions ensure no gaps, jumps, or holes in the function.

Flashcard 21: How do you redefine $f(x) = \frac{x^2 - 16}{x - 4}$ to remove the discontinuity?

Answer: Redefine $f(x)$ at $x = 4$ as $f(x) = x + 4$. Factor $x^2-16=(x-4)(x+4)$ to cancel $(x-4)$ and get $x+4$.

Flashcard 22: When is a discontinuity non-removable?

Answer: If $exists \, \lim_{x \to a} f(x)$; jump or infinite discontinuities. When limits don't exist or are infinite, discontinuities cannot be removed.

Flashcard 23: Describe a jump discontinuity.

Answer: A discontinuity where the left and right limits exist but are unequal. The one-sided limits differ, creating a break in the function.

Flashcard 24: What is a common factor indicating a removable discontinuity?

Answer: A factor that cancels in both numerator and denominator. Canceling common factors reveals removable discontinuities.

Flashcard 25: Find the removable discontinuity in $g(x) = \frac{x^2 - 1}{x - 1}$.

Answer: At $x = 1$, as $g(x)$ can be redefined as $x + 1$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and simplify to $x+1$.

Flashcard 26: How can you remove a discontinuity in $f(x) = \frac{x^2 - 9}{x - 3}$?

Answer: Redefine $f(x)$ at $x = 3$ as $f(x) = x + 3$. Factor $x^2-9=(x-3)(x+3)$ to cancel $(x-3)$ and get $x+3$.

Flashcard 27: What is the simplified form of $h(x) = \frac{x^2 + x - 6}{x - 2}$?

Answer: $x + 3$, after factoring numerator and canceling $(x-2)$. Factor $x^2+x-6=(x-2)(x+3)$ to cancel $(x-2)$ and get $x+3$.

Flashcard 28: Is the function $f(x) = \frac{x^2 - 1}{x - 1}$ continuous at $x = 1$?

Answer: No, it has a removable discontinuity at $x = 1$. The function is undefined at $x=1$ but has a finite limit there.

Flashcard 29: How do you verify a removable discontinuity exists at $x = a$?

Answer: Check if $\lim_{x \to a} f(x)$ exists and $f(a)$ is undefined or $\neq \lim_{x \to a} f(x)$. These conditions define exactly when a discontinuity can be removed.

Flashcard 30: How do you remove a discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$ at $x = 1$?

Answer: Redefine $f(x)$ as $f(x) = x + 1$ at $x = 1$. Factor $x^2-1=(x-1)(x+1)$ to cancel $(x-1)$ and get $x+1=2$.