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: 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: What indicates a removable discontinuity in a function's graph?

Answer: A hole in the graph where a point can be filled. Removable discontinuities appear as holes that can be filled.

Flashcard 9: 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 10: 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 11: 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 12: What is the removable discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$?

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

Flashcard 13: 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 14: 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 15: What is the removable discontinuity in $h(x) = \frac{x^2 - 1}{x + 1}$?

Answer: At $x = -1$, as $h(x)$ can be redefined by simplifying. Factor $x^2-1=(x+1)(x-1)$ to cancel $(x+1)$ and get $x-1$.

Flashcard 16: 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 17: 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 18: 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 19: 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: What is the removable discontinuity in $h(x) = \frac{x^2 - 1}{x + 1}$?

Answer: At $x = -1$, as $h(x)$ can be redefined by simplifying. Factor $x^2-1=(x+1)(x-1)$ to cancel $x+1$ and get $x-1$.

Flashcard 26: 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 27: 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 28: 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 29: 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 30: 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 31: 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 32: 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 33: What is the removable discontinuity in $f(x) = \frac{x^2 - 1}{x - 1}$?

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

Flashcard 34: Identify the removable discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$.

Answer: At $x = 2$, as $f(x)$ can be redefined by simplifying to $x + 2$. Factor $x^2-4=(x-2)(x+2)$ to cancel $(x-2)$ and simplify to $x+2$.

Flashcard 35: 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 36: 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 37: Which step can help identify a removable discontinuity in a rational function?

Answer: Factor the numerator and denominator to find common factors. Common factors that cancel create removable discontinuities at their zeros.

Flashcard 38: 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 39: 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 40: 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 41: State the condition for a removable discontinuity at $x = a$.

Answer: $\lim_{x \to a} f(x)$ exists, but $f(a)$ is not defined or $f(a) \neq \lim_{x \to a} f(x)$. The limit must exist but the function value is either undefined or different.

Flashcard 42: 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 43: 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 44: 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 45: 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 46: 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 47: 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 48: 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 49: Identify the removable discontinuity in $f(x) = \frac{x^2 - 4}{x - 2}$.

Answer: At $x = 2$, as $f(x)$ can be redefined by simplifying to $x + 2$. Factor $x^2-4=(x-2)(x+2)$ to cancel $(x-2)$ and simplify to $x+2$.

Flashcard 50: 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 51: 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 $x+1=2$.

Flashcard 52: 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 53: 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 54: 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 55: 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 56: 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 57: 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 58: 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 59: 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 60: 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 61: 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 62: 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 63: 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 64: State the condition for a removable discontinuity at $x = a$.

Answer: $\lim_{x \to a} f(x)$ exists, but $f(a)$ is not defined or $f(a) \neq \lim_{x \to a} f(x)$. The limit must exist but the function value is either undefined or different.

Flashcard 65: 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 $x+1=2$.

Flashcard 66: 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 67: Which step can help identify a removable discontinuity in a rational function?

Answer: Factor the numerator and denominator to find common factors. Common factors that cancel create removable discontinuities at their zeros.

Flashcard 68: 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 69: 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 70: 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 71: 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 $x-2$.