AP Calculus BC Flashcards: Sketching Graphs Of Functions And Derivatives

What this deck covers

This deck focuses on Sketching Graphs Of Functions And Derivatives, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

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 can be concluded if $f'(x) > 0$?

Answer: $f(x)$ is increasing on that interval. Positive first derivative means function values are getting larger.

Flashcard 2: What does it mean for $f'(x)$ to be undefined?

Answer: Possible critical point or cusp. Undefined derivative often occurs at corners, cusps, or vertical tangents.

Flashcard 3: What is the significance of $f'(x) = 0$?

Answer: Potential local max, min, or saddle point. Zero derivative indicates a horizontal tangent line at that point.

Flashcard 4: What occurs at a local minimum of $f(x)$?

Answer: $f'(x) = 0$ and $f''(x) > 0$. Both conditions ensure a true local minimum exists.

Flashcard 5: What is the relationship between $f''(x)$ and concavity?

Answer: $f''(x) > 0$: concave up; $f''(x) < 0$: concave down. Second derivative test determines the direction of curvature.

Flashcard 6: What occurs at a local minimum of $f(x)$?

Answer: $f'(x) = 0$ and $f''(x) > 0$. Both conditions ensure a true local minimum exists.

Flashcard 7: What does $f''(x) > 0$ imply about $f(x)$?

Answer: Function is concave up. Positive second derivative indicates the function curves upward.

Flashcard 8: What is the relationship between $f(x)$ and $f'(x)$?

Answer: $f'(x)$ gives the slope of $f(x)$. The derivative measures how steeply the function rises or falls.

Flashcard 9: Determine the local maximum of $f(x) = -x^2 + 4x - 3$.

Answer: Local maximum at $x = 2$. Find critical point: $f'(x) = -2x + 4 = 0$ gives $x = 2$.

Flashcard 10: What does $f'(x) < 0$ imply about $f(x)$?

Answer: $f(x)$ is decreasing on that interval. Negative first derivative means function values are getting smaller.

Flashcard 11: What is the critical point of a function $f(x)$?

Answer: Where $f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where the derivative equals zero or doesn't exist.

Flashcard 12: Identify the local minimum of $f(x) = x^2 - 4x + 4$.

Answer: Local minimum at $x = 2$. Complete the square: $f(x) = (x-2)^2$, minimum at vertex.

Flashcard 13: What is the critical point of a function $f(x)$?

Answer: Where $f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where the derivative equals zero or doesn't exist.

Flashcard 14: Calculate $f'(x)$ for $f(x) = x^3 - 3x + 1$.

Answer: $f'(x) = 3x^2 - 3$. Apply power rule: derivative of $x^3$ is $3x^2$, derivative of $-3x$ is $-3$.

Flashcard 15: What is the significance of $f'(x) = 0$?

Answer: Potential local max, min, or saddle point. Zero derivative indicates a horizontal tangent line at that point.

Flashcard 16: What does $f'(x) < 0$ imply about $f(x)$?

Answer: $f(x)$ is decreasing on that interval. Negative first derivative means function values are getting smaller.

Flashcard 17: How do you determine a local maximum using derivatives?

Answer: If $f'(x) = 0$ and $f''(x) < 0$. Second derivative test: negative second derivative confirms a maximum.

Flashcard 18: When does $f(x)$ have a point of inflection?

Answer: When $f''(x)$ changes sign. Sign change in second derivative indicates concavity reversal.

Flashcard 19: Calculate $f''(x)$ for $f(x) = x^3 - 3x^2 + 1$.

Answer: $f''(x) = 6x - 6$. Differentiate twice: $f'(x) = 3x^2 - 6x$, then $f''(x) = 6x - 6$.

Flashcard 20: Find the critical points for $f(x) = x^2 - 4x + 3$.

Answer: Critical points: $x = 2$. Set $f'(x) = 2x - 4 = 0$ to find $x = 2$.

Flashcard 21: What does $f''(x) > 0$ imply about $f(x)$?

Answer: Function is concave up. Positive second derivative indicates the function curves upward.

Flashcard 22: What is the effect of $f'(x) = 0$ on $f(x)$?

Answer: Potential local extremum. Zero derivative creates a horizontal tangent, possibly an extremum.

Flashcard 23: Find inflection points for $f(x) = x^3 - 3x + 1$.

Answer: Inflection point at $x = 0$. Set $f''(x) = 6x = 0$ to find where concavity changes.

Flashcard 24: What does $f'(x) = 0$ imply about $f(x)$?

Answer: Possible local max, min, or point of inflection. Zero derivative indicates a horizontal tangent line.

Flashcard 25: What does $f''(x) = 0$ suggest about $f(x)$?

Answer: Possible inflection point. Zero second derivative may indicate where concavity changes.

Flashcard 26: What does a zero $f'(x)$ indicate?

Answer: Potential local max, min, or saddle point. Zero first derivative creates a horizontal tangent line.

Flashcard 27: Determine the local maximum of $f(x) = -x^2 + 4x - 3$.

Answer: Local maximum at $x = 2$. Find critical point: $f'(x) = -2x + 4 = 0$ gives $x = 2$.

Flashcard 28: What does the second derivative of a function indicate?

Answer: The concavity of the function. $f''(x)$ describes whether the graph curves upward or downward.

Flashcard 29: What does $f''(x) = 0$ suggest about $f(x)$?

Answer: Possible inflection point. Zero second derivative may indicate where concavity changes.

Flashcard 30: Find $f''(x)$ for $f(x) = x^4 - 4x^2 + 6$.

Answer: $f''(x) = 12x^2 - 8$. Differentiate twice using power rule: $(x^4)'' = 12x^2$, $(-4x^2)'' = -8$.

Flashcard 31: What does a zero $f'(x)$ indicate?

Answer: Potential local max, min, or saddle point. Zero first derivative creates a horizontal tangent line.

Flashcard 32: What is the effect of $f'(x) = 0$ on $f(x)$?

Answer: Potential local extremum. Zero derivative creates a horizontal tangent, possibly an extremum.

Flashcard 33: When does $f(x)$ have a point of inflection?

Answer: When $f''(x)$ changes sign. Sign change in second derivative indicates concavity reversal.

Flashcard 34: Compute $f''(x)$ for $f(x) = x^5 - 5x^3 + 10x$.

Answer: $f''(x) = 20x^3 - 30x$. Differentiate twice: $f'(x) = 5x^4 - 15x^2$, then $f''(x) = 20x^3 - 30x$.

Flashcard 35: What does the second derivative of a function indicate?

Answer: The concavity of the function. $f''(x)$ describes whether the graph curves upward or downward.

Flashcard 36: What is the relationship between $f''(x)$ and concavity?

Answer: $f''(x) > 0$: concave up; $f''(x) < 0$: concave down. Second derivative test determines the direction of curvature.

Flashcard 37: Calculate $f'(x)$ for $f(x) = 2x^3 - 5x^2 + 4x - 1$.

Answer: $f'(x) = 6x^2 - 10x + 4$. Apply power rule to each term: $6x^2 - 10x + 4$.

Flashcard 38: Determine where $f(x) = -x^3 + 3x^2$ is decreasing.

Answer: Decreasing on $(0, 2)$. Find where $f'(x) = -3x^2 + 6x = -3x(x-2) < 0$.

Flashcard 39: What does a negative $f''(x)$ indicate?

Answer: The graph is concave down. Negative second derivative means the graph curves downward.

Flashcard 40: What does a positive $f''(x)$ indicate about $f(x)$?

Answer: The graph of $f(x)$ is concave up. Positive second derivative means the graph curves upward like a bowl.

Flashcard 41: What is the first derivative of $f(x)$ used to determine?

Answer: The slope of the tangent line to $f(x)$. $f'(x)$ represents the instantaneous rate of change at any point.

Flashcard 42: Identify the intervals of increase for $f(x) = x^2 - 4x + 3$.

Answer: Increasing on $(2, \text{infinity})$. Find where $f'(x) = 2x - 4 > 0$, so $x > 2$.

Flashcard 43: Calculate $f'(x)$ for $f(x) = x^3 - 3x + 1$.

Answer: $f'(x) = 3x^2 - 3$. Apply power rule: derivative of $x^3$ is $3x^2$, derivative of $-3x$ is $-3$.

Flashcard 44: What does it mean for $f'(x)$ to be undefined?

Answer: Possible critical point or cusp. Undefined derivative often occurs at corners, cusps, or vertical tangents.

Flashcard 45: Determine where $f(x) = -x^3 + 3x^2$ is decreasing.

Answer: Decreasing on $(0, 2)$. Find where $f'(x) = -3x^2 + 6x = -3x(x-2) < 0$.

Flashcard 46: Compute $f''(x)$ for $f(x) = x^5 - 5x^3 + 10x$.

Answer: $f''(x) = 20x^3 - 30x$. Differentiate twice: $f'(x) = 5x^4 - 15x^2$, then $f''(x) = 20x^3 - 30x$.

Flashcard 47: What is the relationship between $f(x)$ and $f'(x)$?

Answer: $f'(x)$ gives the slope of $f(x)$. The derivative measures how steeply the function rises or falls.

Flashcard 48: Calculate $f'(x)$ for $f(x) = 2x^3 - 5x^2 + 4x - 1$.

Answer: $f'(x) = 6x^2 - 10x + 4$. Apply power rule to each term: $6x^2 - 10x + 4$.

Flashcard 49: Find the critical points for $f(x) = x^2 - 4x + 3$.

Answer: Critical points: $x = 2$. Set $f'(x) = 2x - 4 = 0$ to find $x = 2$.

Flashcard 50: State the definition of an inflection point.

Answer: Where $f''(x)$ changes sign. Inflection points occur where concavity changes from up to down or vice versa.

Flashcard 51: What does a positive $f''(x)$ indicate about $f(x)$?

Answer: The graph of $f(x)$ is concave up. Positive second derivative means the graph curves upward like a bowl.

Flashcard 52: State the definition of an inflection point.

Answer: Where $f''(x)$ changes sign. Inflection points occur where concavity changes from up to down or vice versa.

Flashcard 53: Identify the local minimum of $f(x) = x^2 - 4x + 4$.

Answer: Local minimum at $x = 2$. Complete the square: $f(x) = (x-2)^2$, minimum at vertex.

Flashcard 54: Find $f''(x)$ for $f(x) = x^4 - 4x^2 + 6$.

Answer: $f''(x) = 12x^2 - 8$. Differentiate twice using power rule: $(x^4)'' = 12x^2$, $(-4x^2)'' = -8$.

Flashcard 55: Identify the intervals of increase for $f(x) = x^2 - 4x + 3$.

Answer: Increasing on $(2, \text{infinity})$. Find where $f'(x) = 2x - 4 > 0$, so $x > 2$.

Flashcard 56: What does a negative $f''(x)$ indicate?

Answer: The graph is concave down. Negative second derivative means the graph curves downward.

Flashcard 57: What is the first derivative of $f(x)$ used to determine?

Answer: The slope of the tangent line to $f(x)$. $f'(x)$ represents the instantaneous rate of change at any point.

Flashcard 58: What can be concluded if $f'(x) > 0$?

Answer: $f(x)$ is increasing on that interval. Positive first derivative means function values are getting larger.

Flashcard 59: How do you determine a local maximum using derivatives?

Answer: If $f'(x) = 0$ and $f''(x) < 0$. Second derivative test: negative second derivative confirms a maximum.

Flashcard 60: Find inflection points for $f(x) = x^3 - 3x + 1$.

Answer: Inflection point at $x = 0$. Set $f''(x) = 6x = 0$ to find where concavity changes.

Flashcard 61: Calculate $f''(x)$ for $f(x) = x^3 - 3x^2 + 1$.

Answer: $f''(x) = 6x - 6$. Differentiate twice: $f'(x) = 3x^2 - 6x$, then $f''(x) = 6x - 6$.