AP Calculus AB Flashcards: Extreme Value Theorem Extrema Critical Points

What this deck covers

This deck focuses on Extreme Value Theorem Extrema Critical Points, 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: State the Second Derivative Test.

Answer: Use $f''(x)$ to determine concavity and local extrema. Evaluate second derivative at critical points.

Flashcard 2: Explain how to find global extrema on a closed interval.

Answer: Evaluate $f(x)$ at critical points and endpoints. Check critical points and interval endpoints systematically.

Flashcard 3: What does $f''(x) > 0$ at a point indicate?

Answer: The function is concave up; possible local min. Positive second derivative indicates upward concavity.

Flashcard 4: Determine the global extrema of $f(x) = -x^2 + 4x$ on $[0, 4]$.

Answer: Global max at $x = 2$; global min at $x = 0, 4$. Vertex of parabola at $x = 2$ gives maximum.

Flashcard 5: Identify the critical points of $f(x) = \frac{x^3}{3} - x$.

Answer: Critical points at $x = -1, 1$. $f'(x) = x^2 - 1 = 0$ when $x = \pm 1$.

Flashcard 6: What is a global maximum?

Answer: The highest value of a function on its entire domain. Also called the absolute maximum value.

Flashcard 7: Define a critical point of a function.

Answer: A point where $f'(x) = 0$ or $f'(x)$ is undefined. These are the only candidates for local extrema.

Flashcard 8: Identify the critical points of $f(x) = \frac{x^3}{3} - x$.

Answer: Critical points at $x = -1, 1$. $f'(x) = x^2 - 1 = 0$ when $x = \pm 1$.

Flashcard 9: What does $f''(x) > 0$ at a point indicate?

Answer: The function is concave up; possible local min. Positive second derivative indicates upward concavity.

Flashcard 10: What is a local maximum?

Answer: A point higher than all nearby points. Also called a relative maximum.

Flashcard 11: What is verified by the Second Derivative Test?

Answer: Concavity and nature of local extrema. Second derivative test classifies critical point behavior.

Flashcard 12: What is verified by the Second Derivative Test?

Answer: Concavity and nature of local extrema. Second derivative test classifies critical point behavior.

Flashcard 13: What is the impact of a discontinuity on EVT?

Answer: Discontinuity can prevent existence of global extrema. EVT doesn't apply if function isn't continuous.

Flashcard 14: Differentiate between global and local extrema.

Answer: Global is on entire domain; local is on a neighborhood. Global considers entire domain; local considers neighborhoods.

Flashcard 15: What role do endpoints play in finding global extrema?

Answer: Endpoints are evaluated for possible global extrema. Endpoints must be checked for global extrema.

Flashcard 16: What condition must be met for the EVT to apply?

Answer: Function must be continuous on a closed interval. Requires both continuity and a closed, bounded interval.

Flashcard 17: Determine local extrema for $f(x) = x^3 - 3x^2 + 4$.

Answer: Local max at $x = 0$; local min at $x = 2$. $f'(x) = 3x^2 - 6x$ gives critical points at $x = 0, 2$.

Flashcard 18: Determine the global extrema of $f(x) = x^2 - 4$ on $[-2, 2]$.

Answer: Global max at $x = -2, 2$; global min at $x = 0$. Evaluate at critical point $x = 0$ and endpoints.

Flashcard 19: What function property is tested with $f''(x) = 0$?

Answer: Possible inflection point; concavity change. Second derivative zero suggests possible inflection point.

Flashcard 20: What is a global minimum?

Answer: The lowest value of a function on its entire domain. Also called the absolute minimum value.

Flashcard 21: Explain how to find global extrema on a closed interval.

Answer: Evaluate $f(x)$ at critical points and endpoints. Check critical points and interval endpoints systematically.

Flashcard 22: What is the purpose of the Extreme Value Theorem?

Answer: To ensure global extrema exist on a closed interval. Guarantees extrema exist for optimization problems.

Flashcard 23: How can you confirm a global extremum?

Answer: Compare function values at critical points and endpoints. Test all candidates to find absolute extrema.

Flashcard 24: What is a local minimum?

Answer: A point lower than all nearby points. Also called a relative minimum.

Flashcard 25: Why are continuous functions important in EVT?

Answer: Continuity ensures the existence of extrema on $[a, b]$. Discontinuous functions may not have global extrema.

Flashcard 26: Identify the local extrema of $f(x) = x^4 - 4x^3$.

Answer: Local max at $x = 0$; local min at $x = 3$. $f'(x) = 4x^3 - 12x^2$ gives critical points at $x = 0, 3$.

Flashcard 27: How can you confirm a global extremum?

Answer: Compare function values at critical points and endpoints. Test all candidates to find absolute extrema.

Flashcard 28: State the Second Derivative Test.

Answer: Use $f''(x)$ to determine concavity and local extrema. Evaluate second derivative at critical points.

Flashcard 29: Determine the global extrema of $f(x) = -x^2$ on $[-1, 1]$.

Answer: Global max at $x = 0$; global min at $x = -1, 1$. Parabola opens downward with vertex at origin.

Flashcard 30: Determine the global extrema of $f(x) = -x^2 + 4x$ on $[0, 4]$.

Answer: Global max at $x = 2$; global min at $x = 0, 4$. Vertex of parabola at $x = 2$ gives maximum.

Flashcard 31: What is a sufficient condition for a local minimum?

Answer: $f'(x)$ changes from negative to positive. Derivative changes from negative to positive.

Flashcard 32: Identify the global extrema of $f(x) = x^3 - 3x$ on $[-2, 2]$.

Answer: Global max at $x = 2$; global min at $x = -2$. $f(-2) = -2$, $f(1) = -2$, $f(2) = 2$.

Flashcard 33: Determine the global extrema of $f(x) = x^2 - 4$ on $[-2, 2]$.

Answer: Global max at $x = -2, 2$; global min at $x = 0$. Evaluate at critical point $x = 0$ and endpoints.

Flashcard 34: What is the purpose of the Extreme Value Theorem?

Answer: To ensure global extrema exist on a closed interval. Guarantees extrema exist for optimization problems.

Flashcard 35: What does $f''(x) < 0$ at a point indicate?

Answer: The function is concave down; possible local max. Negative second derivative indicates downward concavity.

Flashcard 36: What is a global maximum?

Answer: The highest value of a function on its entire domain. Also called the absolute maximum value.

Flashcard 37: What function property is tested with $f''(x) = 0$?

Answer: Possible inflection point; concavity change. Second derivative zero suggests possible inflection point.

Flashcard 38: What is the Extreme Value Theorem?

Answer: If a function is continuous on [a, b], it attains a global max and min. Guarantees existence of absolute extrema on closed intervals.

Flashcard 39: Find critical points for $f(x) = \frac{1}{3}x^3 - x^2 + 2$.

Answer: Critical points at $x = 0$ and $x = 2$. $f'(x) = x^2 - 2x = x(x-2) = 0$ when $x = 0, 2$.

Flashcard 40: Define a critical point of a function.

Answer: A point where $f'(x) = 0$ or $f'(x)$ is undefined. These are the only candidates for local extrema.

Flashcard 41: How do you find critical points?

Answer: Solve $f'(x) = 0$ or find where $f'(x)$ is undefined. Find where the derivative equals zero or doesn't exist.

Flashcard 42: Find critical points for $f(x) = \frac{1}{3}x^3 - x^2 + 2$.

Answer: Critical points at $x = 0$ and $x = 2$. $f'(x) = x^2 - 2x = x(x-2) = 0$ when $x = 0, 2$.

Flashcard 43: State the First Derivative Test.

Answer: Use $f'(x)$ sign changes to determine local extrema. Check if derivative changes sign around critical points.

Flashcard 44: How does $f'(x)$ determine extrema?

Answer: Sign changes in $f'(x)$ indicate local extrema. Sign changes in first derivative locate extrema.

Flashcard 45: Why are continuous functions important in EVT?

Answer: Continuity ensures the existence of extrema on $[a, b]$. Discontinuous functions may not have global extrema.

Flashcard 46: What is a necessary condition for local extrema?

Answer: $f'(x) = 0$ or $f'(x)$ is undefined. Critical points are necessary but not sufficient conditions.

Flashcard 47: What is a global minimum?

Answer: The lowest value of a function on its entire domain. Also called the absolute minimum value.

Flashcard 48: Identify the critical points of $f(x) = \frac{1}{x}$.

Answer: No critical points; $f'(x)$ never zero. $f'(x) = -\frac{1}{x^2}$ is never zero, always negative.

Flashcard 49: How do you find critical points?

Answer: Solve $f'(x) = 0$ or find where $f'(x)$ is undefined. Find where the derivative equals zero or doesn't exist.

Flashcard 50: Differentiate between global and local extrema.

Answer: Global is on entire domain; local is on a neighborhood. Global considers entire domain; local considers neighborhoods.

Flashcard 51: What is a local maximum?

Answer: A point higher than all nearby points. Also called a relative maximum.

Flashcard 52: What is the significance of critical points?

Answer: Potential locations of local extrema. Only at critical points can local extrema occur.

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

Answer: Critical point at $x = 2$. $f'(x) = 2x - 4 = 0$ when $x = 2$.

Flashcard 54: What is a local minimum?

Answer: A point lower than all nearby points. Also called a relative minimum.

Flashcard 55: What role do endpoints play in finding global extrema?

Answer: Endpoints are evaluated for possible global extrema. Endpoints must be checked for global extrema.

Flashcard 56: What does $f''(x) < 0$ at a point indicate?

Answer: The function is concave down; possible local max. Negative second derivative indicates downward concavity.

Flashcard 57: State the First Derivative Test.

Answer: Use $f'(x)$ sign changes to determine local extrema. Check if derivative changes sign around critical points.

Flashcard 58: What condition must be met for the EVT to apply?

Answer: Function must be continuous on a closed interval. Requires both continuity and a closed, bounded interval.

Flashcard 59: What is the Extreme Value Theorem?

Answer: If a function is continuous on [a, b], it attains a global max and min. Guarantees existence of absolute extrema on closed intervals.

Flashcard 60: Identify the critical points of $f(x) = x^2 + 3x + 2$.

Answer: Critical point at $x = -\frac{3}{2}$. $f'(x) = 2x + 3 = 0$ when $x = -\frac{3}{2}$.

Flashcard 61: How does $f'(x)$ determine extrema?

Answer: Sign changes in $f'(x)$ indicate local extrema. Sign changes in first derivative locate extrema.

Flashcard 62: Identify the local extrema of $f(x) = \frac{x^2}{2} - 2x$.

Answer: Local min at $x = 2$. $f'(x) = x - 2 = 0$ when $x = 2$.

Flashcard 63: Identify the critical points of $f(x) = x^2 - 4x + 4$.

Answer: Critical point at $x = 2$. $f'(x) = 2x - 4 = 0$ when $x = 2$.

Flashcard 64: Identify the local extrema of $f(x) = x^4 - 4x^3$.

Answer: Local max at $x = 0$; local min at $x = 3$. $f'(x) = 4x^3 - 12x^2$ gives critical points at $x = 0, 3$.

Flashcard 65: Identify the global extrema of $f(x) = x^3 - 3x$ on $[-2, 2]$.

Answer: Global max at $x = 2$; global min at $x = -2$. $f(-2) = -2$, $f(1) = -2$, $f(2) = 2$.

Flashcard 66: Identify the critical points of $f(x) = \frac{1}{x}$.

Answer: No critical points; $f'(x)$ never zero. $f'(x) = -\frac{1}{x^2}$ is never zero, always negative.

Flashcard 67: What is a sufficient condition for a local minimum?

Answer: $f'(x)$ changes from negative to positive. Derivative changes from negative to positive.

Flashcard 68: Identify the critical points of $f(x) = x^2 + 3x + 2$.

Answer: Critical point at $x = -\frac{3}{2}$. $f'(x) = 2x + 3 = 0$ when $x = -\frac{3}{2}$.

Flashcard 69: Identify the local extrema of $f(x) = \frac{x^2}{2} - 2x$.

Answer: Local min at $x = 2$. $f'(x) = x - 2 = 0$ when $x = 2$.

Flashcard 70: What is a sufficient condition for a local maximum?

Answer: $f'(x)$ changes from positive to negative. Derivative changes from positive to negative.

Flashcard 71: What is a necessary condition for local extrema?

Answer: $f'(x) = 0$ or $f'(x)$ is undefined. Critical points are necessary but not sufficient conditions.

Flashcard 72: What is the significance of critical points?

Answer: Potential locations of local extrema. Only at critical points can local extrema occur.

Flashcard 73: Determine local extrema for $f(x) = x^3 - 3x^2 + 4$.

Answer: Local max at $x = 0$; local min at $x = 2$. $f'(x) = 3x^2 - 6x$ gives critical points at $x = 0, 2$.

Flashcard 74: What is the impact of a discontinuity on EVT?

Answer: Discontinuity can prevent existence of global extrema. EVT doesn't apply if function isn't continuous.

Flashcard 75: What is a sufficient condition for a local maximum?

Answer: $f'(x)$ changes from positive to negative. Derivative changes from positive to negative.

Flashcard 76: Determine the global extrema of $f(x) = -x^2$ on $[-1, 1]$.

Answer: Global max at $x = 0$; global min at $x = -1, 1$. Parabola opens downward with vertex at origin.