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: What is a sufficient condition for a local maximum?

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

Flashcard 2: State the First Derivative Test.

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

Flashcard 3: 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 4: 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 5: 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 6: 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 7: 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 8: 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 9: 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 10: 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 11: What is a global maximum?

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

Flashcard 12: 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 13: What is the significance of critical points?

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

Flashcard 14: How does $f'(x)$ determine extrema?

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

Flashcard 15: How can you confirm a global extremum?

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

Flashcard 16: 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 17: State the Second Derivative Test.

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

Flashcard 18: 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 19: What is a global minimum?

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

Flashcard 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 26: 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 27: What function property is tested with $f''(x) = 0$?

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

Flashcard 28: 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 29: 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 30: What is a sufficient condition for a local minimum?

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