AP Calculus BC 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 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 is the First Derivative Test?

Answer: Determines local extrema using $f'(x)$ sign change. Analyzes where $f'$ changes from positive to negative or vice versa.

Flashcard 2: Determine if $x=1$ is a critical point for $f(x) = x^3 - 3x$.

Answer: Yes, $x=1$ is a critical point. $f'(1) = 3(1)^2 - 3 = 0$, so derivative equals zero.

Flashcard 3: Find the derivative of $f(x) = 3x^2 - 6x + 1$.

Answer: $f'(x) = 6x - 6$. Power rule applied to each term of the polynomial.

Flashcard 4: Find critical points of $f(x) = \frac{1}{3}x^3 - x^2$.

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

Flashcard 5: Identify the global extrema of $f(x) = x^2 - 4x$ on $[0, 3]$.

Answer: Global min at $x=2$, Global max at $x=3$. Vertex at $x=2$ gives min; endpoint at $x=3$ gives max.

Flashcard 6: Find critical points of $f(x) = x^2 - 4x$.

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

Flashcard 7: Identify extrema of $f(x) = x^2 - 2x + 1$ on $[0, 2]$.

Answer: Global min at $x=1$, Global max at $x=2$. Vertex of parabola at $x=1$ gives min; endpoint gives max.

Flashcard 8: Identify the extrema of $f(x) = -x^2 + 4x - 3$ on $[0, 3]$.

Answer: Global max at $x=2$, Global min at $x=3$. Vertex at $x=2$ gives max; endpoint at $x=3$ gives min.

Flashcard 9: Find global extrema of $f(x) = 2x^2$ on $[-1, 2]$.

Answer: Global min at $x=0$, Global max at $x=2$. Evaluate $f$ at critical point $x=0$ and endpoints $x=-1,2$.

Flashcard 10: What is the derivative test for local extrema?

Answer: Use $f'(x)$ sign changes to identify local extrema. Sign changes in $f'(x)$ indicate transitions between increasing/decreasing.

Flashcard 11: Find global extrema for $f(x) = -x^2$ on $[-1, 1]$.

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

Flashcard 12: Find critical points of $f(x)=x^4-4x^2$.

Answer: Critical points: $x = 0, x = 2, x = -2$. Set $f'(x) = 4x^3 - 8x = 4x(x^2-2) = 0$ to solve.

Flashcard 13: Identify global extrema of $f(x) = x^3 - 6x^2 + 9x + 1$ on $[0, 3]$.

Answer: Global min at $x=0$, Global max at $x=3$. Evaluate at critical points and endpoints to compare values.

Flashcard 14: What conditions are necessary for EVT to apply?

Answer: Function must be continuous on closed interval. Closed interval ensures compactness for guaranteed extrema existence.

Flashcard 15: What is a local maximum?

Answer: $f(c) \geq f(x)$ for all $x$ near $c$. Function value at $c$ exceeds nearby values within some interval.

Flashcard 16: Find critical points of $f(x) = x^2 - 1$.

Answer: Critical point: $x = 0$. $f'(x) = 2x = 0$ only at $x = 0$.

Flashcard 17: What is the definition of a continuous function?

Answer: No gaps, jumps, or holes in domain. Function exists at every point with no discontinuities or breaks.

Flashcard 18: What is the Second Derivative Test?

Answer: Uses $f''(x)$ to determine concavity and local extrema. If $f''(c) > 0$, local min; if $f''(c) < 0$, local max.

Flashcard 19: Identify the critical point for $f(x) = x^2$.

Answer: Critical point: $x = 0$. $f'(x) = 2x = 0$ only when $x = 0$.

Flashcard 20: Is $f(x) = \frac{1}{x}$ on $[1, 3]$ continuous?

Answer: Yes, $f(x)$ is continuous on $[1, 3]$. Rational functions are continuous where denominator is nonzero.

Flashcard 21: Determine if $x=0$ is a critical point for $f(x)=x^4$.

Answer: Yes, $x=0$ is a critical point. $f'(0) = 4(0)^3 = 0$, so derivative equals zero at origin.

Flashcard 22: Is $f(x) = x^3$ on $[-1, 1]$ continuous?

Answer: Yes, $f(x)$ is continuous on $[-1, 1]$. Polynomial functions are continuous everywhere on their domain.

Flashcard 23: Describe how to find global extrema on $[a, b]$.

Answer: Evaluate $f$ at critical points and endpoints. Compare function values at all candidate points to find extrema.

Flashcard 24: What is a critical point?

Answer: Where $f'(x) = 0$ or $f'(x)$ is undefined. Potential locations for local extrema based on derivative behavior.

Flashcard 25: Identify the critical points of $f(x) = x^3 - 3x^2 + 2$.

Answer: Critical points: $x = 0, x = 2$. Set $f'(x) = 3x^2 - 6x = 3x(x-2) = 0$ to find critical points.

Flashcard 26: What is a local minimum?

Answer: $f(c) \leq f(x)$ for all $x$ near $c$. Function value at $c$ is below nearby values within some interval.

Flashcard 27: State Fermat's Theorem.

Answer: If $f$ has local extremum at $c$, $f'(c)=0$ or undefined. Local extrema only occur where the derivative is zero or undefined.

Flashcard 28: Calculate $f'(x)$ for $f(x) = x^3 - 6x^2 + 9x$.

Answer: $f'(x) = 3x^2 - 12x + 9$. Power rule applied: $\frac{d}{dx}[x^3 - 6x^2 + 9x] = 3x^2 - 12x + 9$.

Flashcard 29: Define a global maximum.

Answer: $f(c) \geq f(x)$ for all $x$ in domain of $f$. Function value at $c$ exceeds all others in the entire domain.

Flashcard 30: What is the Extreme Value Theorem?

Answer: If $f$ is continuous on $[a, b]$, it has a max and min. Guarantees absolute extrema exist on closed, bounded intervals.