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 definition of a continuous function?

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

Flashcard 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: Identify the critical point for $f(x) = x^2$.

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

Flashcard 9: 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 10: Define a global minimum.

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

Flashcard 11: Find $f'(x)$ for $f(x) = \frac{x^2}{x+2}$.

Answer: $f'(x) = \frac{2x(x+2)-x^2}{(x+2)^2}$. Quotient rule: $\frac{d}{dx}[\frac{x^2}{x+2}] = \frac{2x(x+2)-x^2}{(x+2)^2}$.

Flashcard 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: 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 20: Find $f'(x)$ for $f(x) = \frac{x}{x+1}$.

Answer: $f'(x) = \frac{1}{(x+1)^2}$. Quotient rule applied: $\frac{d}{dx}[\frac{x}{x+1}] = \frac{1}{(x+1)^2}$.

Flashcard 21: 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 22: 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 23: 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 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: 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 26: 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 27: 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 28: Find $f'(x)$ for $f(x) = \frac{x^2}{x+2}$.

Answer: $f'(x) = \frac{2x(x+2)-x^2}{(x+2)^2}$. Quotient rule: $\frac{d}{dx}[\frac{x^2}{x+2}] = \frac{2x(x+2)-x^2}{(x+2)^2}$.

Flashcard 29: 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 30: 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 31: Find $f'(x)$ for $f(x) = \frac{x}{x+1}$.

Answer: $f'(x) = \frac{1}{(x+1)^2}$. Quotient rule applied: $\frac{d}{dx}[\frac{x}{x+1}] = \frac{1}{(x+1)^2}$.

Flashcard 32: 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 33: 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 34: 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 35: 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 36: 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 37: 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 38: 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 39: 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 40: 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 41: 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.

Flashcard 42: 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 43: 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 44: 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 45: Find critical points of $f(x) = x^2 - 1$.

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

Flashcard 46: 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 47: 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.

Flashcard 48: 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 49: 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 50: 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 51: 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 52: 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 53: 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 54: Find critical points of $f(x) = x^2 - 1$.

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

Flashcard 55: 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 56: Identify the critical point for $f(x) = x^2$.

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

Flashcard 57: 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 58: Define a global minimum.

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

Flashcard 59: 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 60: 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 61: 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 62: 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 63: 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 64: 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 65: 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 66: 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.