AP Calculus BC Flashcards: Candidates Test

What this deck covers

This deck focuses on Candidates Test, 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: Which theorem supports the use of the Candidates Test?

Answer: Extreme Value Theorem. Guarantees continuous functions have absolute extrema on closed intervals.

Flashcard 2: Evaluate $f(x) = -x^3 + 3x^2$ at endpoints $x = 0$ and $x = 2$.

Answer: $f(0) = 0$, $f(2) = 0$. Shows function values at the boundary points of the interval.

Flashcard 3: Evaluate $f(x) = x^2$ at $x = -1, 0, 1$ to find extrema.

Answer: Minimum at $x = 0$, Maximum at $x = -1$ or $x = 1$. $f(-1) = 1$, $f(0) = 0$, $f(1) = 1$, so min at $x = 0$, max at endpoints.

Flashcard 4: What is the absolute maximum of $f(x) = -x^2 + 4$ on $[-2, 2]$?

Answer: 4 at $x = 0$. Critical point at $x = 0$ gives maximum value since $f'(x) = -2x$.

Flashcard 5: What is the final step in the Candidates Test?

Answer: Compare function values to identify absolute extrema. The highest and lowest function values give the absolute max and min.

Flashcard 6: What is the final step in the Candidates Test?

Answer: Compare function values to identify absolute extrema. The highest and lowest function values give the absolute max and min.

Flashcard 7: Identify the absolute extrema for $f(x) = \frac{1}{x^2}$ on $[1, 3]$.

Answer: Minimum at $x = 3$, Maximum at $x = 1$. $f(x) = \frac{1}{x^2}$ decreases on $[1,3]$, so max at $x=1$, min at $x=3$.

Flashcard 8: What is the absolute maximum of $f(x) = x^4 - 4x^2 + 4$ on $[-2, 2]$?

Answer: Maximum at $x = 0$. Critical points at $x = \pm\sqrt{2}$ give local minima, max at $x = 0$.

Flashcard 9: Evaluate $f(x) = x^2 - 4x + 3$ at endpoints $x = 0$ and $x = 3$.

Answer: $f(0) = 3$, $f(3) = -3$. Direct substitution into the function at the boundary points.

Flashcard 10: Evaluate $f(x) = x^2 + x - 6$ at $x = -3, 0, 2$.

Answer: $f(-3) = 0$, $f(0) = -6$, $f(2) = 0$. Direct evaluation by substituting each x-value into the function.

Flashcard 11: Determine the absolute minimum of $f(x) = x^2$ on $[-1, 1]$.

Answer: Minimum at $x = 0$. Critical point $x = 0$ gives $f(0) = 0$, the minimum value.

Flashcard 12: What is a critical point?

Answer: Point where $f'(x) = 0$ or $f'(x)$ is undefined. These are potential locations for local extrema of the function.

Flashcard 13: What condition must a function meet to apply the Candidates Test?

Answer: Function must be continuous on a closed interval. Ensures the function has guaranteed absolute maximum and minimum values.

Flashcard 14: What role do endpoints play in the Candidates Test?

Answer: They are evaluated for potential extrema. Include boundary points as candidates since extrema can occur there.

Flashcard 15: Evaluate $f(x) = -x^3 + 3x^2$ at endpoints $x = 0$ and $x = 2$.

Answer: $f(0) = 0$, $f(2) = 0$. Shows function values at the boundary points of the interval.

Flashcard 16: Find the absolute minimum of $f(x) = x^3 - 3x$ on $[-2, 2]$.

Answer: Minimum at $x = -2$. Critical point $x = 1$ gives local max, compare with endpoint values.

Flashcard 17: Evaluate $f(x) = 3x^2 - 12x$ at $x = 0, 2, 4$.

Answer: $f(0) = 0$, $f(2) = -12$, $f(4) = 0$. Direct substitution shows function values at these candidate points.

Flashcard 18: State the condition for using limits in the Candidates Test.

Answer: Limits are used if endpoints are undefined. When function is undefined at endpoints, use limit values instead.

Flashcard 19: What is the absolute maximum of $f(x) = x^3$ on $[-1, 2]$?

Answer: $f(2) = 8$. Evaluating at endpoints and critical point shows maximum at $x = 2$.

Flashcard 20: Evaluate $f(x) = x^2 + x - 6$ at $x = -3, 0, 2$.

Answer: $f(-3) = 0$, $f(0) = -6$, $f(2) = 0$. Direct evaluation by substituting each x-value into the function.

Flashcard 21: Identify the second step in the Candidates Test.

Answer: Evaluate the function at critical points and endpoints. All candidate points must be tested to find absolute extrema.

Flashcard 22: How do you find critical points?

Answer: Solve $f'(x) = 0$ and check where $f'(x)$ is undefined. Both conditions identify all points where extrema can occur.

Flashcard 23: What is the Extreme Value Theorem?

Answer: A continuous function on a closed interval has absolute extrema. The mathematical foundation that justifies the Candidates Test method.

Flashcard 24: Identify the second step in the Candidates Test.

Answer: Evaluate the function at critical points and endpoints. All candidate points must be tested to find absolute extrema.

Flashcard 25: What is the absolute minimum of $f(x) = 1 - x^2$ on $[-1, 1]$?

Answer: -1 at $x = -1$ or $x = 1$. Parabola opens downward with maximum at $x = 0$, minimum at endpoints.

Flashcard 26: Evaluate $f(x) = x^2 - 4x + 3$ at endpoints $x = 0$ and $x = 3$.

Answer: $f(0) = 3$, $f(3) = -3$. Direct substitution into the function at the boundary points.

Flashcard 27: Identify the absolute extrema for $f(x) = \frac{1}{x^2}$ on $[1, 3]$.

Answer: Minimum at $x = 3$, Maximum at $x = 1$. $f(x) = \frac{1}{x^2}$ decreases on $[1,3]$, so max at $x=1$, min at $x=3$.

Flashcard 28: Why is continuity important for the Candidates Test?

Answer: Ensures extrema exist on closed intervals. Without continuity, absolute extrema might not exist.

Flashcard 29: State the first step in the Candidates Test.

Answer: Find the derivative and solve for critical points. Critical points occur where the derivative equals zero or is undefined.

Flashcard 30: Determine the absolute minimum of $f(x) = x^2$ on $[-1, 1]$.

Answer: Minimum at $x = 0$. Critical point $x = 0$ gives $f(0) = 0$, the minimum value.

Flashcard 31: Why is continuity important for the Candidates Test?

Answer: Ensures extrema exist on closed intervals. Without continuity, absolute extrema might not exist.

Flashcard 32: What is the absolute maximum of $f(x) = x^3$ on $[-1, 2]$?

Answer: $f(2) = 8$. Evaluating at endpoints and critical point shows maximum at $x = 2$.

Flashcard 33: Identify the absolute extrema for $f(x) = x^2 - 4$ on $[-3, 3]$.

Answer: Maximum at $x = 3$, Minimum at $x = 0$. Compare $f(-3) = 5$, $f(0) = -4$, $f(3) = 5$ for extrema locations.

Flashcard 34: What is the Candidates Test used for?

Answer: Determining absolute extrema on a closed interval. Finds global max/min on closed intervals using critical points and endpoints.

Flashcard 35: Find the critical points for $f(x) = x^3 - 3x^2 + 4$.

Answer: $x = 0, x = 2$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, so $x = 0$ and $x = 2$.

Flashcard 36: Evaluate $f(x) = 3x^2 - 12x$ at $x = 0, 2, 4$.

Answer: $f(0) = 0$, $f(2) = -12$, $f(4) = 0$. Direct substitution shows function values at these candidate points.

Flashcard 37: How is the derivative used in the Candidates Test?

Answer: To find critical points. Sets $f'(x) = 0$ to locate potential extrema within the interval.

Flashcard 38: How is the derivative used in the Candidates Test?

Answer: To find critical points. Sets $f'(x) = 0$ to locate potential extrema within the interval.

Flashcard 39: Determine the absolute minimum of $f(x) = x^3 - 3x^2 + 2$ on $[0, 3]$.

Answer: Minimum at $x = 2$. Critical point $x = 2$ gives $f(2) = -2$, the lowest value on the interval.

Flashcard 40: Which theorem supports the use of the Candidates Test?

Answer: Extreme Value Theorem. Guarantees continuous functions have absolute extrema on closed intervals.

Flashcard 41: Identify the absolute extrema for $f(x) = x^2 - 4$ on $[-3, 3]$.

Answer: Maximum at $x = 3$, Minimum at $x = 0$. Compare $f(-3) = 5$, $f(0) = -4$, $f(3) = 5$ for extrema locations.

Flashcard 42: What is the absolute maximum of $f(x) = -x^2 + 4$ on $[-2, 2]$?

Answer: 4 at $x = 0$. Critical point at $x = 0$ gives maximum value since $f'(x) = -2x$.

Flashcard 43: What is the purpose of evaluating endpoints in the Candidates Test?

Answer: To ensure all potential extrema are considered. Extrema can occur at boundary points of the closed interval.

Flashcard 44: What is the absolute maximum of $f(x) = x^4 - 4x^2 + 4$ on $[-2, 2]$?

Answer: Maximum at $x = 0$. Critical points at $x = \pm\sqrt{2}$ give local minima, max at $x = 0$.

Flashcard 45: Find the absolute minimum of $f(x) = x^3 - 3x$ on $[-2, 2]$.

Answer: Minimum at $x = -2$. Critical point $x = 1$ gives local max, compare with endpoint values.

Flashcard 46: What function values are compared in the Candidates Test?

Answer: Values at critical points and endpoints. These are all possible locations where absolute extrema can occur.

Flashcard 47: Find the critical points for $f(x) = x^3 - 3x^2 + 4$.

Answer: $x = 0, x = 2$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, so $x = 0$ and $x = 2$.

Flashcard 48: Evaluate $f(x) = x^2 + 1$ at $x = -2, 0, 2$.

Answer: $f(-2) = 5$, $f(0) = 1$, $f(2) = 5$. Shows function values at critical point and endpoints for comparison.

Flashcard 49: Find the absolute extrema of $f(x) = |x|$ on $[-2, 2]$.

Answer: Minimum at $x = 0$, Maximum at $x = -2$ or $x = 2$. Critical point at $x = 0$ gives minimum, endpoints give maximum values.

Flashcard 50: State the first step in the Candidates Test.

Answer: Find the derivative and solve for critical points. Critical points occur where the derivative equals zero or is undefined.

Flashcard 51: Determine the absolute minimum of $f(x) = x^3 - 3x^2 + 2$ on $[0, 3]$.

Answer: Minimum at $x = 2$. Critical point $x = 2$ gives $f(2) = -2$, the lowest value on the interval.

Flashcard 52: What role do endpoints play in the Candidates Test?

Answer: They are evaluated for potential extrema. Include boundary points as candidates since extrema can occur there.

Flashcard 53: What is the absolute minimum of $f(x) = 1 - x^2$ on $[-1, 1]$?

Answer: $-1$ at $x = -1$ or $x = 1$. Parabola opens downward with maximum at $x = 0$, minimum at endpoints.

Flashcard 54: What is the Candidates Test used for?

Answer: Determining absolute extrema on a closed interval. Finds global max/min on closed intervals using critical points and endpoints.

Flashcard 55: Find the absolute extrema for $f(x) = \frac{1}{x}$ on $[1, 4]$.

Answer: Minimum at $x = 4$, Maximum at $x = 1$. $f(x) = \frac{1}{x}$ decreases on $[1,4]$, so max at $x=1$, min at $x=4$.

Flashcard 56: How do you find critical points?

Answer: Solve $f'(x) = 0$ and check where $f'(x)$ is undefined. Both conditions identify all points where extrema can occur.

Flashcard 57: State the condition for using limits in the Candidates Test.

Answer: Limits are used if endpoints are undefined. When function is undefined at endpoints, use limit values instead.

Flashcard 58: Evaluate $f(x) = x^2 + 1$ at $x = -2, 0, 2$.

Answer: $f(-2) = 5$, $f(0) = 1$, $f(2) = 5$. Shows function values at critical point and endpoints for comparison.

Flashcard 59: What function values are compared in the Candidates Test?

Answer: Values at critical points and endpoints. These are all possible locations where absolute extrema can occur.

Flashcard 60: Find the absolute extrema for $f(x) = \frac{1}{x}$ on $[1, 4]$.

Answer: Minimum at $x = 4$, Maximum at $x = 1$. $f(x) = \frac{1}{x}$ decreases on $[1,4]$, so max at $x=1$, min at $x=4$.

Flashcard 61: Evaluate $f(x) = x^2$ at $x = -1, 0, 1$ to find extrema.

Answer: Minimum at $x = 0$, Maximum at $x = -1$ or $x = 1$. $f(-1) = 1$, $f(0) = 0$, $f(1) = 1$, so min at $x = 0$, max at endpoints.

Flashcard 62: What is the purpose of evaluating endpoints in the Candidates Test?

Answer: To ensure all potential extrema are considered. Extrema can occur at boundary points of the closed interval.

Flashcard 63: Find the absolute extrema of $f(x) = |x|$ on $[-2, 2]$.

Answer: Minimum at $x = 0$, Maximum at $x = -2$ or $x = 2$. Critical point at $x = 0$ gives minimum, endpoints give maximum values.

Flashcard 64: What condition must a function meet to apply the Candidates Test?

Answer: Function must be continuous on a closed interval. Ensures the function has guaranteed absolute maximum and minimum values.