Study Candidates Test in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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AP Calculus BC Flashcards: Candidates Test
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QUESTION
What condition must a function meet to apply the Candidates Test?
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ANSWER
Function must be continuous on a closed interval. Ensures the function has guaranteed absolute maximum and minimum values.
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Flashcard 1: 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 2: 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 3: Evaluate f(x)=x2 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: Evaluate f(x)=3x2−12x 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 5: 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 6: 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 7: Determine the absolute minimum of f(x)=x3−3x2+2 on [0,3].
Answer: Minimum at x=2. Critical point x=2 gives f(2)=−2, the lowest value on the interval.
Flashcard 8: Identify the absolute extrema for f(x)=x2−4 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 9: Evaluate f(x)=x2−4x+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: 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 11: 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 12: 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 13: 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 14: Identify the absolute extrema for f(x)=x21 on [1,3].
Answer: Minimum at x=3, Maximum at x=1. f(x)=x21 decreases on [1,3], so max at x=1, min at x=3.
Flashcard 15: Why is continuity important for the Candidates Test?
Answer: Ensures extrema exist on closed intervals. Without continuity, absolute extrema might not exist.
Flashcard 16: 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 17: Determine the absolute minimum of f(x)=x2 on [−1,1].
Answer: Minimum at x=0. Critical point x=0 gives f(0)=0, the minimum value.
Flashcard 18: 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 19: Evaluate f(x)=x2+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 20: Find the absolute minimum of f(x)=x3−3x on [−2,2].
Answer: Minimum at x=−2. Critical point x=1 gives local max, compare with endpoint values.
Flashcard 21: Evaluate f(x)=−x3+3x2 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 22: What is the absolute minimum of f(x)=1−x2 on [−1,1]?
Answer: -1 at x=−1 or x=1. Parabola opens downward with maximum at x=0, minimum at endpoints.
Flashcard 23: Find the absolute extrema for f(x)=x1 on [1,4].
Answer: Minimum at x=4, Maximum at x=1. f(x)=x1 decreases on [1,4], so max at x=1, min at x=4.
Flashcard 24: 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 25: What is the absolute maximum of f(x)=x4−4x2+4 on [−2,2]?
Answer: Maximum at x=0. Critical points at x=±2 give local minima, max at x=0.
Flashcard 26: What is the absolute maximum of f(x)=x3 on [−1,2]?
Answer: f(2)=8. Evaluating at endpoints and critical point shows maximum at x=2.
Flashcard 27: 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 28: Find the critical points for f(x)=x3−3x2+4.
Answer: x=0,x=2. f′(x)=3x2−6x=3x(x−2), so x=0 and x=2.
Flashcard 29: Which theorem supports the use of the Candidates Test?
Answer: Extreme Value Theorem. Guarantees continuous functions have absolute extrema on closed intervals.
Flashcard 30: Evaluate f(x)=x2+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.