A continuous function on has critical points and values ; where is the absolute minimum?
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AP Calculus BC Quiz
Practice Candidates Test in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A continuous function r on [−4,2] has critical points x=−2,1 and values r(−4)=3,r(−2)=0,r(1)=−1,r(2)=2; where is the absolute minimum?
This quiz focuses on Candidates Test, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
A continuous function r on [−4,2] has critical points x=−2,1 and values r(−4)=3,r(−2)=0,r(1)=−1,r(2)=2; where is the absolute minimum?
Explanation: This problem tests the Candidates Test for finding the absolute minimum on the interval [-4,2]. The Candidates Test tells us that absolute extrema of continuous functions on closed intervals must occur at critical points or endpoints. Our candidates are the critical points x = -2 and x = 1, plus the endpoints x = -4 and x = 2. Comparing function values: r(-4) = 3, r(-2) = 0, r(1) = -1, and r(2) = 2, we see that r(1) = -1 is the smallest value. Students might incorrectly choose x = -2 since it has value 0, but the critical point x = 1 yields the actual minimum. Remember the complete process: list all critical points within the interval, include both endpoints, evaluate at each candidate, and identify the smallest value for the absolute minimum.
For continuous q on [−3,3] with critical points x=−1,2 and values q(−3)=0,q(−1)=4,q(2)=5,q(3)=1, where is the absolute maximum?
Explanation: This question applies the Candidates Test to find the absolute maximum of q on [-3,3]. The Candidates Test guarantees that for a continuous function on a closed interval, the absolute maximum and minimum occur at critical points or endpoints. We must check the critical points x = -1 and x = 2, plus the endpoints x = -3 and x = 3. Evaluating at all candidates: q(-3) = 0, q(-1) = 4, q(2) = 5, and q(3) = 1, we find that q(2) = 5 is the largest value. Some students might select x = -1 because it's the first critical point with a large value, but we must compare all candidates. The foolproof method: identify critical points, add both endpoints, evaluate the function at each candidate, and select the one with the maximum value.
A continuous function g on [0,6] has critical points x=2,5 and values g(0)=3,g(2)=−1,g(5)=2,g(6)=0; where is the absolute minimum?
Explanation: This problem applies the Candidates Test to locate the absolute minimum of a continuous function on [0,6]. The Candidates Test requires checking all critical points (where the derivative is zero or undefined) and endpoints when finding absolute extrema on closed intervals. Our candidates are the critical points x = 2 and x = 5, plus the endpoints x = 0 and x = 6. Comparing the function values: g(0) = 3, g(2) = -1, g(5) = 2, and g(6) = 0, we find that g(2) = -1 is the smallest value. Students might incorrectly choose x = 0 thinking endpoints always give extrema, but critical points can produce more extreme values. The key checklist: identify all critical points, include both endpoints, evaluate the function at each candidate, and select the smallest (for minimum) or largest (for maximum) value.
For continuous u on [2,8] with critical points x=4,7 and values u(2)=1,u(4)=−2,u(7)=0,u(8)=−3, where is the absolute minimum?
Explanation: This question tests the Candidates Test for finding the absolute minimum of u on [2,8]. The Candidates Test tells us that for continuous functions on closed intervals, absolute extrema must occur at critical points or endpoints—these are the only candidates we need to check. We have critical points at x=4 and x=7, and endpoints at x=2 and x=8. Comparing values: u(2)=1, u(4)=−2, u(7)=0, and u(8)=−3, we find that u(8)=−3 is the smallest value. Some might choose x=4 since it's a critical point with a negative value, but the endpoint x=8 has an even smaller value. Always remember: check all critical points AND both endpoints, then select the location with the minimum function value.
For continuous h on [−3,3], critical points are x=−2,1 with h(−3)=4,h(−2)=2,h(1)=6,h(3)=5. Where is the absolute maximum?
Explanation: The Candidates Test is a method to find absolute extrema of a continuous function on a closed interval by evaluating the function at critical points and endpoints. We check critical points because they may represent local maxima that could be the global maximum on the interval. Endpoints need evaluation as the function might reach its highest value at the start or end of the domain. This comparison ensures we don't miss the absolute maximum by overlooking any candidate. One tempting distractor is x=-3 with h(-3)=4, but it fails as it's less than h(1)=6, demonstrating incomplete comparison. Always list and evaluate the function at: both endpoints and all critical points within the interval.
A continuous function t on [−2,6] has critical points x=0,5 and values t(−2)=−1,t(0)=3,t(5)=3,t(6)=2; where is the absolute maximum?
Explanation: This problem applies the Candidates Test to find where the absolute maximum occurs for t on [-2,6]. The Candidates Test guarantees that absolute extrema of continuous functions on closed intervals occur at critical points or endpoints. Our candidates include critical points x = 0 and x = 5, plus endpoints x = -2 and x = 6. Evaluating at all candidates: t(-2) = -1, t(0) = 3, t(5) = 3, and t(6) = 2, we see that both t(0) = 3 and t(5) = 3 equal the maximum value. Students might think only one location can be the maximum, but when multiple candidates achieve the same maximum value, all such locations are correct. The complete method: list critical points and endpoints, evaluate at each, and identify all locations achieving the maximum value.
For continuous f on [−2,4] with critical points x=0,3 and values f(−2)=1,f(0)=5,f(3)=2,f(4)=4, where is the absolute maximum?
Explanation: This problem requires applying the Candidates Test to find the absolute maximum of a continuous function on a closed interval. For any continuous function on a closed interval, the absolute extrema must occur either at critical points (where the derivative is zero or undefined) or at the endpoints of the interval. We must check all candidates: the critical points at x = 0 and x = 3, and the endpoints at x = -2 and x = 4. Evaluating at each candidate: f(-2) = 1, f(0) = 5, f(3) = 2, and f(4) = 4, we see that the largest value is 5, occurring at x = 0. Some students might incorrectly choose an endpoint thinking extrema must occur there, but critical points can also be locations of absolute extrema. The Candidates Test checklist: (1) verify continuity on the closed interval, (2) identify all critical points and endpoints, (3) evaluate the function at each candidate, and (4) compare values to find the absolute maximum and minimum.
For continuous s on [−4,0] with critical points x=−3,−1, given s(−4)=2,s(−3)=2,s(−1)=−5,s(0)=−1, where is the absolute minimum?
Explanation: This question applies the Candidates Test to find the absolute minimum on [-4,0]. The Candidates Test tells us that absolute extrema for continuous functions on closed intervals must occur at critical points or endpoints. We need to check endpoints x = -4 and x = 0, along with critical points x = -3 and x = -1. The function values are: s(-4) = 2, s(-3) = 2, s(-1) = -5, and s(0) = -1. Since s(-1) = -5 is the smallest value, the absolute minimum occurs at x = -1. Choice E incorrectly focuses on x = -3 being a critical point as if that status alone determines extrema, but we must compare actual function values at all candidates. The systematic candidates approach: list all critical points within the interval, add both endpoints, evaluate the function at each candidate, and identify the location with the minimum value.
For continuous u on [−2,6] with critical points x=2,5, given u(−2)=0,u(2)=4,u(5)=3,u(6)=5, where is the absolute maximum?
Explanation: This problem requires applying the Candidates Test to find the absolute maximum of a continuous function on a closed interval. The Candidates Test guarantees that absolute extrema occur at critical points or endpoints for continuous functions on closed intervals. We examine four candidates: endpoints x = -2 and x = 6, and critical points x = 2 and x = 5. Function values at these points are: u(-2) = 0, u(2) = 4, u(5) = 3, and u(6) = 5. The largest value is u(6) = 5, so the absolute maximum occurs at the endpoint x = 6. A common mistake would be assuming the maximum must occur at a critical point like x = 2 where u(2) = 4, but endpoints can also produce absolute extrema. The foolproof candidates checklist is: identify all critical points in the interval, include both endpoints as candidates, evaluate the function at all candidates, then select the location with the maximum value.
A continuous function t on [0,5] has critical points x=1,3 with t(0)=2,t(1)=2,t(3)=−1,t(5)=0; where is the absolute minimum?
Explanation: This problem demonstrates the Candidates Test for locating the absolute minimum of a continuous function on [0, 5]. The Candidates Test principle states that absolute extrema of continuous functions on closed intervals occur at critical points or endpoints—nowhere else. Our candidates are critical points x = 1 and x = 3, plus endpoints x = 0 and x = 5. Computing values: t(0) = 2, t(1) = 2, t(3) = -1, and t(5) = 0, we find the minimum value is -1 at x = 3. Choice E incorrectly focuses on the equal values at x = 0 and x = 1, but equal function values don't determine extrema—we need the smallest value overall. Apply the Candidates Test systematically: list all candidates, evaluate the function at each, then identify where the minimum occurs.
Let f be continuous on [−2,4] with critical points at x=−1,2; f(−2)=1,f(−1)=5,f(2)=3,f(4)=0. Where is the absolute maximum?
Explanation: The Candidates Test is a method to find absolute extrema of a continuous function on a closed interval by evaluating the function at critical points and endpoints. Critical points are locations where the derivative is zero or undefined, potentially indicating local maxima or minima that could be absolute extrema. Endpoints are essential to check because the function's behavior at the boundaries of the interval might yield the highest or lowest values, independent of interior critical points. By comparing the function values at all these points, we identify the absolute maximum as the largest value among them. A tempting distractor might be choosing x=2, where f(2)=3, but this fails because it's lower than f(-1)=5, showing not all candidates were compared properly. Always list and evaluate the function at: both endpoints and all critical points within the interval.
For continuous h on [−3,3] with critical points x=−2,1, given h(−3)=0,h(−2)=−4,h(1)=2,h(3)=1, where is the absolute minimum?
Explanation: This question tests the Candidates Test for finding absolute extrema on the closed interval [-3,3]. According to the Candidates Test, absolute extrema must occur at critical points or endpoints when the function is continuous on a closed interval. We need to check all candidates: endpoints x = -3 and x = 3, and critical points x = -2 and x = 1. The function values are: h(-3) = 0, h(-2) = -4, h(1) = 2, and h(3) = 1. Since h(-2) = -4 is the smallest value, the absolute minimum occurs at x = -2. Choice E incorrectly suggests the minimum is at x = -3 simply because it's the left endpoint, but endpoints are not automatically extrema—we must compare all candidate values. The candidates checklist for finding absolute extrema: locate all critical points in the interval, include both endpoints as candidates, evaluate the function at all candidates, and identify the location with the extreme value.
For continuous g on [0,6] with critical points x=1,5, given g(0)=2,g(1)=−1,g(5)=3,g(6)=0, where is the absolute minimum?
Explanation: This problem requires applying the Candidates Test to find the absolute minimum of a continuous function on [0,6]. The Candidates Test states that absolute extrema of a continuous function on a closed interval must occur at critical points or endpoints. Our candidates are the endpoints x = 0 and x = 6, plus the critical points x = 1 and x = 5. Evaluating the function at each candidate gives us: g(0) = 2, g(1) = -1, g(5) = 3, and g(6) = 0. The smallest value is g(1) = -1, making x = 1 the location of the absolute minimum. Choice E incorrectly claims that endpoints cannot be minima, which is false—endpoints are valid candidates for absolute extrema. Always check your complete candidates list: all critical points within the interval plus both endpoints, then select the location with the smallest (for minimum) or largest (for maximum) function value.
For continuous p on [1,7] with critical points x=3,6, given p(1)=−2,p(3)=0,p(6)=−5,p(7)=−1, where is the absolute minimum?
Explanation: This problem assesses the Candidates Test for finding absolute extrema on a closed interval. To find the absolute minimum, we must evaluate the continuous function at the endpoints and critical points because the Extreme Value Theorem guarantees that a minimum exists, and these are the potential locations where it can occur. The candidates are x = 1, 3, 6, 7 with p-values -2, 0, -5, -1, so the minimum is -5 at x = 6. Checking all these points ensures we don't miss the global extremum, which might not be at an endpoint. A tempting distractor is x = 1 with value -2, but it fails because -2 is greater than -5 at the critical point x = 6. Remember, the checklist for candidates is: list endpoints and critical points inside the interval, evaluate p at each, and compare the values to identify the extrema.
Continuous h on [−1,5] has critical points x=1,4 with h(−1)=0,h(1)=3,h(4)=6,h(5)=6; where is the absolute maximum?
Explanation: This problem assesses the Candidates Test for finding absolute extrema on a closed interval. To find the absolute maximum, we must evaluate the continuous function at the endpoints and critical points because the Extreme Value Theorem guarantees that a maximum exists, and these are the potential locations where it can occur. The candidates are x = -1, 1, 4, 5 with h-values 0, 3, 6, 6, so the maximum is 6 at x = 4 (and also at x = 5). Checking all these points ensures we don't miss the global extremum, which might not be at an endpoint. A tempting distractor is x = 1 with value 3, but it fails because 3 is less than 6 at the critical point x = 4. Remember, the checklist for candidates is: list endpoints and critical points inside the interval, evaluate h at each, and compare the values to identify the extrema.
For continuous f on [−2,4] with critical points x=0,3, given f(−2)=1,f(0)=5,f(3)=2,f(4)=4, where is the absolute maximum?
Explanation: This problem assesses the Candidates Test for finding absolute extrema on a closed interval. To find the absolute maximum, we must evaluate the continuous function at the endpoints and critical points because the Extreme Value Theorem guarantees that a maximum exists, and these are the potential locations where it can occur. The candidates are x = -2, 0, 3, 4 with f-values 1, 5, 2, 4, so the maximum is 5 at x = 0. Checking all these points ensures we don't miss the global extremum, which might not be at an endpoint. A tempting distractor is x = 4 with value 4, but it fails because 4 is less than 5 at the critical point x = 0. Remember, the checklist for candidates is: list endpoints and critical points inside the interval, evaluate f at each, and compare the values to identify the extrema.
A continuous function s on [2,8] has critical points x=3,7 and values s(2)=5,s(3)=2,s(7)=1,s(8)=4; where is the absolute minimum?
Explanation: This problem assesses the Candidates Test for finding absolute extrema on a closed interval. To find the absolute minimum, we must evaluate the continuous function at the endpoints and critical points because the Extreme Value Theorem guarantees that a minimum exists, and these are the potential locations where it can occur. The candidates are x = 2, 3, 7, 8 with s-values 5, 2, 1, 4, so the minimum is 1 at x = 7. Checking all these points ensures we don't miss the global extremum, which might not be at an endpoint. A tempting distractor is x = 3 with value 2, but it fails because 2 is greater than 1 at the critical point x = 7. Remember, the checklist for candidates is: list endpoints and critical points inside the interval, evaluate s at each, and compare the values to identify the extrema.
For continuous t on [−2,3] with critical points x=−1,2, given t(−2)=0,t(−1)=4,t(2)=−1,t(3)=3, where is the absolute maximum?
Explanation: This problem assesses the Candidates Test for finding absolute extrema on a closed interval. To find the absolute maximum, we must evaluate the continuous function at the endpoints and critical points because the Extreme Value Theorem guarantees that a maximum exists, and these are the potential locations where it can occur. The candidates are x = -2, -1, 2, 3 with t-values 0, 4, -1, 3, so the maximum is 4 at x = -1. Checking all these points ensures we don't miss the global extremum, which might not be at an endpoint. A tempting distractor is x = 3 with value 3, but it fails because 3 is less than 4 at the critical point x = -1. Remember, the checklist for candidates is: list endpoints and critical points inside the interval, evaluate t at each, and compare the values to identify the extrema.
A continuous function p on [1,7] has critical points x=3,6 and values p(1)=−2,p(3)=1,p(6)=−1,p(7)=0; where is the absolute minimum?
Explanation: This problem requires using the Candidates Test to find where the absolute minimum occurs for a continuous function on [1,7]. The Candidates Test states that for continuous functions on closed intervals, absolute extrema must occur at critical points or endpoints—nowhere else. Our candidate locations are the critical points x = 3 and x = 6, along with the endpoints x = 1 and x = 7. Checking all candidates: p(1) = -2, p(3) = 1, p(6) = -1, and p(7) = 0, we see that p(1) = -2 is the smallest value. Students might mistakenly choose x = 6 since it's a critical point with a negative value, but the endpoint x = 1 has an even smaller value. The systematic approach: list all critical points and endpoints, evaluate the function at each, and identify the minimum value among all candidates.
For continuous h on [−1,5] with critical points x=1,4, given h(−1)=0,h(1)=2,h(4)=−3,h(5)=−1, where is the absolute minimum?
Explanation: This problem requires applying the Candidates Test to find the absolute minimum of a continuous function on a closed interval. The Candidates Test guarantees finding absolute extrema by systematically checking all critical points and endpoints where extrema can occur. We must examine four candidates: the endpoints x = -1 and x = 5, and the critical points x = 1 and x = 4. Computing function values at each candidate: h(-1) = 0, h(1) = 2, h(4) = -3, and h(5) = -1. The smallest value is h(4) = -3, so the absolute minimum occurs at the critical point x = 4. A tempting error would be choosing x = 5 since h(5) = -1 is negative, but we must compare all candidate values, not just look for negative values. The candidates checklist for finding absolute extrema is: identify all critical points in the open interval, add both endpoints to the candidate list, evaluate the function at every candidate, then select the location with the extreme value.