Candidates Test - AP Calculus BC

1. 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?
   Answer: $x=1$ 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.
2. For continuous $v$ on $[3,9]$ with critical points $x=4,8$, given $v(3)=-2,v(4)=0,v(8)=0,v(9)=-1$, where is the absolute maximum?
   Answer: $x=4$ and $x=8$ This problem applies the Candidates Test to find the absolute maximum on [3,9]. For continuous functions on closed intervals, the Candidates Test requires checking critical points and endpoints for absolute extrema. Our candidates are endpoints x = 3 and x = 9, plus critical points x = 4 and x = 8. Evaluating the function: v(3) = -2, v(4) = 0, v(8) = 0, and v(9) = -1. Both v(4) = 0 and v(8) = 0 equal the largest value, so the absolute maximum occurs at both x = 4 and x = 8. This illustrates that absolute extrema can occur at multiple locations when the function achieves its extreme value at more than one point. The complete candidates checklist: find all critical points in the interval, include both endpoints, evaluate at all candidates, and identify all locations where the maximum value is attained.
3. For continuous $s$ on $[0,5]$ with critical points $x=1,4$ and values $s(0)=2,s(1)=2,s(4)=-3,s(5)=-1$, where is the absolute minimum?
   Answer: $x=4$ This question requires applying the Candidates Test to locate the absolute minimum of s on [0,5]. The Candidates Test states that for continuous functions on closed intervals, absolute extrema must occur at critical points (where the derivative equals zero or doesn't exist) or at endpoints. We check the critical points x = 1 and x = 4, along with endpoints x = 0 and x = 5. Evaluating: s(0) = 2, s(1) = 2, s(4) = -3, and s(5) = -1, we find that s(4) = -3 is the smallest value. A tempting error is choosing x = 5 because it's negative and an endpoint, but the critical point x = 4 has an even smaller value. The systematic checklist: identify all critical points, add both endpoints, compute function values at each candidate, and select the minimum.
4. 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?​
   Answer: At $x=0$ 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.
5. For continuous $h$ on $[-1,5]$ with critical points $x=1,4$ and $h(-1)=2,h(1)=0,h(4)=-3,h(5)=-1$, where is the absolute minimum?​
   Answer: At $x=4$ This question applies the Candidates Test to locate the absolute minimum of a continuous function on the interval [-1, 5]. The Candidates Test states that for a continuous function on a closed interval, absolute extrema occur at critical points or endpoints—nowhere else. Our candidates are the critical points x = 1 and x = 4, plus the endpoints x = -1 and x = 5. Computing function values: h(-1) = 2, h(1) = 0, h(4) = -3, and h(5) = -1, we see the smallest value is -3 at x = 4. Choice E temptingly suggests both critical points, but only x = 4 gives the absolute minimum since h(4) < h(1). The systematic approach: list all candidates (critical points and endpoints), evaluate the function at each, then identify where the minimum value occurs.
6. For continuous $r$ on $[-4,0]$ with critical points $x=-3,-1$, given $r(-4)=-2,r(-3)=1,r(-1)=0,r(0)=2$, where is the absolute maximum?
   Answer: $x=0$ 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 = -4, -3, -1, 0 with r-values -2, 1, 0, 2, so the maximum is 2 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 = -3 with value 1, but it fails because 1 is less than 2 at the endpoint x = 0. Remember, the checklist for candidates is: list endpoints and critical points inside the interval, evaluate r at each, and compare the values to identify the extrema.
7. For continuous $f$ on $[-2,4]$ with critical points $x=-1,2$, given $f(-2)=3,f(-1)=5,f(2)=1,f(4)=4$, where is the absolute maximum?
   Answer: $x=-1$ This problem tests the Candidates Test for finding absolute extrema on a closed interval. For a continuous function on a closed interval, the absolute maximum and minimum must occur at either critical points (where the derivative is zero or undefined) or at the endpoints of the interval. We must check all candidates: the endpoints x = -2 and x = 4, and the critical points x = -1 and x = 2. Evaluating at each candidate: f(-2) = 3, f(-1) = 5, f(2) = 1, and f(4) = 4. The largest value is f(-1) = 5, so the absolute maximum occurs at x = -1. Choice E incorrectly assumes that extrema must occur at critical points, but endpoints can also be absolute extrema. Remember the candidates checklist: identify all critical points in the interval, evaluate the function at all critical points AND both endpoints, then compare all values to find the absolute maximum and minimum.
8. A continuous function $r$ on $[-4,2]$ has critical points $x=-3,-1$ with $r(-4)=-2,r(-3)=1,r(-1)=-3,r(2)=0$. Where is the absolute minimum?
   Answer: At $x=-1$ 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 key as they may mark the lowest points where the function dips. Endpoints must be assessed because the minimum could occur at the interval's edge. This systematic comparison identifies the true absolute minimum. One tempting option is x=-3 with r(-3)=1, but it fails as it's higher than r(-1)=-3, showing a critical point oversight. Always list and evaluate the function at: both endpoints and all critical points within the interval.
9. For continuous $u$ on $[-5,1]$ with critical points $x=-4,-2$, given $u(-5)=0,u(-4)=3,u(-2)=-1,u(1)=2$, where is the absolute minimum?
   Answer: $x=-2$ This question tests the Candidates Test for finding the absolute minimum on [-5,1]. The Candidates Test states that for continuous functions on closed intervals, absolute extrema must occur at critical points or endpoints. We must check all candidates: endpoints x = -5 and x = 1, and critical points x = -4 and x = -2. The function values are: u(-5) = 0, u(-4) = 3, u(-2) = -1, and u(1) = 2. Since u(-2) = -1 is the smallest value, the absolute minimum occurs at x = -2. Choice E incorrectly assumes x = -4 should be the minimum simply because it's a critical point, ignoring the actual function values that must be compared. The candidates process never changes: identify all critical points within the interval, add both endpoints to the list, evaluate at every candidate, and select the location with the minimum function value.
10. 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?
    Answer: $x=7$ 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.
11. 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?​
    Answer: $x=4$ 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.
12. 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?​
    Answer: At $x=3$ 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.
13. For continuous $v$ on $[1,4]$ with critical points $x=2,3$, given $v(1)=2,v(2)=-2,v(3)=-1,v(4)=0$, where is the absolute minimum?​
    Answer: $x=2$ This problem requires applying the Candidates Test to find the absolute minimum of a continuous function on a closed interval. The Candidates Test states that absolute extrema of continuous functions on closed intervals must occur at critical points or endpoints. Our candidates for the minimum are: endpoints x = 1 and x = 4, and critical points x = 2 and x = 3. Evaluating at each candidate: v(1) = 2, v(2) = -2, v(3) = -1, and v(4) = 0. The smallest value is v(2) = -2, which means the absolute minimum occurs at the critical point x = 2. Students might incorrectly choose x = 3 since it's also negative at v(3) = -1, but we need the smallest value among all candidates. The complete candidates checklist is: find all critical points within the interval, list both endpoints, evaluate the function at every candidate location, then identify where the minimum value occurs.
14. For continuous $r$ on $[0,5]$ with critical points $x=1,4$, given $r(0)=-2,r(1)=0,r(4)=3,r(5)=1$, where is the absolute maximum?​
    Answer: $x=4$ This problem requires applying the Candidates Test to find the absolute maximum of a continuous function on a closed interval. The Candidates Test guarantees finding absolute extrema by checking all critical points and endpoints where such extrema must occur for continuous functions. Our candidate locations are: endpoints x = 0 and x = 5, and critical points x = 1 and x = 4. Evaluating at each candidate: r(0) = -2, r(1) = 0, r(4) = 3, and r(5) = 1. Since r(4) = 3 is the largest value among all candidates, the absolute maximum occurs at the critical point x = 4. Students might incorrectly eliminate negative values like r(0) = -2 from consideration, but we must evaluate all candidates before comparing. The complete candidates checklist for extrema is: identify critical points within the interval, add both endpoints to your list, evaluate the function at every candidate, then select the location producing the maximum value.
15. 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?
    Answer: $x=8$ 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.
16. For continuous $s$ on $[2,10]$, critical points are $x=4,9$ with $s(2)=7,s(4)=5,s(9)=8,s(10)=6$. Where is the absolute maximum?
    Answer: At $x=9$ 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 evaluate critical points to capture potential peaks that could be the maximum. Endpoints are checked as the function might rise to its highest at the boundaries. Comparing these reveals the absolute maximum reliably. A tempting distractor is x=2 with s(2)=7, but it's less than s(9)=8, illustrating why all candidates matter. Always list and evaluate the function at: both endpoints and all critical points within the interval.
17. 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?
    Answer: $x=4$ 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.
18. Continuous $q$ on $[0,5]$ has critical points $x=1,4$ with $q(0)=2,q(1)=-1,q(4)=0,q(5)=-3$; where is the absolute minimum?
    Answer: $x=5$ 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 = 0, 1, 4, 5 with q-values 2, -1, 0, -3, so the minimum is -3 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 -1, but it fails because -1 is greater than -3 at the endpoint x = 5. Remember, the checklist for candidates is: list endpoints and critical points inside the interval, evaluate q at each, and compare the values to identify the extrema.
19. 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?
    Answer: $x=-1$ 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.
20. A continuous function $r$ on $[2,10]$ has critical points $x=4,9$ with $r(2)=6,r(4)=2,r(9)=1,r(10)=3$; where is the absolute minimum?​
    Answer: At $x=9$ This problem applies the Candidates Test to find the absolute minimum of function r on the interval [2, 10]. According to the Candidates Test, absolute extrema of continuous functions on closed intervals must occur at critical points or endpoints—no other locations are possible. Our candidates include critical points x = 4 and x = 9, plus endpoints x = 2 and x = 10. Evaluating at each candidate: r(2) = 6, r(4) = 2, r(9) = 1, and r(10) = 3, we find the smallest value is 1 at x = 9. Choice E incorrectly lists both critical points, but only x = 9 yields the absolute minimum since r(9) < r(4). The systematic Candidates Test approach: list all critical points and endpoints, evaluate the function at each, then identify the location with the minimum value.

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