Candidates Test - AP Calculus AB

1. For continuous $v$ on $[1,5]$ with critical points $x=2,4$ and $v(1)=-2,v(2)=0,v(4)=-3,v(5)=-1$, where is the absolute maximum?
   Answer: $x=2$ This problem demonstrates the Candidates Test for finding the absolute maximum on [1, 5]. The Candidates Test requires examining critical points (x = 2, 4) and endpoints (x = 1, 5) of continuous function v. With v(1) = -2, v(2) = 0, v(4) = -3, and v(5) = -1, we compare all four values to find the largest. The maximum value is 0, which occurs at x = 2, making this critical point the location of the absolute maximum. A common mistake is thinking the maximum must be at an endpoint or choosing x = 5 because it has the largest value among endpoints, but the Candidates Test requires comparing all candidates equally. Follow the systematic approach: list critical points and endpoints, evaluate the function at each, and identify the location with the maximum value.
2. A continuous function $h$ on $[0,9]$ has critical points $x=4,6$. If $h(0)=1,h(4)=3,h(6)=0,h(9)=2$, where is the absolute minimum?
   Answer: At $x=6$ This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=4, but h(4)=3 is greater than h(6)=0, so it's not the minimum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.
3. A continuous function $t$ on $[-1,3]$ has candidates $x=-1,0,2,3$ with $t(-1)=1,t(0)=0,t(2)=-1,t(3)=-1$. Where is the absolute minimum?
   Answer: $x=2$ and $x=3$ This problem applies the Candidates Test to find the absolute minimum on [-1,3], where multiple points achieve the minimum value. The candidates include endpoints (x = -1, 3) and critical points (x = 0, 2). Evaluating at all candidates: t(-1) = 1, t(0) = 0, t(2) = -1, and t(3) = -1. The minimum value is -1, occurring at both x = 2 and x = 3, making both locations absolute minima. Choices D and E incorrectly identify only one minimum location, but the Candidates Test reveals that both x = 2 and x = 3 achieve the minimum value. Remember the complete candidates process: find all critical points and endpoints, evaluate at each, then report all locations achieving the extreme values.
4. A continuous function $h$ on $[0,2]$ has critical points $x=\tfrac12,\tfrac32$. If $h(0)=1,h(\tfrac12)=2,h(\tfrac32)=0,h(2)=3$, where is the absolute maximum?
   Answer: At $x=2$ The Candidates Test is a method in calculus to find the absolute maximum and minimum values of a continuous function on a closed interval. To apply this test, first identify all critical points within the interval where the derivative is zero or undefined. Then, evaluate the function at these critical points and at the endpoints of the interval. By comparing these function values, you can determine the absolute extrema, as they must occur at one of these candidate points. A tempting distractor is choice C, at x=1/2, because h(1/2)=2 is high but the maximum is h(2)=3 at the endpoint. Always remember the transferable candidates checklist: identify the closed interval, find critical points, evaluate the function at endpoints and critical points, and compare all values to locate the extrema.
5. A continuous function $h$ on $[0,3]$ has critical points $x=1,2$. If $h(0)=2,h(1)=3,h(2)=1,h(3)=0$, where is the absolute maximum?
   Answer: At $x=1$ This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=2, but h(2)=1 is less than h(1)=3, so it's not the maximum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.
6. A continuous function $p$ on $[-2,1]$ has critical points $x=-1,0$. If $p(-2)=0,p(-1)=1,p(0)=-2,p(1)=-1$, where is the absolute minimum?
   Answer: At $x=0$ The Candidates Test is a method in calculus to find the absolute maximum and minimum values of a continuous function on a closed interval. To apply this test, first identify all critical points within the interval where the derivative is zero or undefined. Then, evaluate the function at these critical points and at the endpoints of the interval. By comparing these function values, you can determine the absolute extrema, as they must occur at one of these candidate points. A tempting distractor is choice C, at x=1, because p(1)=-1 is low but the minimum is p(0)=-2 at a critical point. Always remember the transferable candidates checklist: identify the closed interval, find critical points, evaluate the function at endpoints and critical points, and compare all values to locate the extrema.
7. A continuous function $f$ on $[2,7]$ has critical points $x=4,6$. If $f(2)=1,f(4)=0,f(6)=2,f(7)=3$, where is the absolute minimum?
   Answer: At $x=4$ The Candidates Test is a method in calculus to find the absolute maximum and minimum values of a continuous function on a closed interval. To apply this test, first identify all critical points within the interval where the derivative is zero or undefined. Then, evaluate the function at these critical points and at the endpoints of the interval. By comparing these function values, you can determine the absolute extrema, as they must occur at one of these candidate points. A tempting distractor is choice D, at x=2, because f(2)=1 is an endpoint but the minimum is f(4)=0 at a critical point. Always remember the transferable candidates checklist: identify the closed interval, find critical points, evaluate the function at endpoints and critical points, and compare all values to locate the extrema.
8. A continuous function $u$ on $[-5,1]$ has critical points at $x=-4,-2$ with $u(-5)=0,u(-4)=-2,u(-2)=-1,u(1)=3$. 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 ensures we find absolute extrema by checking all possible locations: endpoints and critical points within the interval. For function u on [-5,1], the candidates include endpoints x = -5 and x = 1, plus critical points x = -4 and x = -2. Evaluating at each candidate: u(-5) = 0, u(-4) = -2, u(-2) = -1, and u(1) = 3. The smallest value is u(-4) = -2, so the absolute minimum occurs at x = -4. Students might mistakenly choose x = -5 thinking the minimum must be at the left endpoint. To find extrema using the Candidates Test: identify all critical points within the interval, add both endpoints to your candidate list, evaluate the function at every candidate, then select the location yielding the extreme value.
9. A continuous function $f$ on $[ -2,6 ]$ has critical points $x=0,4$. If $f(-2)=3,f(0)=1,f(4)=2,f(6)=0$, where is the absolute minimum?
   Answer: At $x=6$ The Candidates Test is a method in calculus to find the absolute maximum and minimum values of a continuous function on a closed interval. To apply this test, first identify all critical points within the interval where the derivative is zero or undefined. Then, evaluate the function at these critical points and at the endpoints of the interval. By comparing these function values, you can determine the absolute extrema, as they must occur at one of these candidate points. A tempting distractor is choice A, at x=0, because f(0)=1 is less than f(-2)=3 and f(4)=2, but this fails to check the endpoint x=6 where f(6)=0 is smaller. Always remember the transferable candidates checklist: identify the closed interval, find critical points, evaluate the function at endpoints and critical points, and compare all values to locate the extrema.
10. A continuous function $f$ on $[2,9]$ has critical points $x=4,8$. If $f(2)=1,f(4)=6,f(8)=5,f(9)=2$, where is the absolute maximum?
    Answer: At $x=4$ This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=8, but f(8)=5 is less than f(4)=6, so it's not the maximum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.
11. A continuous function $p$ on $[0,8]$ has critical points $x=2,7$. If $p(0)=5,p(2)=4,p(7)=6,p(8)=1$, where is the absolute minimum?
    Answer: At $x=8$ This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=2, but p(2)=4 is greater than p(8)=1, so it's not the minimum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.
12. A continuous function $t$ on $[2,8]$ has critical points at $x=5,7$ with $t(2)=6,t(5)=4,t(7)=5,t(8)=3$. Where is the absolute minimum?
    Answer: $x=8$ 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 locations where they can occur: critical points and endpoints. For function t on [2,8], we must evaluate at endpoints x = 2 and x = 8, and at critical points x = 5 and x = 7. The function values are: t(2) = 6, t(5) = 4, t(7) = 5, and t(8) = 3. Since t(8) = 3 is the smallest value among all candidates, the absolute minimum occurs at x = 8. A student might incorrectly choose x = 5 by only comparing critical points without considering endpoints. The candidates method for extrema: list all critical points in the open interval, include both endpoints of the closed interval, evaluate the function at each candidate, then select the location with the minimum value.
13. A continuous function $h$ on $[-2,3]$ has critical points $x=-1,2$. If $h(-2)=4,h(-1)=1,h(2)=0,h(3)=2$, where is the absolute minimum?
    Answer: At $x=2$ This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=-1, but h(-1)=1 is greater than h(2)=0, so it's not the minimum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.
14. A continuous function $h$ on $[-1,5]$ has candidates $x=-1,0,4,5$ with $h(-1)=1,h(0)=3,h(4)=3,h(5)=2$. Where is the absolute maximum?
    Answer: $x=0$ and $x=4$ This question demonstrates the Candidates Test when multiple points achieve the same extreme value. For continuous functions on closed intervals, absolute extrema must occur at candidates: endpoints or critical points. Checking all candidates: h(-1) = 1, h(0) = 3, h(4) = 3, and h(5) = 2. The maximum value is 3, which occurs at both x = 0 and x = 4, making both locations absolute maxima. Choices D and E incorrectly suggest only one location can be an absolute maximum, but multiple points can share the same extreme value. The candidates algorithm is foolproof: list all endpoints and critical points, evaluate the function at each, then identify all locations achieving the extreme values.
15. A continuous function $r$ on $[-4,0]$ has candidates $x=-4,-2,-1,0$ with $r(-4)=6,r(-2)=6,r(-1)=5,r(0)=7$. Where is the absolute maximum?
    Answer: $x=0$ This problem demonstrates the Candidates Test on [-4,0] where we seek the absolute maximum. The candidates include endpoints (x = -4, 0) and critical points (x = -2, -1). Evaluating at all candidates: r(-4) = 6, r(-2) = 6, r(-1) = 5, and r(0) = 7. The largest value is r(0) = 7, so the absolute maximum occurs at x = 0. Choice A incorrectly identifies both x = -4 and x = -2 where r = 6, but this is not the maximum value since r(0) = 7 is larger. The candidates algorithm ensures accuracy: check all endpoints and critical points, evaluate the function at each, then compare to find the true extrema.
16. A continuous function $u$ on $[0,5]$ has candidates $x=0,1,4,5$ with $u(0)=3,u(1)=2,u(4)=4,u(5)=1$. Where is the absolute maximum?
    Answer: $x=4$ This question demonstrates the Candidates Test for finding the absolute maximum on [0,5]. The candidates for absolute extrema are endpoints (x = 0, 5) and critical points (x = 1, 4). Evaluating the function at all candidates: u(0) = 3, u(1) = 2, u(4) = 4, and u(5) = 1. The largest value is u(4) = 4, so the absolute maximum occurs at x = 4. Choice E incorrectly focuses only on the endpoint x = 0, missing that critical points in the interior can yield larger values. The candidates algorithm is systematic: identify all endpoints and critical points, evaluate the function at each candidate, then select the location(s) with the most extreme values.
17. A continuous function $g$ on $[0,6]$ has critical points $x=1,5$. If $g(0)=3,g(1)=4,g(5)=2,g(6)=5$, where is the absolute maximum?
    Answer: At $x=6$ This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=1, but g(1)=4 is less than g(6)=5, so it's not the maximum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.
18. A continuous function $p$ on $[4,9]$ has critical points $x=6,8$. If $p(4)=2,p(6)=0,p(8)=1,p(9)=3$, where is the absolute minimum?
    Answer: At $x=6$ The Candidates Test is a method in calculus to find the absolute maximum and minimum values of a continuous function on a closed interval. To apply this test, first identify all critical points within the interval where the derivative is zero or undefined. Then, evaluate the function at these critical points and at the endpoints of the interval. By comparing these function values, you can determine the absolute extrema, as they must occur at one of these candidate points. A tempting distractor is choice A, at x=4, because p(4)=2 is an endpoint but not the minimum, which is p(6)=0 at a critical point. Always remember the transferable candidates checklist: identify the closed interval, find critical points, evaluate the function at endpoints and critical points, and compare all values to locate the extrema.
19. A continuous function $p$ on $[0,10]$ has critical points $x=3,9$. If $p(0)=2,p(3)=7,p(9)=1,p(10)=4$, where is the absolute maximum?
    Answer: At $x=3$ The Candidates Test is a method in calculus to find the absolute maximum and minimum values of a continuous function on a closed interval. To apply this test, first identify all critical points within the interval where the derivative is zero or undefined. Then, evaluate the function at these critical points and at the endpoints of the interval. By comparing these function values, you can determine the absolute extrema, as they must occur at one of these candidate points. A tempting distractor is choice D, at x=0, because p(0)=2 might seem like a starting point, but the maximum is p(3)=7 at a critical point. Always remember the transferable candidates checklist: identify the closed interval, find critical points, evaluate the function at endpoints and critical points, and compare all values to locate the extrema.
20. A continuous function $s$ on $[2,10]$ has candidates $x=2,4,9,10$ with $s(2)=-2,s(4)=-2,s(9)=1,s(10)=0$. Where is the absolute minimum?
    Answer: $x=2$ and $x=4$ This question uses the Candidates Test to find the absolute minimum on [2,10], demonstrating that multiple points can achieve the same extreme value. The candidates are endpoints (x = 2, 10) and critical points (x = 4, 9). Evaluating at all candidates: s(2) = -2, s(4) = -2, s(9) = 1, and s(10) = 0. The minimum value is -2, which occurs at both x = 2 and x = 4, so both are locations of the absolute minimum. Choices D and E incorrectly suggest only one location can be an absolute minimum, missing that extrema can occur at multiple points. The candidates method handles all cases: evaluate at endpoints and critical points, then identify all locations achieving the extreme values.

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