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AP Calculus AB Quiz

AP Calculus AB Quiz: Candidates Test

Practice Candidates Test in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A continuous function uuu on [0,5][0,5][0,5] has candidates x=0,1,4,5x=0,1,4,5x=0,1,4,5 with u(0)=3,u(1)=2,u(4)=4,u(5)=1u(0)=3,u(1)=2,u(4)=4,u(5)=1u(0)=3,u(1)=2,u(4)=4,u(5)=1. Where is the absolute maximum?

Select an answer to continue

What this quiz covers

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 AB.

How to use this quiz

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.

All questions

Question 1

A continuous function uuu on [0,5][0,5][0,5] has candidates x=0,1,4,5x=0,1,4,5x=0,1,4,5 with u(0)=3,u(1)=2,u(4)=4,u(5)=1u(0)=3,u(1)=2,u(4)=4,u(5)=1u(0)=3,u(1)=2,u(4)=4,u(5)=1. Where is the absolute maximum?

  1. x=0x=0x=0
  2. x=1x=1x=1
  3. x=5x=5x=5
  4. x=4x=4x=4 (correct answer)
  5. x=0x=0x=0 (endpoint)

Explanation: 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.

Question 2

A continuous function vvv on [−2,2][-2,2][−2,2] has candidates x=−2,−1,1,2x=-2,-1,1,2x=−2,−1,1,2 with v(−2)=0,v(−1)=−4,v(1)=−3,v(2)=−5v(-2)=0,v(-1)=-4,v(1)=-3,v(2)=-5v(−2)=0,v(−1)=−4,v(1)=−3,v(2)=−5. Where is the absolute minimum?

  1. x=−1x=-1x=−1
  2. x=−2x=-2x=−2
  3. x=1x=1x=1
  4. x=2x=2x=2 (correct answer)
  5. x=−1x=-1x=−1 (critical point)

Explanation: This problem uses the Candidates Test to find the absolute minimum on [-2,2]. The candidates include endpoints (x = -2, 2) and critical points (x = -1, 1). Evaluating at all candidates: v(-2) = 0, v(-1) = -4, v(1) = -3, and v(2) = -5. The smallest value is v(2) = -5, so the absolute minimum occurs at x = 2. Choice E incorrectly assumes the minimum must occur at a critical point where v'(x) = 0, but endpoints can be extrema without having horizontal tangents. The candidates method ensures we find all absolute extrema: list endpoints and critical points, evaluate at each, then identify the most extreme values.

Question 3

A continuous function hhh on [−5,5][-5,5][−5,5] has critical points x=−4,1x=-4,1x=−4,1. If h(−5)=0,h(−4)=3,h(1)=2,h(5)=4h(-5)=0,h(-4)=3,h(1)=2,h(5)=4h(−5)=0,h(−4)=3,h(1)=2,h(5)=4, where is the absolute maximum?

  1. At x=5x=5x=5 (correct answer)
  2. At x=−4x=-4x=−4
  3. At x=−5x=-5x=−5
  4. At x=1x=1x=1
  5. At x=−5x=-5x=−5 only

Explanation: 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 B, at x=-4, because h(-4)=3 is high but not the maximum, which is h(5)=4 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.

Question 4

A continuous function ppp on [−2,5][-2,5][−2,5] has critical points x=1,4x=1,4x=1,4. If p(−2)=0,p(1)=2,p(4)=6,p(5)=3p(-2)=0,p(1)=2,p(4)=6,p(5)=3p(−2)=0,p(1)=2,p(4)=6,p(5)=3, where is the absolute maximum?

  1. At x=5x=5x=5
  2. At x=−2x=-2x=−2
  3. At x=1x=1x=1
  4. At x=−2x=-2x=−2 or x=5x=5x=5 only
  5. At x=4x=4x=4 (correct answer)

Explanation: 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 p(1)=2 is less than p(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.

Question 5

A continuous function ggg on [−6,−2][-6,-2][−6,−2] has critical points x=−5,−3x=-5,-3x=−5,−3. If g(−6)=2,g(−5)=1,g(−3)=0,g(−2)=3g(-6)=2,g(-5)=1,g(-3)=0,g(-2)=3g(−6)=2,g(−5)=1,g(−3)=0,g(−2)=3, where is the absolute minimum?

  1. At x=−6x=-6x=−6
  2. At x=−3x=-3x=−3 (correct answer)
  3. At x=−5x=-5x=−5
  4. At x=−2x=-2x=−2
  5. At x=−6x=-6x=−6 or x=−2x=-2x=−2 only

Explanation: 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=-5, but g(-5)=1 is greater than g(-3)=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.

Question 6

A continuous function hhh on [0,5][0,5][0,5] has critical points x=1,4x=1,4x=1,4. If h(0)=2,h(1)=0,h(4)=3,h(5)=1h(0)=2,h(1)=0,h(4)=3,h(5)=1h(0)=2,h(1)=0,h(4)=3,h(5)=1, where is the absolute maximum?

  1. At x=1x=1x=1
  2. At x=0x=0x=0
  3. At x=4x=4x=4 (correct answer)
  4. At x=5x=5x=5
  5. At x=0x=0x=0 only

Explanation: 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)=0 is less than h(4)=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.

Question 7

A continuous function fff on [2,8][2,8][2,8] has critical points x=3,7x=3,7x=3,7. If f(2)=5,f(3)=4,f(7)=6,f(8)=1f(2)=5,f(3)=4,f(7)=6,f(8)=1f(2)=5,f(3)=4,f(7)=6,f(8)=1, where is the absolute minimum?

  1. At x=8x=8x=8 (correct answer)
  2. At x=7x=7x=7
  3. At x=2x=2x=2
  4. At x=3x=3x=3
  5. At x=2x=2x=2 or x=8x=8x=8 only

Explanation: 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=3, but f(3)=4 is greater than f(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.

Question 8

For continuous vvv on [1,5][1,5][1,5] with critical points x=2,4x=2,4x=2,4 and v(1)=−2,v(2)=0,v(4)=−3,v(5)=−1v(1)=-2,v(2)=0,v(4)=-3,v(5)=-1v(1)=−2,v(2)=0,v(4)=−3,v(5)=−1, where is the absolute maximum?

  1. x=4x=4x=4
  2. x=5x=5x=5
  3. x=1x=1x=1
  4. x=2x=2x=2 (correct answer)
  5. x=1x=1x=1 and x=5x=5x=5

Explanation: 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.

Question 9

A continuous function ggg on [0,6][0,6][0,6] has critical points x=2,5x=2,5x=2,5 and values g(0)=1,g(2)=4,g(5)=0,g(6)=3g(0)=1,g(2)=4,g(5)=0,g(6)=3g(0)=1,g(2)=4,g(5)=0,g(6)=3; where is the absolute minimum?

  1. x=2x=2x=2
  2. x=0x=0x=0
  3. x=6x=6x=6
  4. x=5x=5x=5 (correct answer)
  5. There is no absolute minimum on [0,6][0,6][0,6]

Explanation: This problem applies the Candidates Test with given function values to find the absolute minimum. The Candidates Test requires checking all critical points (x = 2, 5) and endpoints (x = 0, 6) for a continuous function on [0, 6]. We have g(0) = 1, g(2) = 4, g(5) = 0, and g(6) = 3, so we compare these four values to find the smallest. The minimum value is 0, which occurs at x = 5, making this the location of the absolute minimum. Students might incorrectly choose x = 0 thinking the endpoint with the second-smallest value is the answer, but the Candidates Test requires finding the actual minimum value among all candidates. Remember the Candidates Test checklist: identify all critical points, include both endpoints, evaluate the function at all candidates, and compare to find extrema.

Question 10

A continuous function hhh on [−4,0][-4,0][−4,0] has critical points x=−3,−1x=-3,-1x=−3,−1. If h(−4)=0,h(−3)=2,h(−1)=1,h(0)=3h(-4)=0,h(-3)=2,h(-1)=1,h(0)=3h(−4)=0,h(−3)=2,h(−1)=1,h(0)=3, where is the absolute maximum?

  1. At x=−3x=-3x=−3
  2. At x=0x=0x=0 (correct answer)
  3. At x=−1x=-1x=−1
  4. At x=−4x=-4x=−4
  5. At x=−4x=-4x=−4 only

Explanation: 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=−3x=-3x=−3, but h(−3)=2h(-3)=2h(−3)=2 is less than h(0)=3h(0)=3h(0)=3, so it's not the maximum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate fff at each, and compare the values to find the max or min.

Question 11

A continuous function hhh on [0,9][0,9][0,9] has critical points x=4,6x=4,6x=4,6. If h(0)=1,h(4)=3,h(6)=0,h(9)=2h(0)=1,h(4)=3,h(6)=0,h(9)=2h(0)=1,h(4)=3,h(6)=0,h(9)=2, where is the absolute minimum?

  1. At x=0x=0x=0
  2. At x=9x=9x=9
  3. At x=4x=4x=4
  4. At x=6x=6x=6 (correct answer)
  5. At x=0x=0x=0 or x=9x=9x=9 only

Explanation: 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.

Question 12

A continuous function hhh on [−3,1][-3,1][−3,1] has critical points x=−2,0x=-2,0x=−2,0. If h(−3)=2,h(−2)=1,h(0)=3,h(1)=0h(-3)=2,h(-2)=1,h(0)=3,h(1)=0h(−3)=2,h(−2)=1,h(0)=3,h(1)=0, where is the absolute minimum?

  1. At x=−3x=-3x=−3
  2. At x=1x=1x=1 (correct answer)
  3. At x=−2x=-2x=−2
  4. At x=0x=0x=0
  5. At x=−3x=-3x=−3 or x=1x=1x=1 only

Explanation: 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=-2, because h(-2)=1 is less than h(-3)=2 and h(0)=3, but this overlooks the endpoint x=1 where h(1)=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.

Question 13

A continuous function ppp on [−3,2][-3,2][−3,2] has critical points x=−1,1x=-1,1x=−1,1. If p(−3)=2,p(−1)=4,p(1)=0,p(2)=3p(-3)=2,p(-1)=4,p(1)=0,p(2)=3p(−3)=2,p(−1)=4,p(1)=0,p(2)=3, where is the absolute minimum?

  1. At x=−1x=-1x=−1
  2. At x=2x=2x=2
  3. At x=−3x=-3x=−3
  4. At x=1x=1x=1 (correct answer)
  5. At x=−3x=-3x=−3 or x=2x=2x=2 only

Explanation: 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 p(-1)=4 is greater than p(1)=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.

Question 14

A continuous function ggg on [−1,6][-1,6][−1,6] has critical points x=0,5x=0,5x=0,5. If g(−1)=3,g(0)=2,g(5)=−1,g(6)=0g(-1)=3,g(0)=2,g(5)=-1,g(6)=0g(−1)=3,g(0)=2,g(5)=−1,g(6)=0, where is the absolute minimum?

  1. At x=5x=5x=5 (correct answer)
  2. At x=−1x=-1x=−1
  3. At x=0x=0x=0
  4. At x=6x=6x=6
  5. At x=−1x=-1x=−1 or x=6x=6x=6 only

Explanation: 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=0, but g(0)=2 is greater than g(5)=-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.

Question 15

A continuous function fff on [−2,4][-2,4][−2,4] has candidates x=−2,1,3,4x=-2,1,3,4x=−2,1,3,4 with f(−2)=0,f(1)=5,f(3)=2,f(4)=−1f(-2)=0,f(1)=5,f(3)=2,f(4)=-1f(−2)=0,f(1)=5,f(3)=2,f(4)=−1. Where is the absolute maximum?

  1. x=1x=1x=1 (correct answer)
  2. x=3x=3x=3
  3. x=−2x=-2x=−2
  4. x=4x=4x=4
  5. x=3x=3x=3 (since f′(3)=0f'(3)=0f′(3)=0)

Explanation: 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 endpoints or critical points (where f'(x) = 0 or f'(x) is undefined). We check all candidates: f(-2) = 0, f(1) = 5, f(3) = 2, and f(4) = -1. The largest value is f(1) = 5, so the absolute maximum occurs at x = 1. Choice E incorrectly assumes the maximum must occur where f'(x) = 0, but endpoints can also be extrema even without a horizontal tangent. Remember the candidates checklist: (1) find all critical points in the interval, (2) include both endpoints, (3) evaluate f at all candidates, (4) compare values to identify absolute extrema.

Question 16

A continuous function sss on [2,10][2,10][2,10] has candidates x=2,4,9,10x=2,4,9,10x=2,4,9,10 with s(2)=−2,s(4)=−2,s(9)=1,s(10)=0s(2)=-2,s(4)=-2,s(9)=1,s(10)=0s(2)=−2,s(4)=−2,s(9)=1,s(10)=0. Where is the absolute minimum?

  1. x=9x=9x=9
  2. x=10x=10x=10
  3. x=2x=2x=2 and x=4x=4x=4 (correct answer)
  4. x=4x=4x=4 only
  5. x=2x=2x=2 only

Explanation: 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.

Question 17

A continuous function hhh on [−1,5][-1,5][−1,5] has candidates x=−1,0,4,5x=-1,0,4,5x=−1,0,4,5 with h(−1)=1,h(0)=3,h(4)=3,h(5)=2h(-1)=1,h(0)=3,h(4)=3,h(5)=2h(−1)=1,h(0)=3,h(4)=3,h(5)=2. Where is the absolute maximum?

  1. x=5x=5x=5
  2. x=−1x=-1x=−1
  3. x=0x=0x=0 and x=4x=4x=4 (correct answer)
  4. x=0x=0x=0 only
  5. x=4x=4x=4 only

Explanation: 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.

Question 18

A continuous function fff on [−2,5][-2,5][−2,5] has critical points at x=0,3x=0,3x=0,3 with f(−2)=1,f(0)=4,f(3)=2,f(5)=0f(-2)=1,f(0)=4,f(3)=2,f(5)=0f(−2)=1,f(0)=4,f(3)=2,f(5)=0. Where is the absolute maximum?

  1. x=0x=0x=0 (correct answer)
  2. x=3x=3x=3
  3. x=5x=5x=5
  4. x=−2x=-2x=−2
  5. Cannot be determined from given information

Explanation: This problem requires applying the Candidates Test to find the absolute maximum of a continuous function on a closed interval. The Candidates Test states that for a continuous function on a closed interval, the absolute extrema must occur either at critical points or at endpoints. We must check all candidates: the endpoints x = -2 and x = 5, and the critical points x = 0 and x = 3. Evaluating at each candidate: f(-2) = 1, f(0) = 4, f(3) = 2, and f(5) = 0. Since f(0) = 4 is the largest value among all candidates, the absolute maximum occurs at x = 0. A common error would be assuming the maximum must be at an endpoint without checking critical points. Remember the candidates checklist: identify all critical points in the interval, evaluate the function at all critical points and both endpoints, then select the location with the largest (or smallest) value.

Question 19

A continuous function fff on [2,9][2,9][2,9] has critical points x=4,8x=4,8x=4,8. If f(2)=1,f(4)=6,f(8)=5,f(9)=2f(2)=1,f(4)=6,f(8)=5,f(9)=2f(2)=1,f(4)=6,f(8)=5,f(9)=2, where is the absolute maximum?

  1. At x=8x=8x=8
  2. At x=2x=2x=2
  3. At x=9x=9x=9
  4. At x=2x=2x=2 or x=9x=9x=9 only
  5. At x=4x=4x=4 (correct answer)

Explanation: 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.

Question 20

A continuous function ppp on [−2,1][-2,1][−2,1] has critical points x=−1,0x=-1,0x=−1,0. If p(−2)=0,p(−1)=1,p(0)=−2,p(1)=−1p(-2)=0,p(-1)=1,p(0)=-2,p(1)=-1p(−2)=0,p(−1)=1,p(0)=−2,p(1)=−1, where is the absolute minimum?

  1. At x=−1x=-1x=−1
  2. At x=−2x=-2x=−2
  3. At x=1x=1x=1
  4. At x=−2x=-2x=−2 or x=1x=1x=1 only
  5. At x=0x=0x=0 (correct answer)

Explanation: 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.