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.

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.

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.

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.

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.

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.

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.

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.

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.

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.