This question tests the Extreme Value Theorem (EVT) and the impact of critical points on extrema existence. Continuity of r on [-3,3] guarantees, by EVT, both an absolute maximum and an absolute minimum on the closed interval. The single critical point at x=0 is a candidate for a local extremum, but absolute extrema could occur there or at endpoints x=-3 or x=3. Evaluating function values at the critical point and endpoints is required to determine the actual max and min. Choice B tempts by assuming the critical point is the absolute maximum, but it fails as endpoints might yield higher values depending on the function. EVT checklist: ensure continuity on closed bounded interval, locate critical points (f'=0 or undefined), compare values at those points and endpoints to identify extrema.

This question examines the Extreme Value Theorem (EVT) and properties of extrema for differentiable functions. Continuity on the closed interval [1,5] ensures, by EVT, that p attains both an absolute maximum and an absolute minimum value on the interval. Differentiability on (1,5) implies critical points where p'(x)=0 are possible interior candidates for extrema, but absolute extrema could also be at endpoints x=1 or x=5. The theorem guarantees existence without specifying locations, emphasizing evaluation at both critical points and boundaries. Choice B is a common distractor but incorrect because absolute maxima can occur at endpoints even if p'(x)=0 points exist interiorly. For EVT, follow this checklist: verify continuity on a closed bounded interval, locate critical points (where derivative is zero or undefined), and compare function values at those points and the endpoints.

This question evaluates the Extreme Value Theorem (EVT) and the role of critical points in finding extrema. Since g is continuous on the closed interval [0,4], the EVT ensures it has both an absolute maximum and an absolute minimum on that interval. The critical point at x=2 where g'(2)=0 is a candidate for an extremum, but the absolute max and min could also occur at the endpoints x=0 or x=4, especially given g(0)=g(4). Evaluating all candidates—endpoints and critical points—is necessary to determine the actual extrema. Choice B is a tempting distractor but fails because the absolute maximum might occur at an endpoint if g(2) is not the largest value, as the critical point only guarantees a local extremum possibility. For EVT applications, use this checklist: confirm continuity, verify closed bounded interval, find critical points where derivative is zero or undefined, and compare values at those points and endpoints.

This question assesses the Extreme Value Theorem (EVT) and consequences of no zero-derivative points on extrema. Since m is continuous on [2,7], the EVT guarantees absolute maximum and minimum values are attained on the interval. With m'(x) ≠ 0 everywhere in (2,7), there are no critical points where m' = 0, implying m is strictly monotonic, so extrema must be at the endpoints. Continuity ensures these extrema exist, and the lack of interior critical points confines them to the boundaries. A tempting distractor is choice B, which says there is at least one critical point in (2,7), but this fails because m' exists and is nonzero, so no such points. For applying the EVT, confirm continuity on a closed interval; if there are no interior critical points, compare values at the endpoints to find the absolute max and min.

This question tests the Extreme Value Theorem (EVT) and the role of a single critical point in determining extrema. The function f is continuous on [0,3], so by the EVT, it attains both an absolute maximum and an absolute minimum on the interval. The single critical point c in (0,3) is a candidate for a local extremum, but absolute extrema could also occur at the endpoints. Continuity on the closed interval ensures the existence of these global extrema, regardless of whether the critical point is a max, min, or neither. A tempting distractor is choice A, which claims a local extremum at c, but this fails because f'(c) = 0 does not guarantee one, as in $x^3$ where it's an inflection point. To apply the EVT, check continuity on a closed, bounded interval; locate all critical points and endpoints, then evaluate the function to determine the absolute extrema.

This question probes understanding of the Extreme Value Theorem (EVT), extrema, and critical points for a specific function. For f(x) = |x-1| on [-2,4], which is continuous on the closed interval, the EVT guarantees absolute extrema exist. The function has a critical point at x=1 where f' does not exist (due to the corner), and evaluating shows f(1)=0 is the global minimum, while maxima are at the endpoints. Critical points and endpoints must be checked to locate these extrema, with the minimum occurring at the critical point in this case. A tempting distractor is choice B, which claims a global maximum at x=1, but this fails because x=1 is actually the minimum, as the function increases away from it on both sides. For the EVT checklist, confirm continuity on a closed interval; identify critical points (f' = 0 or undefined) and endpoints, then compare function values to find absolute max and min.

This problem combines derivative sign analysis with extrema identification. Since s is continuous on [2,8] with s'(x)<0 on (2,5) (decreasing) and s'(x)>0 on (5,8) (increasing), the function changes from decreasing to increasing at x=5, making it a local minimum. Moreover, since s decreases until x=5 and then increases afterward, this local minimum at x=5 must also be the absolute minimum on the entire interval [2,8]. Option A incorrectly identifies this as a maximum rather than minimum—when derivatives change from negative to positive, we have a minimum. The sign change rule: negative to positive derivative = minimum point.

This question explores the Extreme Value Theorem (EVT), extrema at endpoints, and function behavior on a closed interval. The function r is continuous on [0,4], so the EVT guarantees absolute extrema exist, with the minimum value of 2 attained at both endpoints x=0 and x=4, since r(x) > 2 inside. There must be critical points inside where r' = 0 or undefined, as the function rises above 2 and returns, indicating local maxima or minima. The absolute minimum occurs at the endpoints, highlighting that extrema can be at boundaries even without critical points there. A tempting distractor is choice C, which claims no absolute minimum because r is never less than 2, but this fails as it attains 2 at the endpoints, which is the minimum. For the EVT checklist, verify continuity on a closed interval; find critical points in the interior, evaluate at them and endpoints, and identify the highest and lowest values.

This question tests EVT application when only discrete function values are known. Since f is continuous on the closed interval [0,4], the Extreme Value Theorem guarantees that f must attain both an absolute maximum and an absolute minimum somewhere on this interval. From the given values, we can see that 5 is the largest and 1 is the smallest, but we cannot determine where the actual absolute extrema occur because f might attain even larger values between the given points or the same extreme values at multiple locations. Option A incorrectly assumes the given values are the only possible values f can take. The EVT guarantee: extrema exist, but their exact locations require complete function information.

This problem tests the relationship between local extrema and EVT guarantees. Since m is continuous on the closed interval [0,2], the Extreme Value Theorem guarantees that m must have both an absolute maximum and an absolute minimum somewhere on this interval. The local maximum at x=1 tells us about behavior near x=1 but doesn't determine whether this is the absolute maximum or where other extrema occur. Option D incorrectly suggests local maxima prevent absolute minima from existing, but EVT's guarantee is independent of local behavior. EVT reminder: continuous on closed interval always means both absolute max and min exist, regardless of local extrema.