This problem focuses on correctly distinguishing between local and absolute extrema for continuous functions on closed intervals. The fundamental distinction is that local extrema compare function values only within a small neighborhood of a point, while absolute (global) extrema compare function values across the entire interval. An absolute minimum is the smallest value that the function attains over the complete interval $[a,b]$, not just in a local neighborhood. Local maxima need not be greater than all values on the interval, and local minima can occur at interior points as well as endpoints. Choice A incorrectly describes local maxima as global, while choices C, D, and E make various incorrect claims about extrema locations and relationships. The crucial distinction: local extrema involve neighborhood comparisons, while absolute extrema involve interval-wide comparisons.

This problem examines the relationship between critical points and absolute extrema for continuous functions on closed intervals. Since $p$ is continuous on $[1,4]$, the EVT guarantees absolute extrema exist, and these can occur at endpoints or critical points in the interior. The key insight is that if an absolute extremum occurs at an interior point of the interval where the function is differentiable, then that point must be a critical point (where $p'(x) = 0$ or $p'(x)$ doesn't exist). The corner at $x=2$ represents a point where $p'(2)$ doesn't exist, making it a critical point by definition. Choice A incorrectly assumes the corner must be the absolute maximum without sufficient information. Choice C wrongly excludes endpoints from being absolute extrema. The critical EVT principle: absolute extrema on closed intervals occur at endpoints or critical points in the interior.

This question applies the Extreme Value Theorem to determine what must be true about extrema. Since q is continuous on the closed interval [0,2], the EVT guarantees that q attains both an absolute maximum and minimum somewhere on [0,2]. Looking at the given values, q(1)=1 is the smallest, so the absolute minimum is at most 1, confirming choice C is correct. Choice A is tempting but incorrect because we don't know if q dips below 1 elsewhere in the interval. The EVT principle: on a closed interval, a continuous function must reach its extreme values, though we may need to check endpoints and critical points to find them.

This question applies the Extreme Value Theorem to continuous functions on closed intervals, focusing on how corners (points of non-differentiability) relate to absolute extrema. Since $f$ is continuous on the closed interval $[0,2]$, the EVT unconditionally guarantees that $f$ must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. The sharp corner at $x=1$ represents a point where $f'(1)$ doesn't exist, but this doesn't prevent the existence of absolute extrema - corners can actually be locations where absolute extrema occur. The continuity of $f$ is what matters for EVT, not differentiability. Choice A makes specific claims about the corner's role that aren't guaranteed, while choice B incorrectly suggests corners prevent absolute extrema. The key EVT principle: continuity on closed interval guarantees absolute extrema exist, regardless of corners or non-differentiable points.

This question applies the Extreme Value Theorem to determine what must be true when a continuous function on a closed interval has exactly one critical point in the interior. Since $f$ is continuous on the closed interval $[-1,2]$, the EVT unconditionally guarantees that $f$ must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. The presence of exactly one critical point in $(-1,2)$ provides information about the function's structure but doesn't change the fundamental EVT guarantee. We cannot determine from this information alone whether the critical point corresponds to an absolute maximum, minimum, or neither, as the absolute extrema could occur at endpoints or at the critical point. Choice A incorrectly suggests only one absolute extremum exists. The core EVT truth: continuous function on closed interval always has both absolute maximum and minimum, regardless of the number of critical points.

This question examines how missing an endpoint affects the Extreme Value Theorem's guarantees. The function s is continuous on [-1,1), but this interval is half-open (missing the right endpoint at x=1). Without a closed interval, the EVT doesn't apply, so neither an absolute maximum nor minimum is guaranteed. For example, s(x)=1/(1-x) is continuous on [-1,1) but approaches infinity as x→1⁻, having no maximum. Choice D incorrectly claims continuity alone guarantees extrema, but the EVT specifically requires a closed interval. The critical distinction: continuous on [a,b] → extrema guaranteed; continuous on [a,b) or (a,b] → no guarantees.

This question tests the Extreme Value Theorem's guarantee about absolute extrema on closed intervals. Since r is continuous on the closed interval [2,7], the EVT guarantees that r must attain both an absolute maximum and an absolute minimum somewhere on [2,7]. The existence of exactly one critical point provides information about where extrema might occur, but doesn't change the EVT's guarantee. Choice D incorrectly assumes the critical point must be a minimum, but it could be a maximum or neither. The EVT reminder: continuity on a closed interval always guarantees both absolute extrema exist, regardless of the number or nature of critical points.

This question examines how the Extreme Value Theorem applies when the interval is not closed. The EVT requires a function to be continuous on a closed interval [a,b] to guarantee absolute extrema. Since g is continuous on (0,4] but the interval is half-open (missing the left endpoint), the EVT does not apply. Without the closed interval condition, neither an absolute maximum nor minimum is guaranteed, though they might still exist. Choice A is tempting but incorrect because the missing endpoint means g could approach infinity as x approaches 0. The key insight: EVT requires both continuity AND a closed interval; missing either condition means no guarantees about extrema.

This problem tests the Extreme Value Theorem's application to differentiable functions. Since p is continuous on the closed interval [-1,3], the EVT guarantees that p must attain both a global maximum and a global minimum value on this interval. The additional information that p is differentiable on the open interval (-1,3) doesn't change this guarantee—it just means that any critical points in the interior must satisfy p'(x)=0. Choice A is incorrect because global extrema can also occur at endpoints x=-1 or x=3, where p' might not even be defined. EVT checklist: continuous on closed interval = both global extrema must exist, with candidates at critical points and endpoints.

This question examines the nature of the set of points where a continuous function on a closed interval attains its absolute maximum value. Since $f$ is continuous on the closed interval $[0,3]$, the EVT guarantees that $f$ attains its absolute maximum value at least once. The set of absolute maximizers (points where $f$ achieves its maximum value) is always non-empty for continuous functions on closed intervals - it contains at least one point where the maximum is achieved. This set could contain a single point, multiple discrete points, or even an entire interval if the function is constant at its maximum value over some region. The set cannot be empty because EVT guarantees the maximum is attained. Choice A incorrectly suggests the set could be empty, while choices C, D, and E make incorrect restrictions on the set's composition. The key principle: absolute maximizers set is always non-empty for continuous functions on closed intervals.