This problem tests understanding of the Extreme Value Theorem applied to a continuous function on a closed interval. Since $g$ is continuous on the closed interval $[0,6]$, the EVT guarantees that $g$ must attain both an absolute maximum value and an absolute minimum value somewhere within this interval. The given function values help us see the function's behavior but don't determine the specific locations of the absolute extrema without additional information about the function between these points. We cannot conclude that any particular point is a critical point or that specific values represent absolute extrema based solely on the discrete values shown. Choice E incorrectly suggests that non-differentiability prevents absolute extrema, but continuity alone is sufficient for EVT. The key EVT principle: continuous function + closed interval = guaranteed absolute maximum and minimum exist somewhere.

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 examines how the absence of critical points in the interior affects the location of absolute extrema for continuous functions on closed intervals. Since $f$ is continuous on $[2,7]$ and has no critical points in $(2,7)$, this means $f'(x)

eq 0$ throughout the interior (where the derivative exists), so the function is strictly monotonic on the interior. The EVT still guarantees that absolute extrema exist, and since there are no critical points in the interior, any absolute maximum or minimum must occur at the endpoints $x=2$ or $x=7$. The function cannot be constant because that would create critical points (where $f'(x)=0$). Choice A incorrectly denies the existence of absolute extrema, contradicting EVT. The key insight: no interior critical points forces absolute extrema to occur at endpoints of the closed interval.

This problem analyzes absolute extrema when a continuous function on a closed interval has equal endpoint values and a single local maximum in the interior. Since $h$ is continuous on $[0,5]$ with $h(0)=h(5)=1$ and has a single local maximum in $(0,5)$, we can determine the absolute extrema locations. The local maximum in the interior must be the absolute maximum because it's higher than the equal endpoint values, and since there are no other local extrema, the minimum value must occur at the endpoints. The EVT guarantees both absolute maximum and minimum exist, with the interior local maximum being absolute and the endpoints sharing the absolute minimum value. Choice B incorrectly suggests that equal endpoints prevent an absolute minimum. The principle: when endpoints are equal and there's a single interior local maximum, that maximum is absolute and the endpoints provide the absolute minimum.

This question applies the Extreme Value Theorem to determine what must be guaranteed for a continuous function on a closed interval with specific endpoint values. Since $g$ is continuous on the closed interval $[-4,2]$, the EVT unconditionally guarantees that $g$ must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. The given endpoint values $g(-4)=0$ and $g(2)=5$ provide information about the function but don't determine where the absolute extrema occur, as the maximum or minimum could occur at interior points. We cannot conclude specific extrema locations without knowing the function's behavior throughout the entire interval. Choice A incorrectly assumes the minimum occurs at $x=-4$, and choice C incorrectly assumes the maximum occurs at $x=2$. The fundamental EVT guarantee: continuous function on closed interval ensures both absolute maximum and minimum exist somewhere.

This problem analyzes the conditions under which an interior point with a global minimum must be a critical point. Since $g$ is continuous on $[1,5]$ and has a global minimum at $x=3$, we need to determine when this interior point must be a critical point. If $x=3$ lies in the interior $(1,5)$ and $g$ is differentiable at $x=3$, then by Fermat's theorem, we must have $g'(3)=0$, making it a critical point. However, if $g$ is not differentiable at $x=3$, then $x=3$ is still a critical point because critical points include both points where the derivative equals zero and points where the derivative doesn't exist. Choice C omits the differentiability condition, while choice A incorrectly assumes all global minima are critical points regardless of location. The precise principle: interior absolute extrema of continuous functions are critical points (either $f'(x)=0$ or $f'(x)$ doesn't exist).

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 problem tests the fundamental guarantee of the Extreme Value Theorem for any continuous function on a closed interval. Since $h$ is continuous on the closed interval $[2,9]$, the EVT provides an unconditional guarantee that $h$ must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. This guarantee holds regardless of the specific formula for $h$, the presence or absence of critical points, or any other characteristics of the function - only continuity on the closed interval is required. The other choices make specific claims about critical points, extrema locations, or function behavior that are not guaranteed by EVT alone. Choice A incorrectly suggests critical points are required, while choices C and D incorrectly restrict extrema to endpoints. The fundamental EVT truth: continuous function on closed interval always has absolute maximum and minimum values.

This problem analyzes the relationship between local minima in the interior of closed intervals and critical points. When continuous function $g$ on $[-2,2]$ has a local minimum at $x=0$, which is an interior point of the interval, this point must be a critical point by definition. A critical point is either a point where the derivative equals zero or where the derivative doesn't exist. Since $x=0$ is a local minimum in the interior, if $g$ is differentiable at $x=0$, then $g'(0)=0$ by Fermat's theorem; if $g$ is not differentiable at $x=0$, then $g'(0)$ doesn't exist. Either way, $x=0$ satisfies the definition of a critical point. Choice A makes a specific claim about absolute minima that isn't necessarily true. The key principle: interior local extrema of continuous functions are always critical points.

This problem applies the Extreme Value Theorem to continuous functions on closed intervals when given specific endpoint conditions. Since $f$ is continuous on the closed interval $[0,1]$, the EVT unconditionally guarantees that $f$ must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. The condition that $f(0)=f(1)$ (equal endpoint values) provides additional information about the function but doesn't change the fundamental EVT guarantee. Equal endpoint values don't prevent the existence of absolute extrema - they simply mean the endpoints have the same function value. The absolute extrema could occur at the endpoints, in the interior, or both, depending on the function's behavior. Choice E incorrectly suggests equal endpoint values prevent absolute extrema. The key EVT truth: continuous function on closed interval always has absolute maximum and minimum, regardless of endpoint value relationships.