Extreme Value Theorem, Extrema, Critical Points - AP Calculus AB

1. Let $h$ be continuous on $[2,9]$; which statement is always true regardless of the formula for $h$?
   Answer: $h$ attains both an absolute maximum and an absolute minimum on $[2,9]$. 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.
2. A continuous function $t$ on $[0,3]$ has a local maximum at $x=1$. Which statement must be true?
   Answer: $t$ has an absolute maximum value somewhere on $[0,3]$. This question connects local extrema with the Extreme Value Theorem's guarantee of absolute extrema. Since t is continuous on the closed interval [0,3], the EVT guarantees that t must attain an absolute maximum value somewhere on [0,3]. The local maximum at x=1 might or might not be this absolute maximum (we'd need to compare t(1) with values at endpoints and other critical points). Choice E incorrectly assumes a local maximum must be the absolute maximum, but the absolute max could occur elsewhere. The EVT principle: continuity on a closed interval guarantees absolute extrema exist, though local extrema are just candidates for these absolute extrema.
3. Continuous $g$ on $[-2,2]$ has a local minimum at $x=0$; which statement must be true?
   Answer: $x=0$ is a critical point of $g$ (i.e., $g'(0)=0$ or $g'(0)$ does not exist). 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.
4. A continuous function $h$ on $[0,2]$ has $h(0)<h(1)$ and $h(2)<h(1)$; which statement must be true?
   Answer: $h$ has a local maximum at some $c$ in $(0,2)$. This question analyzes what must be true when a continuous function on a closed interval has a specific pattern of values that creates an interior peak. Since $h$ is continuous on $[0,2]$ with $h(0)<h(1)$ and $h(2)<h(1)$, the value at $x=1$ is greater than both endpoint values. By the continuity of $h$ and the intermediate value theorem, since $h$ increases from $h(0)$ to $h(1)$ and then decreases from $h(1)$ to $h(2)$, there must be at least one point in $(0,2)$ where $h$ attains a local maximum. This could be exactly at $x=1$ or at some other interior point, but the given inequalities guarantee the existence of an interior local maximum. Choice A assumes the global maximum is at $x=1$, which isn't necessarily true. The key principle: when interior values exceed both endpoints, continuity guarantees an interior local maximum exists.
5. Let $f$ be continuous on $[0,1]$ and satisfy $f(0)=f(1)$; which statement must be true?
   Answer: $f$ must have at least one absolute maximum and at least one absolute minimum on $[0,1]$. 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.
6. A continuous function $g$ on $[-4,2]$ has $g(-4)=0$ and $g(2)=5$; which statement must be true?
   Answer: $g$ must have at least one absolute maximum and at least one absolute minimum on $[-4,2]$. 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.
7. Let $q$ be continuous on $[-3,1]$ with a sharp corner at $x=-1$; which statement is necessarily true?​
   Answer: $x=-1$ is a critical point of $q$ because $q'$ does not exist there. This problem tests understanding of critical points at non-differentiable locations. A critical point is defined as a point in the domain where either the derivative equals zero or the derivative does not exist. Since q has a sharp corner at x=-1, the derivative q'(-1) does not exist, making x=-1 a critical point by definition. The continuity of q on [-3,1] combined with the EVT still guarantees that both global extrema exist on this interval, despite the corner. Choice C incorrectly claims the corner prevents a global maximum, but non-differentiable points are valid locations for extrema. Critical point checklist: check where f'(x)=0 OR where f'(x) doesn't exist (corners, cusps, vertical tangents).
8. The graph of continuous $p$ on $[1,6]$ has a highest point at $x=6$; which statement is necessarily true?
   Answer: The absolute maximum can occur at an endpoint. This problem examines where absolute extrema can occur when a continuous function on a closed interval has its highest point at an endpoint. When continuous function $p$ on $[1,6]$ has its highest point at $x=6$, this means the absolute maximum occurs at the endpoint $x=6$. This demonstrates that absolute extrema can indeed occur at endpoints of closed intervals, which is a key principle of the EVT. The absolute maximum doesn't need to occur at a critical point since it's at an endpoint, and endpoints are not required to be critical points. Choice A incorrectly assumes we know $x=6$ is a critical point, while choice C incorrectly restricts absolute maxima to interior critical points. Choice D incorrectly excludes endpoints from being absolute extrema locations. The important principle: absolute extrema can occur at endpoints without those endpoints being critical points.
9. A function $g$ is continuous on $[1,3]$ and has a local maximum at $x=2$; which statement must be true?
   Answer: $x=2$ is a critical point of $g$. This question examines the relationship between local maxima in the interior of closed intervals and critical points. When function $g$ is continuous on $[1,3]$ and has a local maximum at $x=2$, which is an interior point, this point must be a critical point by the definition of critical points and Fermat's theorem. At an interior local maximum, either the derivative equals zero (if the function is differentiable there) or the derivative doesn't exist (if there's a corner or cusp). Either case makes $x=2$ a critical point. We cannot conclude that this local maximum is also the absolute maximum without additional information, as the absolute extrema could occur at endpoints or other interior points. Choice A makes assumptions about function values, while choice C incorrectly assumes the local maximum is absolute. The fundamental principle: interior local extrema are always critical points.
10. A continuous function $p$ on $[1,4]$ has a local maximum at $x=1$ (endpoint) and no other local extrema; which statement must be true?
    Answer: $p$ attains an absolute maximum and an absolute minimum on $[1,4]$. This problem applies the Extreme Value Theorem to continuous functions on closed intervals when given information about local extrema at endpoints. Since $p$ is continuous on the closed interval $[1,4]$, the EVT unconditionally guarantees that $p$ must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. The fact that $p$ has a local maximum at the endpoint $x=1$ and no other local extrema provides information about the function's behavior but doesn't change the fundamental EVT guarantee. A local maximum at an endpoint means the function value at $x=1$ is at least as large as nearby values, and since there are no other local extrema, the function exhibits monotonic behavior in the interior. Choice A makes claims about which specific extremum occurs where, while choice C incorrectly suggests EVT doesn't apply. The core EVT principle: continuous function on closed interval always has absolute maximum and minimum values.
11. Let $q$ be continuous on $[0,2]$ with $q(0)=3$, $q(1)=1$, and $q(2)=4$. Which statement must be true?
    Answer: $q$ attains an absolute minimum value somewhere in $[0,2]$. 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.
12. A function $g$ is continuous on $[0,2]$ and differentiable on $(0,2)$; which statement must be true if $g$ has an absolute maximum in $(0,2)$?
    Answer: That point must satisfy $g'(x)=0$. This problem examines where absolute maxima can occur in the interior of closed intervals for continuous, differentiable functions. When function $g$ is continuous on $[0,2]$ and differentiable on $(0,2)$, and if $g$ has an absolute maximum at some interior point, then by Fermat's theorem, that point must satisfy $g'(x)=0$. This is because at an interior point where a differentiable function attains its maximum, the derivative must equal zero (the tangent line must be horizontal). This doesn't apply to endpoints, which can be absolute extrema without being critical points. Choice B incorrectly suggests absolute maxima must occur at endpoints, while choice C incorrectly suggests they must occur where the derivative doesn't exist. The key principle: interior absolute extrema of differentiable functions occur where the derivative equals zero.
13. If $f$ is continuous on $[-2,5]$, which statement must be true about global extrema on this interval?​
    Answer: The global maximum and global minimum values of $f$ on $[-2,5]$ both exist. This question tests understanding of the Extreme Value Theorem (EVT) and how it guarantees global extrema. The EVT states that if a function is continuous on a closed interval [a,b], then it must attain both a global maximum and a global minimum value on that interval. Since f is continuous on the closed interval [-2,5], the EVT directly applies, guaranteeing that both global extrema exist. Choice A is incorrect because endpoints are valid locations for global extrema even though they're not critical points in the traditional sense. Choice D fails because global extrema can occur at endpoints or where f'(x) doesn't exist, not just where f'(x)=0. EVT checklist: continuous function + closed interval = guaranteed global max and min.
14. Let $g$ be continuous on $[1,5]$ and have a global minimum at $x=3$; which statement must be true?
    Answer: If $x=3$ lies in $(1,5)$ and $g$ is differentiable at $3$, then $g'(3)=0$. 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).
15. A continuous function $p$ on $[1,4]$ has a corner at $x=2$ and no other non-smooth points; which statement must be true?
    Answer: If $p$ has an absolute extremum in $(1,4)$, then it occurs at a critical point. 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.
16. A continuous function $r$ on $[0,10]$ has a local maximum at $x=4$; which statement must be true?​
    Answer: $x=4$ must be a critical point of $r$ (either $r'(4)=0$ or $r'(4)$ does not exist). This question tests the relationship between local extrema and critical points. If r has a local maximum at x=4, and x=4 is in the interior of the domain [0,10], then x=4 must be a critical point—meaning either r'(4)=0 or r'(4) does not exist. This is a fundamental theorem: local extrema at interior points must be critical points. Choice A is tempting but incorrect because a local maximum isn't necessarily the global maximum—there could be higher values elsewhere on [0,10]. EVT principle for local extrema: interior local max/min → critical point (derivative = 0 or DNE).
17. A function $g$ is continuous on $[0,1]$ and has no local extrema in $(0,1)$; which statement must be true?
    Answer: Any absolute maximum and absolute minimum occur at $x=0$ or $x=1$. This question examines where absolute extrema must occur when a continuous function has no local extrema in the interior of a closed interval. Since $g$ is continuous on $[0,1]$ and has no local extrema in $(0,1)$, this means the function is strictly monotonic on the interior - either strictly increasing or strictly decreasing from one endpoint to the other. The EVT guarantees that absolute extrema exist, and since there are no interior local extrema (which would be candidate points for absolute extrema), both the absolute maximum and absolute minimum must occur at the endpoints $x=0$ or $x=1$. The function cannot have critical points in the interior since local extrema occur at critical points. Choice A incorrectly denies the existence of absolute extrema, contradicting EVT. The key principle: no interior local extrema forces absolute extrema to occur at endpoints.
18. Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$; which statement correctly describes where absolute extrema can occur?
    Answer: At endpoints or at critical points in $(a,b)$. This problem examines where absolute extrema can occur for continuous functions that are differentiable on the interior of closed intervals. For a continuous function $f$ on $[a,b]$ that is differentiable on $(a,b)$, absolute extrema can occur at two types of locations: the endpoints $a$ and $b$, or at critical points in the interior $(a,b)$ where $f'(x)=0$ or $f'(x)$ doesn't exist. Since $f$ is differentiable throughout $(a,b)$, interior critical points must satisfy $f'(x)=0$. The EVT guarantees absolute extrema exist, and they must occur at these candidate points. Choice A incorrectly restricts to only points where $f'(x)=0$, ignoring endpoints. Choice B incorrectly restricts to only endpoints. The complete principle: absolute extrema occur at endpoints or interior critical points (where derivative equals zero or doesn't exist).
19. Let $f$ be continuous on $[-4,4]$ and suppose $f$ has no critical points in $(-4,4)$; which statement is true?
    Answer: Any global maximum or global minimum of $f$ on $[-4,4]$ must occur at $x=-4$ or $x=4$. This question tests understanding of where global extrema must occur when there are no interior critical points. Since f is continuous on [-4,4] with no critical points in the open interval (-4,4), any global extrema must occur at the endpoints. This is because global extrema can only occur at critical points or endpoints, and we're told there are no critical points in the interior. The EVT still guarantees that both global extrema exist since f is continuous on the closed interval. Choice A incorrectly claims no extrema exist, but EVT guarantees they do—they just must be at the endpoints. Extrema location principle: global max/min occur at critical points or endpoints; no interior critical points means extrema are at endpoints.
20. Let $g$ be continuous on $[0,5]$ and shown increasing everywhere except a flat segment; which statement about critical points is true?
    Answer: A flat segment guarantees infinitely many critical points, but absolute extrema may still occur at endpoints. This question examines how flat segments in continuous functions relate to critical points and absolute extrema on closed intervals. When continuous function $g$ on $[0,5]$ is increasing everywhere except for a flat segment, the flat segment indeed contains infinitely many critical points because $g'(x)=0$ at every point in the flat region. However, this doesn't prevent absolute extrema from existing - the EVT still guarantees that $g$ attains absolute maximum and minimum values, and these could occur at endpoints, on the flat segment, or at other critical points. The presence of many critical points doesn't affect the EVT guarantee, and absolute extrema can certainly occur at endpoints regardless of interior critical point behavior. Choice B incorrectly suggests flat segments prevent absolute maxima, while other choices make various incorrect claims about critical points and extrema. The key principle: flat segments create multiple critical points but don't prevent absolute extrema from existing, including at endpoints.

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