AP Calculus AB Quiz: Extreme Value Theorem Extrema Critical Points

What this quiz covers

This quiz focuses on Extreme Value Theorem Extrema Critical Points, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. A local maximum must be greater than all values on $[a,b]$.
2. An absolute minimum is the least value of $f$ on the entire interval, not just nearby. (correct answer)
3. A local minimum can occur only at endpoints.
4. Absolute maxima occur only where $f'(x)$ does not exist.
5. If $f$ has a critical point, it must be an absolute extremum.

Explanation: 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.

Question 2

1. The corner at $x=2$ is an absolute maximum.
2. If $p$ has an absolute extremum in $(1,4)$, then it occurs at a critical point. (correct answer)
3. Endpoints cannot be absolute extrema for $p$.
4. Because $p$ is continuous, it must be differentiable at $x=2$.
5. If $p'(2)$ does not exist, then $p$ has no absolute extrema.

Explanation: 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.

Question 3

1. $q$ has an absolute minimum at $x=1$.
2. $q$ has an absolute maximum at $x=2$.
3. $q$ attains an absolute minimum value somewhere in $[0,2]$. (correct answer)
4. $q$ has a critical point in $(0,2)$.
5. $q$ cannot have any local maxima in $(0,2)$.

Explanation: 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.

Question 4

1. The corner at $x=1$ must be an absolute minimum.
2. Because of the corner, $f$ has no absolute extrema.
3. $f$ attains an absolute maximum and an absolute minimum on $[0,2]$. (correct answer)
4. The corner is not a critical point since $f'(1)$ does not exist.
5. Absolute extrema must occur where $f'(x)$ exists.

Explanation: 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.

Question 5

1. $f$ has exactly one absolute extremum.
2. $f$ has no absolute extrema because there is a critical point.
3. $f$ attains an absolute maximum and an absolute minimum on $[-1,2]$. (correct answer)
4. The critical point must be the absolute maximum.
5. The critical point must be the absolute minimum.

Explanation: 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.

Question 6

1. $s$ must attain an absolute maximum on $[-1,1)$.
2. $s$ must attain an absolute minimum on $[-1,1)$.
3. Neither an absolute maximum nor an absolute minimum is guaranteed on $[-1,1)$. (correct answer)
4. Both absolute extrema are guaranteed because $s$ is continuous.
5. Any absolute maximum must occur at $x=-1$.

Explanation: 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.

Question 7

1. That critical point is a local extremum of $r$.
2. The absolute maximum of $r$ occurs at an endpoint.
3. $r$ attains both an absolute maximum and an absolute minimum on $[2,7]$. (correct answer)
4. The absolute minimum of $r$ occurs at the critical point.
5. $r$ is differentiable at every point of $[2,7]$.

Explanation: 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.

Question 8

1. $g$ must attain an absolute maximum on $(0,4]$.
2. $g$ must attain both an absolute maximum and minimum on $(0,4]$.
3. No absolute extrema can occur because $x=0$ is excluded.
4. Neither an absolute maximum nor an absolute minimum is guaranteed on $(0,4]$. (correct answer)
5. If $g$ has a local minimum, it must be at $x=4$.

Explanation: 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.

Question 9

1. If $p$ has a global maximum, it must occur where $p'(x)=0$.
2. A global minimum cannot occur at an endpoint because $p'(x)$ is undefined there.
3. The global maximum and global minimum values occur only at critical points in $(-1,3)$.
4. Both a global maximum value and a global minimum value of $p$ on $[-1,3]$ must exist. (correct answer)
5. If $p'(x)\neq 0$ for all $x$ in $(-1,3)$, then $p$ has no global extrema.

Explanation: 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.

Question 10

1. The absolute minimum at $x=3$ must be a critical point.
2. An absolute minimum can occur at an endpoint without being a critical point. (correct answer)
3. Absolute minima cannot occur at endpoints.
4. If the minimum is absolute, it must also be a local minimum in the two-sided sense.
5. Endpoints are excluded from EVT, so no absolute minimum exists.

Explanation: This question examines the relationship between absolute minima at endpoints and critical points for continuous functions on closed intervals. When continuous function $p$ on $[0,3]$ has its lowest point (absolute minimum) at the endpoint $x=3$, this absolute extremum occurs at a boundary point of the interval. Importantly, endpoints can be absolute extrema without being critical points, since critical points are defined as interior points where the derivative equals zero or doesn't exist. An endpoint is simply a boundary of the domain and doesn't require any derivative condition to be an absolute extremum. Choice A incorrectly assumes the endpoint must be a critical point, while choices C and D make incorrect claims about endpoint exclusions. The key principle: absolute extrema can occur at endpoints without those endpoints being critical points.

Question 11

1. It is empty unless $f'(x)=0$ somewhere.
2. It contains at least one point in $[0,3]$. (correct answer)
3. It contains only interior points.
4. It contains only endpoints.
5. It contains exactly one point.

Explanation: 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.

Question 12

1. $f$ must attain an absolute maximum and an absolute minimum on $[1,2]$. (correct answer)
2. $f$ attains absolute extrema only if $f'(x)=0$ somewhere.
3. $f$ cannot attain an absolute maximum on a closed interval.
4. Absolute extrema are guaranteed only if $f$ is linear.
5. Continuity implies no critical points, so no absolute extrema.

Explanation: This problem tests the fundamental guarantee of the Extreme Value Theorem for continuous functions on closed intervals. Since $f$ is continuous on the closed interval $[1,2]$, the EVT provides an unconditional guarantee that $f$ must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. This guarantee is independent of whether $f$ has critical points, whether $f'(x)=0$ anywhere, the linearity of $f$, or any other specific properties of the function. The EVT requires only continuity and a closed, bounded interval to guarantee the existence of absolute extrema. Choices B, C, D, and E incorrectly add additional requirements that are not necessary for the EVT guarantee. The core EVT principle: continuous function on closed interval unconditionally guarantees absolute maximum and minimum exist.

Question 13

1. A flat segment guarantees infinitely many critical points, but absolute extrema may still occur at endpoints. (correct answer)
2. A flat segment prevents any absolute maximum from existing.
3. If $g$ is increasing, it cannot have any critical points.
4. Any point on a flat segment must be an absolute minimum.
5. Critical points can occur only at corners, not on flat segments.

Explanation: 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.

Question 14

1. $f$ has no absolute extrema on $[2,7]$.
2. $f$ must be constant on $[2,7]$.
3. Any absolute maximum or absolute minimum must occur at $x=2$ or $x=7$. (correct answer)
4. $f$ must have a local maximum in $(2,7)$.
5. The absence of critical points implies $f$ is discontinuous.

Explanation: 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.

Question 15

1. $p$ has a local maximum at $x=1$.
2. $p$ has either a local maximum or local minimum at $x=1$.
3. $p$ has an absolute extremum at $x=1$.
4. It is possible that $p$ has neither a local maximum nor a local minimum at $x=1$. (correct answer)
5. $p'(1)$ must exist and equal $0$.

Explanation: This question tests understanding of critical points and their relationship to local extrema. A critical point is where f'(c)=0 or f'(c) doesn't exist, but having a critical point doesn't guarantee a local extremum. The function could have an inflection point at x=1 (like y=x³ at x=0), where the derivative equals zero but there's no local max or min. Choice B incorrectly assumes all critical points are local extrema, but critical points can also be inflection points. The key distinction: critical point → possible extremum, but not all critical points are extrema (check second derivative or sign changes in f').

Question 16

1. The absolute maximum and absolute minimum of $f$ on $[-2,5]$ both exist. (correct answer)
2. If $f'(-2)$ and $f'(5)$ exist, then $f$ has no absolute extrema on $[-2,5]$.
3. If $f$ has a local maximum, then it is also the absolute maximum on $[-2,5]$.
4. The absolute maximum of $f$ must occur at an endpoint because the interval is closed.
5. If $f$ has an absolute maximum, then $f'(x)=0$ at that $x$.

Explanation: This question tests understanding of the Extreme Value Theorem (EVT) for finding absolute extrema. The EVT states that if a function is continuous on a closed interval [a,b], then it must attain both an absolute maximum and an absolute minimum on that interval. Since f is continuous on the closed interval [-2,5], the EVT guarantees both absolute extrema exist, making choice A correct. Choice E incorrectly assumes extrema only occur where f'(x)=0, but extrema can also occur at endpoints or where f' doesn't exist. The EVT checklist: (1) Is the function continuous? (2) Is the interval closed? If both yes, then absolute extrema are guaranteed.

Question 17

1. $h$ has a global maximum at $x=1$.
2. $h$ has a local maximum at some $c$ in $(0,2)$. (correct answer)
3. $h$ has no critical points in $(0,2)$.
4. $h$ must be differentiable at $x=1$.
5. The absolute minimum must occur at $x=1$.

Explanation: 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.

Question 18

1. Every critical point is an absolute extremum.
2. Every absolute extremum occurs at a critical point.
3. An absolute extremum occurs either at an endpoint or at a critical point in $(0,4)$. (correct answer)
4. If $f'(x)=0$, then $x$ must be an absolute maximum.
5. Absolute extrema exist only if $f$ is not differentiable.

Explanation: This question examines the relationship between critical points and absolute extrema for continuous functions that are differentiable on the interior of closed intervals. For continuous function $f$ on $[0,4]$ that is differentiable on $(0,4)$, absolute extrema can occur at two types of locations: endpoints ($x=0$ or $x=4$) or interior critical points (where $f'(x)=0$ since $f$ is differentiable throughout the interior). Not every critical point is necessarily an absolute extremum, and not every absolute extremum occurs at a critical point (since endpoints can be absolute extrema without being critical points). However, any absolute extremum that occurs in the interior must happen at a critical point. Choice B incorrectly suggests all absolute extrema occur at critical points, ignoring endpoints. The key principle: absolute extrema occur at endpoints or interior critical points.

Question 19

1. $v$ has an absolute maximum at an interior point of $(-4,4)$.
2. $v$ has an absolute minimum at an interior point of $(-4,4)$.
3. $v$ has at least one absolute maximum and at least one absolute minimum on $[-4,4]$. (correct answer)
4. $v$ must have a critical point in $(-4,4)$.
5. $v$ cannot have any local extrema because the endpoint values are equal.

Explanation: This question applies the Extreme Value Theorem to a function with equal endpoint values. Since v is continuous on the closed interval [-4,4], the EVT guarantees that v must attain both an absolute maximum and an absolute minimum somewhere on [-4,4]. The fact that v(-4)=v(4)=2 doesn't prevent extrema from existing; v could rise above 2 or dip below 2 in the interior. Choice D incorrectly assumes equal endpoints require an interior critical point, but v could be constant at 2 throughout. The EVT guarantee: continuous function + closed interval = both absolute extrema exist, regardless of endpoint values or behavior.

Question 20

1. $u$ is constant on $[a,b]$.
2. $u$ has no absolute extrema on $[a,b]$.
3. Any absolute maximum or minimum of $u$ on $[a,b]$ occurs at $x=a$ or $x=b$. (correct answer)
4. $u$ must have at least one local maximum in $(a,b)$.
5. $u'(x)$ exists for all $x\in(a,b)$.

Explanation: This question explores where absolute extrema must occur when there are no interior critical points. If u has no critical points in (a,b), then u has no points where u'(x)=0 or u'(x) doesn't exist in the open interval. By the EVT, u (being continuous on [a,b]) must have absolute extrema, and since these can only occur at critical points or endpoints, they must occur at x=a or x=b. Choice A incorrectly concludes u must be constant, but u could be strictly increasing or decreasing. The extrema location principle: absolute extrema occur at critical points or endpoints; no interior critical points means extrema must be at endpoints.