AP Calculus BC 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 BC.

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. $f$ has a local extremum at $c$.
2. $f$ has exactly one absolute extremum on $[0,3]$.
3. $f$ attains both an absolute maximum and an absolute minimum on $[0,3]$. (correct answer)
4. Both absolute extrema must occur at $c$.
5. Neither absolute extremum can occur at an endpoint.

Explanation: 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.

Question 2

1. $s$ has an absolute maximum at $x=5$.
2. $s$ has a local minimum at $x=5$, but it may not be absolute.
3. $s$ has an absolute minimum at $x=5$. (correct answer)
4. $s$ has no extrema because $s'(x)$ is never $0$ on $(2,8)$.
5. $s$ must be constant on $[2,8]$ since the derivative changes sign.

Explanation: 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.

Question 3

1. $r$ has a local maximum at $x=0$.
2. $r$ has an absolute minimum at both endpoints $x=0$ and $x=4$. (correct answer)
3. $r$ has no absolute minimum because it is never less than $2$.
4. $r$ must have a critical point at $x=0$.
5. $r$ must have an absolute maximum at an endpoint.

Explanation: 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.

Question 4

1. $f$ must have an absolute maximum value $5$ and an absolute minimum value $1$.
2. $f$ must have an absolute maximum at $x=1$ and an absolute minimum at $x=2$.
3. $f$ cannot have absolute extrema without knowing $f'(x)$.
4. $f$ has no absolute extrema because the given data are only at five points.
5. EVT guarantees absolute extrema, but their $x$-locations cannot be determined from the data. (correct answer)

Explanation: 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.

Question 5

1. $f$ has both an absolute maximum and an absolute minimum on $[-2,3]$. (correct answer)
2. $f$ has an absolute maximum at $x=-1$ because $f'(-1)=0$.
3. $f$ has no absolute extrema because $f'(1)$ is undefined.
4. $f$ has an absolute minimum at an interior critical point, not at an endpoint.
5. $f$ has exactly two critical points, so it has exactly two absolute extrema.

Explanation: This question tests the Extreme Value Theorem (EVT), which guarantees absolute extrema for continuous functions on closed intervals. Since f is continuous on the closed interval [-2,3], EVT ensures that f must attain both an absolute maximum and an absolute minimum somewhere on this interval. The critical points at x=-1 (where f'(-1)=0) and x=1 (where f'(1) is undefined) are candidates for these extrema, along with the endpoints x=-2 and x=3. Option B incorrectly assumes the absolute maximum must occur at x=-1 just because f'(-1)=0, but we'd need to compare values at all critical points and endpoints. Remember the EVT checklist: continuous function + closed interval = guaranteed absolute max and min.

Question 6

1. $f(2)$ is the absolute maximum value of $f$ on $[0,6]$.
2. $f(5)$ is the absolute minimum value of $f$ on $[0,6]$.
3. The only candidates for absolute extrema are $x=2$ and $x=5$.
4. $f$ must have both an absolute maximum value and an absolute minimum value on $[0,6]$. (correct answer)
5. Endpoints cannot be where absolute extrema occur for a continuous function.

Explanation: This question examines the Extreme Value Theorem (EVT) and distinguishing local from absolute extrema. Continuity of f on [0,6] ensures, by EVT, both an absolute maximum and an absolute minimum exist on the interval. Local extrema at x=2 (max) and x=5 (min) are critical points, but absolute extrema require comparing values at these, plus endpoints x=0 and x=6. The graph might show absolute max or min at endpoints if they exceed local ones. Choice A tempts by equating local max to absolute max, but fails as an endpoint could be higher. EVT checklist: check continuity on closed interval, find critical points for local extrema, evaluate at critical points and endpoints, and select the global max/min from those values.

Question 7

1. $r$ has both an absolute maximum value and an absolute minimum value on $[-3,3]$. (correct answer)
2. $r$ has an absolute maximum at $x=0$.
3. $r$ has an absolute minimum at $x=0$.
4. $r$ has no absolute extrema because it has only one critical point.
5. All absolute extrema must occur at critical points, not endpoints.

Explanation: 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.

Question 8

1. If $p'(x)$ exists for all $x$ in $(1,5)$, $p$ has no absolute extrema on $[1,5]$.
2. Any absolute maximum of $p$ must occur at a point where $p'(x)=0$.
3. If $p$ has an absolute minimum, it must occur at $x=1$.
4. $p$ has both an absolute maximum value and an absolute minimum value on $[1,5]$. (correct answer)
5. If $p'(c)=0$, then $p(c)$ is an absolute extremum value.

Explanation: 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.

Question 9

1. $g$ has an absolute maximum value on $[0,4]$ and an absolute minimum value on $[0,4]$. (correct answer)
2. The absolute maximum value of $g$ must occur at $x=2$.
3. The absolute minimum value of $g$ must occur at $x=2$.
4. Because $g(0)=g(4)$, $g$ has no absolute extrema on $[0,4]$.
5. Since $g'(2)=0$, $x=2$ is an absolute extremum of $g$.

Explanation: 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.

Question 10

1. $m$ has no absolute extrema on $[2,7]$.
2. $m$ has at least one critical point in $(2,7)$.
3. $m$ attains an absolute maximum and an absolute minimum, both at endpoints. (correct answer)
4. $m$ must have a local maximum in $(2,7)$.
5. $m$ is constant on $[2,7]$.

Explanation: 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.

Question 11

1. $f$ has no critical points because $f'(x)$ exists for all $x$.
2. $f$ has a global maximum at $x=1$.
3. $f$ has a global minimum at $x=1$, and $x=1$ is a critical point. (correct answer)
4. $f$ has a local maximum at $x=1$ but no global minimum.
5. Endpoints cannot be candidates for global extrema for $f$.

Explanation: 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.

Question 12

1. Both absolute extrema must occur at $x=0$ because it is the only critical point.
2. Any absolute extremum of $r$ on $[-1,1]$ occurs at $x=-1,0,$ or $1$. (correct answer)
3. $r$ has no absolute extrema since $r'(x)$ is $0$ only once.
4. If $r$ has an absolute maximum, it cannot occur at an endpoint.
5. $r$ must have a local maximum at $x=0$ by Fermat’s Theorem.

Explanation: This question tests understanding of where absolute extrema can occur for differentiable functions. Since r is continuous on [-1,1] and differentiable on (-1,1) with only one critical point at x=0, any absolute extrema must occur at either the critical point (x=0) or the endpoints (x=-1 or x=1). This is because extrema can only occur where f'(x)=0, where f'(x) doesn't exist, or at endpoints of the domain. Option A incorrectly claims both extrema must be at x=0, but one or both could occur at endpoints even if they're not critical points. The extrema location rule: check critical points and endpoints only.

Question 13

1. $m$ has an absolute maximum at $x=1$.
2. $m'(1)=0$.
3. $m$ has both an absolute maximum and an absolute minimum on $[0,2]$. (correct answer)
4. $m$ has no absolute minimum because local maxima prevent minima.
5. The absolute minimum must occur at $x=1$ since it is a local maximum.

Explanation: 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.

Question 14

1. $f$ attains both an absolute maximum and an absolute minimum on $[-2,3]$. (correct answer)
2. The absolute maximum value of $f$ occurs at $x=-1$ or $x=2$.
3. The points $x=-1$ and $x=2$ are absolute extrema of $f$ on $[-2,3]$.
4. If $f'(-1)=0$, then $f$ has a local maximum at $x=-1$.
5. Because $f'$ exists at $x=-2$ and $x=3$, the endpoints cannot be absolute extrema.

Explanation: This question tests understanding of the Extreme Value Theorem (EVT) and how critical points relate to absolute extrema. Since f is continuous on the closed interval [-2,3], EVT guarantees that f attains both an absolute maximum and an absolute minimum somewhere on this interval. These extrema can occur at critical points (where f'(x)=0, like x=-1 and x=2) or at endpoints (x=-2 and x=3). Choice B incorrectly assumes extrema must occur at critical points, but they could occur at endpoints instead. Choice E is false because endpoints can absolutely be extrema regardless of whether f' exists there. The EVT checklist: (1) Is the function continuous? (2) Is the interval closed? If both yes, then absolute extrema exist.

Question 15

1. $q$ has an absolute maximum at $x=1$ because $q'(1)=0$.
2. $q$ has a local maximum at $x=-1$ and a local minimum at $x=1$. (correct answer)
3. $q$ has no absolute extrema on $[-2,2]$ because it is a cubic.
4. All local extrema of $q$ must occur at the endpoints $x=-2$ and $x=2$.
5. Since $q$ is differentiable, its absolute maximum cannot occur at an endpoint.

Explanation: This question distinguishes between local and global (absolute) extrema for a specific polynomial function. For q(x)=x³-3x, we find q'(x)=3x²-3=3(x²-1), which equals zero at x=±1. Using the second derivative test: q''(x)=6x, so q''(-1)=-6<0 (local max) and q''(1)=6>0 (local min). Evaluating: q(-1)=2, q(1)=-2, q(-2)=-2, q(2)=2. The absolute maximum is 2 (at x=±2 and x=-1) and absolute minimum is -2 (at x=±1 and x=2). Choice A incorrectly identifies x=1 as a maximum when it's actually a minimum. Remember: local extrema are determined by derivative tests, while absolute extrema require comparing all critical points and endpoints.

Question 16

1. The absolute maximum must occur at one of $x=-2,0,2$.
2. $f$ must have exactly three local extrema at $x=-2,0,2$.
3. Any absolute extremum of $f$ on $[-3,3]$ occurs at $x=-3$, $x=3$, or at a critical point. (correct answer)
4. Because $f'$ is zero at three points, endpoints cannot be absolute extrema.
5. If $f'(0)=0$, then $f(0)$ is the absolute minimum value on $[-3,3]$.

Explanation: This question tests comprehensive understanding of where absolute extrema can occur when multiple critical points exist. Since f is continuous on [-3,3], EVT guarantees absolute extrema exist. These extrema must occur either at critical points (where f'(x)=0, namely x=-2,0,2) or at endpoints (x=±3). No other locations are possible. Choice A incorrectly limits extrema to only critical points. Choice B wrongly assumes all critical points must be local extrema—some could be inflection points. Choice D falsely claims endpoints can't be extrema when critical points exist. The complete extrema-finding algorithm: identify all critical points and endpoints, evaluate the function at each, then compare to find absolute extrema.

Question 17

1. If $f$ has a local maximum, then $f'(x)=0$ at that $x$.
2. $f$ has at least one critical point in $(1,5)$.
3. $f$ attains an absolute maximum and an absolute minimum on $[1,5]$. (correct answer)
4. The absolute maximum of $f$ cannot occur at $x=1$ or $x=5$.
5. If $f'(x)=0$ at some $x$, then $f$ has an absolute extremum there.

Explanation: This question tests direct application of the Extreme Value Theorem without specific function details. Since f is continuous on the closed interval [1,5], EVT guarantees that f attains both an absolute maximum and an absolute minimum somewhere on this interval. These extrema could occur at endpoints or at interior critical points. Choice A is false because local maxima can occur where f' doesn't exist. Choice B isn't guaranteed—f could be monotonic with no interior critical points. Choice D is false because endpoints can certainly be locations of absolute extrema. The key insight: continuity on a closed interval is sufficient for EVT to guarantee existence of absolute extrema.

Question 18

1. $g$ has no absolute minimum because $g'(0)=0$.
2. $g$ has an absolute maximum at $x=0$ because $g'(0)=0$.
3. $g$ has an absolute minimum at $x=0$ and an absolute maximum at $x=\pm2$. (correct answer)
4. $g$ has absolute extrema only at critical points, not at endpoints.
5. $g$ is not continuous on $[-2,2]$, so EVT does not apply.

Explanation: This question requires applying EVT to find extrema of a specific function using critical points and endpoint analysis. The function g(x)=x²/(x²+1) is continuous everywhere (denominator never zero), so it's continuous on [-2,2]. To find extrema, we check critical points: g'(x)=2x/(x²+1)², which equals zero only at x=0. Evaluating: g(0)=0, g(±2)=4/5. Since 0<4/5, the absolute minimum is 0 at x=0, and the absolute maximum is 4/5 at both x=-2 and x=2. Choice A incorrectly claims no minimum exists because of the critical point. The EVT process: find critical points, evaluate function at critical points and endpoints, then compare values.

Question 19

1. Any absolute extremum must occur at a critical point in $(-1,2)$.
2. $h$ has no absolute extrema unless $h'(x)=0$ somewhere.
3. $h$ must attain an absolute maximum and an absolute minimum on $[-1,2]$. (correct answer)
4. If $h$ is increasing on $(-1,2)$, then $h$ has no absolute minimum.
5. If $h$ has a corner in $(-1,2)$, then $h$ is not continuous on $[-1,2]$.

Explanation: This question assesses the application of the Extreme Value Theorem (EVT) to functions with specified continuity and differentiability, focusing on absolute extrema. Since h is continuous on the closed interval [-1,2], the EVT ensures that h attains both an absolute maximum and an absolute minimum on that interval. Differentiability on (-1,2) means critical points where h' = 0 or undefined in the open interval are candidates for extrema, but absolute extrema may also be at the endpoints. The key is that continuity alone guarantees the existence of these absolute extrema, irrespective of the behavior in the open interval. A tempting distractor is choice A, which claims any absolute extremum must occur at a critical point in (-1,2), but this fails because extrema can occur at endpoints without critical points nearby. For using the EVT, confirm the function is continuous on a closed, bounded interval; if yes, find critical points in the interior and compare function values at those points and the endpoints.

Question 20

1. $p$ must attain an absolute maximum on $[0,4)$ by EVT.
2. $p$ must attain an absolute minimum on $[0,4)$ by EVT.
3. $p$ may fail to attain an absolute maximum on $[0,4)$ because the interval is not closed. (correct answer)
4. $p$ attains an absolute maximum at $x=4$ since the limit equals 10.
5. Because $p(x)<10$ for $x<4$, $p$ has no absolute minimum on $[0,4)$.

Explanation: This question explores what happens when EVT conditions aren't met, specifically when the interval isn't closed. The function p is continuous on [0,4) but this is not a closed interval—it doesn't include the right endpoint. Since p(x)<10 for all x<4 and approaches 10 as x→4⁻, the function gets arbitrarily close to 10 but never reaches it, meaning there's no absolute maximum on [0,4). EVT doesn't apply because the interval isn't closed. Choice A incorrectly applies EVT to a non-closed interval. The critical distinction: EVT requires both continuity AND a closed interval; missing either condition means extrema aren't guaranteed.