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

Let $f$ be continuous on $[-1,2]$ and differentiable on $(-1,2)$ with exactly one critical point $c\in(-1,2)$ . Which is true?
Answer: Any absolute extremum of $f$ on $[-1,2]$ occurs at $x=-1$ , $x=2$ , or $x=c$ . This question tests understanding that absolute extrema must occur at critical points or endpoints, with no other possibilities. Since f is continuous on [-1,2], EVT guarantees absolute extrema exist. The only places these can occur are: the endpoints x=-1 and x=2, or the critical point x=c. There are no other candidates. Choice A incorrectly assumes the extremum must be at the critical point, but it could be at an endpoint. Choice B makes the same error for minima. Choice D is false—a function with one critical point could still be monotonic if that critical point is an inflection point. The fundamental principle: absolute extrema occur only at critical points or interval endpoints.
Let $h(x)=|x-1|$ on $[0,3]$ . Which statement about critical points or extrema is true?
Answer: $h$ has an absolute minimum at $x=1$ , where $h'$ does not exist. This question tests understanding that critical points include places where the derivative doesn't exist, not just where it equals zero. The function h(x)=|x-1| is continuous on [0,3] but not differentiable at x=1 (sharp corner). At x=1, we have h(1)=0, while h(0)=1 and h(3)=2, making x=1 the location of the absolute minimum. The derivative doesn't exist at x=1, but this point is still a critical point and can be an extremum. Choice D incorrectly suggests EVT doesn't apply to non-differentiable functions—EVT only requires continuity. Remember: critical points occur where f'(x)=0 OR where f'(x) doesn't exist.
A function $f$ is continuous on $[1,5]$ and differentiable on $(1,5)$ . Which statement must be true?
Answer: $f$ attains an absolute maximum and an absolute minimum on $[1,5]$ . 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.
A continuous function $g$ is defined on $[0,4]$ and satisfies $g'(2)=0$ and $g(0)=g(4)$ . Which is true?
Answer: $g$ has an absolute maximum value on $[0,4]$ and an absolute minimum value on $[0,4]$ . 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.
Let $f(x)=|x|$ on $[-1,2]$ . Which statement about extrema and critical points is true?
Answer: $f$ has an absolute minimum at $x=0$ , and $x=0$ is a critical point. This question explores the Extreme Value Theorem (EVT), extrema, and critical points for non-differentiable functions. For f(x)=|x| continuous on [-1,2], EVT guarantees absolute extrema, with the minimum at x=0 where f(0)=0 is the smallest value. The point x=0 is a critical point because f'(x) is undefined there, despite not being zero, fitting the definition of critical points as places where the derivative is zero or does not exist. Evaluating at endpoints f(-1)=1, f(2)=2, and the critical point confirms the minimum at x=0 and maximum at x=2. Choice A tempts by claiming no critical points due to undefined derivative, but it fails as undefined derivatives precisely define critical points for potential extrema. Use this EVT checklist: check continuity on closed interval, find critical points (f'=0 or undefined), evaluate at those and endpoints, and identify max/min from values.
A function $p$ is continuous on $[1,5]$ and differentiable on $(1,5)$ . Which statement is true?
Answer: $p$ has both an absolute maximum value and an absolute minimum value on $[1,5]$ . 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.
Let $f(x)=|x|$ on $[-2,1]$ . Which statement about extrema or critical points is true?
Answer: $f$ has an absolute minimum at $x=0$ , and $x=0$ is a critical point. This question examines extrema and critical points for the absolute value function. The function f(x)=|x| on [-2,1] is continuous everywhere but has a corner at x=0 where the derivative doesn't exist, making x=0 a critical point. Since |x| decreases from x=-2 to x=0 (where f(0)=0) and then increases from x=0 to x=1, the point x=0 is both a critical point and the location of the absolute minimum. Option A incorrectly claims x=0 isn't a critical point, but points where f'(x) doesn't exist are indeed critical points. Critical point definition: where f'(x)=0 OR where f'(x) doesn't exist (but f is continuous).
Let $r$ be continuous on $[0,4]$ and satisfy $r(0)=2$ , $r(4)=2$ , and $r(x)>2$ for $0<x<4$ ; which statement is true?
Answer: $r$ has an absolute minimum at both endpoints $x=0$ and $x=4$ . 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.
Let $f$ be continuous on $[-2,3]$ with $f'(-1)=0$ and $f'(1)=0$ . Which statement must be true?
Answer: $f$ has both an absolute maximum value and an absolute minimum value on $[-2,3]$ . This question assesses understanding of the Extreme Value Theorem (EVT) and extrema on closed intervals. The EVT states that if a function is continuous on a closed, bounded interval like [-2,3], it must attain both an absolute maximum and an absolute minimum value somewhere on that interval. The presence of critical points where f'(-1)=0 and f'(1)=0 indicates potential locations for local extrema, which are candidates for absolute extrema, but the theorem guarantees their existence regardless of where they occur. These critical points must be evaluated along with the endpoints x=-2 and x=3 to find the actual absolute extrema. A tempting distractor like choice A fails because the absolute maximum could occur at an endpoint rather than at the critical points x=-1 or x=1, depending on the function's values. To apply EVT effectively, always check: (1) continuity on the interval, (2) that the interval is closed and bounded, and (3) evaluate function values at critical points and endpoints.
Let $r$ be differentiable on $(-1,1)$ and continuous on $[-1,1]$ , with $r'(x)=0$ only at $x=0$ . Which is true?
Answer: Any absolute extremum of $r$ on $[-1,1]$ occurs at $x=-1,0,$ or $1$ . 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.
Let $f$ be continuous on $[-2,3]$ with $f'(-1)=0$ and $f'(2)=0$ . Which statement must be true?
Answer: $f$ attains both an absolute maximum and an absolute minimum on $[-2,3]$ . 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.
A function $g$ is continuous on $[0,6]$ and differentiable on $(0,6)$ ; which statement is always true?
Answer: $g$ attains both an absolute maximum and an absolute minimum on $[0,6]$ . This question evaluates knowledge of the Extreme Value Theorem (EVT) and how continuity and differentiability relate to extrema. The function g is continuous on the closed interval [0,6], so by the EVT, it must attain both an absolute maximum and an absolute minimum on that interval. While g is differentiable on (0,6), critical points where g' = 0 or undefined are potential sites for local extrema, but absolute extrema can also occur at the endpoints where the derivative may not be defined. The guarantee of absolute extrema comes from continuity on the closed interval, not from the presence or absence of critical points in the open interval. A tempting distractor is choice A, which states that g'(c) = 0 implies a local extremum at c, but this fails because it could be a point of inflection where the derivative changes sign without a max or min. To apply the EVT, verify continuity on a closed, bounded interval; then, identify critical points in the interior and evaluate the function at those points and the endpoints to find the extrema.
A function $r$ is continuous on $[-3,3]$ and has exactly one critical point at $x=0$ . Which statement must be true?
Answer: $r$ has both an absolute maximum value and an absolute minimum value on $[-3,3]$ . 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.
A continuous function $s$ on $[2,8]$ has $s'(x)<0$ on $(2,5)$ and $s'(x)>0$ on $(5,8)$ . Which is true?
Answer: $s$ has an absolute minimum at $x=5$ . 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.
Let $f$ be continuous on $[-2,3]$ with $f'(-1)=0$ and $f'(1)$ undefined. Which statement must be true?
Answer: $f$ has both an absolute maximum and an absolute minimum on $[-2,3]$ . 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.
Let $m$ be continuous on $[2,7]$ with $m'(x)=0$ for all $x$ in $(2,7)$ ; which statement must be true?
Answer: $m$ attains an absolute maximum and an absolute minimum, both at endpoints. 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.
A function $m$ is continuous on $[0,2]$ and has a local maximum at $x=1$ . Which statement must be true?
Answer: $m$ has both an absolute maximum and an absolute minimum on $[0,2]$ . 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.
Let $f(x)=|x-1|$ on $[-2,4]$ ; which statement about extrema and critical points is true?
Answer: $f$ has a global minimum at $x=1$ , and $x=1$ is a critical point. 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.
Let $q$ be continuous on $[0,2]$ and differentiable on $(0,2)$ with $q'(x)>0$ for all $x$ . Which is true?
Answer: $q$ has an absolute minimum at $x=0$ and an absolute maximum at $x=2$ . This question assesses the Extreme Value Theorem (EVT) and extrema for strictly increasing functions. Continuity on [0,2] ensures q has absolute extrema by EVT, and with q'(x)>0 everywhere in (0,2), q is strictly increasing, so the minimum is at x=0 and maximum at x=2. No critical points exist where q'(x)=0, but endpoints serve as locations for extrema in this case. The positive derivative confirms no interior extrema, directing attention to boundaries for absolute max and min. Choice D is tempting but fails because a strictly positive derivative means no point where q'(c)=0, yet extrema still exist at endpoints. For EVT, checklist: confirm continuity and closed interval, identify any critical points, evaluate function at critical points and endpoints, and compare to find extrema.
A function $g$ is continuous on $[0,5]$ and has critical points only at $x=2$ and $x=4$ . Which is true?
Answer: $g$ must have an absolute maximum and absolute minimum somewhere on $[0,5]$ . This problem applies the Extreme Value Theorem to identify where absolute extrema must exist. Since g is continuous on the closed interval [0,5], EVT guarantees that g must have both an absolute maximum and an absolute minimum somewhere on this interval. These extrema can only occur at critical points (x=2 and x=4) or at endpoints (x=0 and x=5), since g has no other critical points. Option A incorrectly restricts the absolute maximum to only critical points, ignoring that endpoints can also be locations of absolute extrema even when they're not critical points. The EVT checklist confirms: continuous on closed interval → must have absolute max and min, occurring at critical points or endpoints.

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