📝 Absolute extrema on open intervals (24 MCQs)

📖 Explanation: The combination of a left‑hand limit of −∞ and a right‑hand limit of +∞ forces the function to rise without bound near the right endpoint while dropping without bound near the left. Consequently the function can attain a highest finite value inside the interval, giving an absolute maximum, but it cannot attain a lowest value, so no absolute minimum exists.

📖 Explanation: An absolute maximum is a location inside the interval at which the function’s value is greater than or equal to its value at every other point of the interval. The definition does not involve derivatives or closed endpoints; it simply compares the function’s values throughout the open interval.

📖 Explanation: Since the function blows up to +∞ near the left endpoint and drops to −∞ near the right endpoint, the graph must descend from arbitrarily large values to arbitrarily small values. Hence a lowest finite value is attained somewhere inside, giving an absolute minimum, while no highest finite value can exist.

📖 Explanation: Differentiating gives f'(x)=−(2x−1)/(x^2−x)^2, which is zero only at x=½. Because the limits at the endpoints are both −∞, the only candidate for an absolute extremum is this interior critical point, and evaluating f(½)=−4 shows it is indeed the absolute maximum.

📖 Explanation: The function f approaches −∞ at both ends of (0,1) and possesses a single interior critical point at x=½ where it reaches its highest finite value, giving an absolute maximum. In contrast, g(x) tends to −∞ as x→½⁻ and +∞ as x→½⁺, so it never attains a finite highest or lowest value on the interval.

📖 Explanation: When the function diverges to +∞ at both ends, it may dip down to a finite low point inside the interval, but the theorem does not guarantee such a dip. Therefore no conclusion about the existence of extrema can be drawn; the function might have none.

📖 Explanation: Both one‑sided limits head to negative infinity, so the function is unbounded below near each endpoint. It can never achieve a greatest finite value (maximum) because values become arbitrarily large negative, and it likewise cannot achieve a least finite value (minimum). Hence neither extremum is guaranteed.

📖 Explanation: The limits indicate the function grows without bound at both ends, but this does not force the existence of a finite lowest or highest value inside the interval. The function could dip to a finite low point or could stay above every finite bound, so no extremum is guaranteed.

📖 Explanation: On an open interval the only places where a continuous function can achieve a highest or lowest value are interior points where the derivative vanishes or fails to exist. Since endpoints are excluded, any absolute extremum that does exist must be located at a critical point.

📖 Explanation: The sine function reaches its highest value 1 at x=π/2, which lies inside (0,π). Near the endpoints the function approaches 0 but never attains it, so there is no absolute minimum within the open interval.

📖 Explanation: As x approaches −π/2⁺, tan x → −∞, and as x approaches π/2⁻, tan x → +∞. Because the function is unbounded in both directions and attains every real value, it cannot have a greatest or least finite value; thus it has no absolute extrema.

📖 Explanation: The limit as x→0⁺ is −∞, so no absolute minimum can exist. As x→e⁻ the function approaches ln e = 1, but the endpoint e is not included, so the supremum 1 is never attained. Hence the function has neither an absolute maximum nor an absolute minimum on the interval.

📖 Explanation: Finite one‑sided limits only give the behavior near the endpoints; the function could dip below L inside the interval or stay above L. Therefore the existence of an absolute minimum is not guaranteed solely by the limits, and it may occur at an interior point or not at all.

📖 Explanation: The derivative vanishes at x=±1; only x=1 lies in (−1,2) and gives f(1)=½, the highest value. As x→−1⁺, f approaches −½ but never attains it, so there is no absolute minimum. Hence the function has an absolute maximum at x=1 and no absolute minimum.

📖 Explanation: When the limits at both ends are +∞, the function grows without bound toward each endpoint. It may still dip to a finite low point, but the theorem does not guarantee such a dip, and any candidate for a maximum would be surpassed by values arbitrarily close to the endpoints. Hence no absolute extremum is forced.

📖 Explanation: The combination of a left limit of −∞ and a right limit of +∞ forces the function to be arbitrarily low near the left endpoint and arbitrarily high near the right endpoint. Consequently a highest finite value can be attained inside the interval (giving an absolute maximum), while no lowest finite value can exist, precluding an absolute minimum.

📖 Explanation: The function is increasing on (0,1); as x→0⁺ it approaches 0, and at x=1 it equals e^{−1}≈0.368. Thus the greatest value is attained at x=1, giving an absolute maximum, while the infimum 0 is never reached, so there is no absolute minimum.

📖 Explanation: The table shows four cases: (i) max only (limits −∞,+∞), (ii) min only (limits +∞,−∞), (iii) neither (both limits +∞ or both −∞), (iv) both (finite limits with appropriate ordering). For f(x)=1/(x^{2}−x), the limits at both ends are −∞, so the function falls into case (iii) where only an absolute maximum exists (at the interior critical point) and no absolute minimum.

📖 Explanation: Endpoints are not part of an open interval, so an absolute extremum cannot occur at an endpoint. The false statement is that an absolute minimum may occur at an endpoint; in fact, endpoints are excluded, making this claim incorrect.

📖 Explanation: A critical point is any interior point of the domain where the derivative is zero or fails to exist. Such points are the only candidates for absolute extrema on an open interval because the interval has no endpoints to consider.

📖 Explanation: The Extreme Value Theorem states that a continuous function on a closed, bounded interval [a,b] must attain both an absolute maximum and an absolute minimum. When the interval is open, the theorem no longer guarantees any extrema because the function may approach but never reach extreme values at the omitted endpoints.

📖 Explanation: The cubic function is odd and strictly increasing; as x→1⁻, f approaches 1 but never attains it, and as x→−1⁺, f approaches −1 but never attains it. Hence the function has no absolute maximum or minimum on the open interval.

📖 Explanation: When the left limit diverges to +∞ and the right limit to −∞, the function must descend from arbitrarily large positive values to arbitrarily large negative values, guaranteeing the existence of a lowest finite value (absolute minimum) somewhere inside, while no highest finite value can exist.

📖 Explanation: The derivative of f is always positive, so the function is strictly increasing. As x approaches −2⁺, f approaches −2/√5≈−0.894, and as x approaches 3⁻, f approaches 3/√10≈0.949. Neither endpoint value is attained, and there are no interior critical points, so the function possesses no absolute maximum or minimum on the interval.