Consider the function f(x) = sin(x) on the interval x ∈ [π/4, 7π/4]. The…
2012
Consider the function f(x) = sin(x) on the interval x ∈ [π/4, 7π/4]. The number and location(s) of the local minima of this function are
Answer: D. Two, at π/4 and 3π/2 — ConceptFor a function whose domain is a closed interval, a local extremum is defined relative to that domain. Interior stationary points can be classified…
- A.
One, at π/2
- B.
One, at 3π/2
- C.
Two, at π/2 and 3π/2
- D.
Two, at π/4 and 3π/2
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Correct answer: D
Concept
For a function whose domain is a closed interval, a local extremum is defined relative to that domain. Interior stationary points can be classified with derivative tests, while an endpoint must be checked using nearby domain points on its one available side.
Application
Differentiate: f′(x) = cos(x). The interior stationary points satisfy cos(x) = 0, giving x = π/2 and x = 3π/2 in the interval.
Use f″(x) = −sin(x). Since f″(π/2) = −1, π/2 is an interior local maximum. Since f″(3π/2) = 1, 3π/2 is an interior local minimum.
Check the left endpoint. Because f′(π/4) = √2/2 > 0, f increases immediately to the right of π/4; therefore π/4 is a one-sided local minimum relative to the stated domain.
Check the right endpoint. The function is increasing as it approaches 7π/4 from the left, so nearby domain values are smaller; hence 7π/4 is not a local minimum.
Cross-check
The sine curve rises from f(π/4) = √2/2 to its peak at π/2, falls to f(3π/2) = −1, and then rises again toward 7π/4. This independently confirms the one-sided minimum at π/4 and the interior minimum at 3π/2.
Therefore, there are two local minima, at x = π/4 and x = 3π/2.