concavity of cos^2 x - 2sin x Solution (a) First find critical points (points wh
ID: 3188504 • Letter: C
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concavity of cos^2 x - 2sin xExplanation / Answer
(a) First find critical points (points where f ' = 0) f(x) = (cosx)^2 - 2sinx on [0, 2pi] f ' (x) = 2cosx * (-sinx) - 2cosx f ' (x) = 0 = -2sinxcosx - 2cosx -2cosx (sinx + 1) = 0 -2cosx = 0 or sinx+1=0 cosx = 0 or sinx = -1 x = pi/2, 3pi/2 or x = 3pi/2 x = pi/2 or 3pi/2 Now we must check when f ' is positive or negative on the follow intervals: [0, pi/2] and [pi/2, 3pi/2] and [3pi/2, 2pi] To check the sign we pick a point in each interval the sign of f ' at that point tells us the sign of f ' throughout that interval. Start with [0, pi/2]. Pick x = pi/4 f ' (pi/4) = -2(sqrt(2)/2)(sqrt(2)/2) - 2(sqrt(2))/2 = -1-sqrt(2) < 0. Thus, f ' < 0 on [0, pi/2] Next is [pi/2, 3pi/2]. Pick x = pi f ' (pi) = -2*(0)*(-1) - 2*(-1) = 2 > 0. Thus, f ' > 0 on [pi/2, 3pi/2] Finally is [3pi/2, 2pi]. Check x = 7pi/4 f ' (7pi/4) = -2*(-sqrt(2)/2)*(sqrt(2)/2) - 2(sqrt(2)/2) = 1-sqrt(2) < 0. Thus, f ' < 0 on [3pi/2, 2pi] also. Remember: If f ' > 0 (f ' < 0) on [a,b], then f is increasing (decreasing) on [a,b]. Thus, we have found (a) and (b) (a) f is increasing on [pi/2, 3pi/2] and f is decreasing on [0, pi/2] U [3pi/2, 2pi] (b) Local minimum at x = pi/2 and x = 2pi; Local maximum at x = 0 and x = 3pi/2 Now apply these steps to f '' to find (c). We know f ' (x) = -2sinxcosx - 2cosx. Then f '' (x) = -2(cosx)^2 + 2(sinx)^2 + 2sinx 0 = -2(cosx)^2 + 2(sinx)^2 + 2sinx Remember (cosx)^2 = 1 - (sinx)^2 0 = -2(1 - (sinx)^2) + 2(sinx)^2 + 2sinx 0 = -2 + 2(sinx)^2 + 2(sinx)^2 + 2sinx 0 = 2(2(sinx)^2 + sinx - 1) 0 = 2(sinx)^2 + sinx - 1 Let y = sinx for a moment. Then 0 = 2y^2 + y - 1 0 = 2y^2 + (2y - y) - 1 since y = 2y-y 0 = (2y^2+2y) + (-y-1) 0 = (2y)(y+1) + -1(y+1) 0 = (2y-1)(y+1) 2y-1 = 0 or y+1 = 0 y = 1/2 or y = -1 Now replace y with sinx sinx = 1/2 or sin x = -1 x = pi/6 or 5pi/6 or x = 3pi/2 x = pi/6, 5pi/6, 3pi/2 Check the signs like before on the following intervals: [0, pi/6) (pi/6, 5pi/6) (5pi/6, 3pi/2) (3pi/2, 2pi]Related Questions
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