F(theta) = 2cos(theta) - 2cos(2*theta); Interval...

Barack Lasagna

From: US
Posts: 25
Votes: 100
F(theta) = 2cos(theta) - 2cos(2*theta); interval [0,2*pi] (a) Where is fxn increasing and decreasing? (b) Where are local max & min's? (c) When concave up? Down? Where are theinflection points? (d) Sketch graph. **I apologize, this here is the correct stated question. When I first posted this question, I forgot the 2 in front ofthe second term in the expression: it correctly reads"2cos(2theta)". Please answer this one.** ***I am having trouble finding the critical #'s for first andsecond derivatives... I will rate you highly for your help. Thanks.***

 

Answer No.1

Anonymous

From: -
Posts: -
Votes: -
Y = 2*cos(x) - 2*cos(2*x) dy/dx = -2*sin(x) +4*sin(2x) = -2*sin(x) +8*sin(x)*cos(x) Made a mistake in above. I stated sin(2x) = sin(x)*cos(x). That iswrong. sin(2x) = 2*sin(x)*cos(x) Set dy/dx = 0 = -2sin(x) + 8*sin(x)*cos(x) cos(x) = 1/4 x = 1.318, 4.965 d2y/dx2 = -2*cos(x) + 8*cos(2x) = -2cos(x) + 16*cos(x) ^2 -8 =0 8*cos(x)^2 - cos(x) - 8 = 0 cos(x) = {1 +/- sqrt(1^2 -4*8*(-8))}/(2*8) cos(x) = {1 +/- sqrt(257))/16 = -0.939 or 1.06 (reject second one,because cos(x) <= +/- 1) x = arcos (-0.939) = 2.792 or 3.491

Answer No.2

SuperBob5

From: US
Posts: 52
Votes: 156
Derivative= -2*sin(x) - 2*sin(2x) sin(2x) = sin(x)*cos(x) derivative = -2*(sin(x) + sin(x)*cos(x)) Set derivative to 0 sin(x) = -sin(x)*cos(x) -1 = cos(x) x = 3*pi/2 For second derivatives: -2*(cos(x) + cos(x)^2 -sin(x)^2) = 0 sin(x)^2 = 1 - cos(x)^2 cos(x) + 2*cos(x)^2 - 1 = 0 cos(x)^2 + 0.5*cos(x) - 0.5 =0 use quadratic formula: cos(x) = (-0.5 +/- sqrt(0.5^2-4*1*(-0.5)))/2 cos(x) = -1/4 +/- sqrt(2.25)/2 = 0.5 or -1 x = -pi and +/- pi/3 (+/- 60 degrees)

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