Question

f(x) = 2sin(x) + cos(2x),  [0, 2pi]      Find all relative extrema, use the second derivative test where applicable.

Accepted Answer

Given the function f(x)=2sinx+cos(2x) in the interval [0,2pi] Taking the first derivative we get, f'(x)=2cosx-2sin(2x) Inorder to find the critical points we have to equate f'(x)=0. So we get, 2cosx-2sin(2x)=0 i.e. cosx-2sinxcosx=0 cosx(1-2sinx)=0 implies, cosx=0 and 1-2sinx=0 So cosx=0 implies x=pi/2, 3pi/2 in the interval [0,2pi]. 1-2sinx=0 implies, x=pi/6, 5pi/6. Therefore the critical points are: pi/2, (3pi)/2,pi/6,(5pi)/6 Now taking the second derivative of f(x) we get, f''(x)=-2sinx-4cos(2x) Applying second derivative test we have, f''(pi/2)=4>0 f''(3pi/2)=4>0 f''(pi/6)=-3<0 f''(5pi/6)=-3<0 Therefore the function has relative minima at x=pi/2, (3pi)/2 and the minimum value is f(x)=1 The function has relative maxima at x=pi/6,(5pi)/6 and the maximum value is f(x)=1.5

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`f(x) = 2sin(x) + cos(2x), [0, 2pi]` Find all relative extrema, use the second derivative test where applicable.

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