Find the critical points of the function f(x)=−3 cos 2(x) contained in the interval (0,2π).

Answer 1

f(x) = -3cos(2x)

f'(x) = 6sin(2x)

f'(x) = 0 when 2x = kπ, k = 0, 1, -1, 2, -2, ..

x = kπ/2

If k = 0, then x = 0

If k = 1, then x = π/2

If k = 2, then x = π

If k = 3, then x = 3σ/2 etc

Solutions in the interval (0, 2π) are π/2, π, and 3π/2

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Answer 2

f=-3cos2x ... f'=6sin2x ... f'=0 @2x=0, 2x= pi/2, 2x=pi, 2x=3pi/2, 2x=2pi or (0,-3), (pi,3), (3pi/4,0), (2pi,-3)