CALCULUS… let f(x)=(3)^(1/2)-2cos(x) for 0< or = to x < or = to 2pi.?
Find all intervals where f is decreasing. Find local maximum or minimum values. Find the intervals of concavity and the inflection points. Please show your work, especially when solving for max and min.
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f(x)=(3)^(1/2)-2cos(x)
f'(x) = 2sin(x)
So when would it be increasing [f(x) is positive] and decreasing [f(x) is negative] , and where are the relative maximums and minimums [f'(x) = 0]?
f”(x) = 2cos(x)
So find the intervals of concavity and the inflection points.
I hope this helps!
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