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+1 vote
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in Applications of Differential Calculus by (48.7k points)
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Find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = x – 2 cos x, [-π, π]

1 Answer

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Best answer

f(x) = x – 2 cos x 

f'(x) = 1 + 2 sin x 

f'(x) = 0 ⇒ 1 + 2 sin x = 0

From the above values, absolute maximum is π + 2 and the absolute minimum is (-π/6) - √3.

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