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in Limit, continuity and differentiability by (50.2k points)

Find the absolute max or min of f(x) = sinx + 1/2cos2x, ∀ x∈ [ 0, π/2]

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Given f(x) = sinx + 1/2cos2x 

 f'(x) = cos x – sin2x

Now, f'(x) = 0 gives cosx – sin2x = 0

⇒ cos x (1 – 2 sinx) = 0

⇒ cos x = 0, (1 – 2 sinx) = 0

 cos x = 0, sinx = 1/2

 x = π/6 , π/2

Now, f(0) = 1/2,

f(π/6) = 1/2 + 1/4 = 3/4,

f(π/2) = 1 – 1/2 = 1/2

Therefore, the absolute max value = 3/4 and absolute min = 1/2.

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