Question

Find the absolute extrema of the following functions on the given closed interval.

f(x) = `2 cos x + sin 2x; [0, pi/2]`

Answer 1

f'(x) = – 2 sin x + 2 cos 2x

f'(x) = 0

⇒ `2sin^2x + sinx - 1` = 0

`(2 sin x - 1)(sin  x + 1)` = 0

sin x = `1/2`, sin x = – 1

⇒ x = `pi/6`, x = `- pi/2 ∈ (0, pi/2)`

∴ Critical point is x = `pi/6 ∈ (0, pi/2)`

Now, f(0) = 2(1) + 0 = 2

`"f"(pi/6) = 2(sqrt(3)/2) + sqrt(3)/2 = (3sqrt(3))/2`

`"f"(pi/2)` = 2(0) + 0 = 0

∴ Absolute maximum `"f"(pi/6) = (3sqrt(3))/2`

Absolute minimum `"f"(pi/2)` = 0