penyelesaian titik minimum dan titik maksimum dari fungsi e^-x/(2+sin(2x))

jawab

y =  ( e⁻ˣ ) / ( 2 + sin 2x)

y' =  0

[ ⁻e⁻ˣ / (sin 2x + 2) ] - [ (2e⁻ˣ. cos 2x)/ (sin 2x +  2)²]

- e⁻ˣ / (sin 2x + 2 ) = (2. e⁻ˣ . cos 2x / (sin 2x + 2)²

[ (sin(2x) + 2 )² / (sin 2x + 2)]   = ( 2 e⁻ˣ. cos 2x )/ (- e⁻ˣ)

sin (2x) + 2 =  2 cos 2x
2 cos 2x  - sin 2x = 2
a= 2
b = - 1
c = 2

k = √(a²+b²)
k = √5

2 cos 2x - sin 2x = 2 → k cos (2x - α)  = 2
√5 cos (2x - α) - 2 = 0

nilai cos (2x -α) maksimun = 1
y = √5 (1) - 2
y = -2 +√5
y maks = -2 + √5

nilai cos(2x - α) = - 1
y = -√5 - 2
y = - 2 - √5
y minimum = - 2 - √5