Question

1. koordinat titik balik fungsi y= x²+4x+4
2. nilai dari cos 60°+ sin 45° - tan 30°

Answer 1

1.) Misalkan y = f(x) = x² + 4x + 4
Turunkan, maka y' = f'(x) = 2x + 4
Karena syarat stasioner y' = 0, maka:
=> 2x + 4 = 0
=> 2(x + 2) = 0
x = -2. =>Titik stasioner
Jika x = -2, maka y = f(x) = 0. => Nilai stasioner
Jadi koordinat titik baliknya adalah (-2, 0)
2.) cos60 + sin45 - tan30
=> 1/2 + (√2)/2 - (√3)/3
=> (3 + 3√2 - √3)/3

Answer 2

1. koordinat titik balik = (Xp,Yp)
Xp = -b/2a = -4/2.1 = -2
Yp = -(b²-4ac/4a) = -(4²-4.1.4/4.1) = -(12/4) = -3
jadi, (Xp, Yp) = (-2 , -3)

2. cos 60 + sin 45 - tan 30 = 1/2 + 1/2 √2 - 1/3 √3 = = 0.629