Mohon bantuannya..Para Master.. Tentukan persamaan garis singgung kurva f(x) untuk setiap nilai absis yang diberikan. a.f(x)=cos 2x di x=π b..2sin (2x-π)di x= c.f(x)=tan(3x-3)di x=1
ekamiawwa. f(x) = cos 2x di x = π m = f'(x) = -2 sin 2x = -2 sin 2π = -2 . 0 = 0 f(π) = cos 2π = 1 pers garis singgung dgn m = 0 di titik (π,1): y - y₁ = m (x - x₁) y - 1 = 0 (x - π) y - 1 = 0 y = 1
2) f(x) = 2 sin (2x - π) di x = π/2 m = f'(x) = 2.2 cos (2x - π) = 4 cos (2.π/2 - π) = 4 cos 0 = 4 . 1 = 4 f(π/2) = 2 sin (2.π/2 - π) = 2 sin 0 = 0 pers garis singgung dgn m = 4 di titik (π/2,0) y - y₁ = m (x - x₁) y - 0 = 4 (x - π/2) y = 4x - 2π
3) f(x) = tan (3x - 3) di x = 1 m = f'(x) = 3 sec² (3x - 3) = 3 sec² (3.1 - 3) = 3 sec² 0 = 3 . 1² = 3 f(1) = tan (3.1 - 3) = tan 0 = 0 pers garis singgung dgn m = 3 di titik (1,0): y - y₁ = m (x - x₁) y - 0 = 3 (x - 1) y = 3x - 3
m = f'(x) = -2 sin 2x
= -2 sin 2π
= -2 . 0
= 0
f(π) = cos 2π = 1
pers garis singgung dgn m = 0 di titik (π,1):
y - y₁ = m (x - x₁)
y - 1 = 0 (x - π)
y - 1 = 0
y = 1
2) f(x) = 2 sin (2x - π) di x = π/2
m = f'(x) = 2.2 cos (2x - π)
= 4 cos (2.π/2 - π)
= 4 cos 0
= 4 . 1
= 4
f(π/2) = 2 sin (2.π/2 - π)
= 2 sin 0
= 0
pers garis singgung dgn m = 4 di titik (π/2,0)
y - y₁ = m (x - x₁)
y - 0 = 4 (x - π/2)
y = 4x - 2π
3) f(x) = tan (3x - 3) di x = 1
m = f'(x) = 3 sec² (3x - 3)
= 3 sec² (3.1 - 3)
= 3 sec² 0
= 3 . 1²
= 3
f(1) = tan (3.1 - 3)
= tan 0
= 0
pers garis singgung dgn m = 3 di titik (1,0):
y - y₁ = m (x - x₁)
y - 0 = 3 (x - 1)
y = 3x - 3
semoga membantu :)