f(x) =2x²+3 dibagi x-6,maka f(0)+f'(0) adalah

f(x) = (2x² + 3)/(x - 6) => f(0) = (2.0² + 3)/(0 - 6) = 3/(-6) = -1/2
f(x) = u/v
u = 2x² + 3 => u' = 4x
v = x - 6 => v' = 1
f'(x) = (u' v - v' u) / v²
f'(x) = (4x (x - 6) - 1(2x² + 3)) / (x - 6)²
f'(0) = (4.0 (0 - 6) - 1(2.0² + 3)) / (0 - 6)²
f'(0) = (0(-6) - 1(3)) / (-6)²
f'(0) = -3/36 = -1/12

f(0) + f'(0) = -1/2 + (-1/12) = (-6 - 1)/12 = -7/12

Bab Differensial
Matematika SMA Kelas XI

f(x) = (2x² + 3) / (x - 6)

f(0) = (2 . 0 + 3) / (0 - 6)
f(0) = 3/-6 = -1/2

f'(x) = (2 . 2x (x - 6) - (2x² + 3) . 1) / (x - 6)²
f'(x) = (4x² - 24x - 2x² - 3) / (x - 6)²
f'(x) = (2x² - 24x - 3) / (x - 6)² 

f'(0) = (0 - 0 - 3) / (0 - 6)²
f'(0) = -3/36
f'(0) = -1/12

f(0) + f'(0) = -1/2 - 1/12
f(0) + f'(0) = -6/12 - 1/12
f(0) + f'(0) = -7/12