f(x) = 2/3x^3 – 3x² + 1,
apa bila f'(x) = 4
nb : ^ = pangkat
2/3x^3 = dua per tiga x pangkat 3
Carilah nilai x dari turunan fungsi :
F(x) = 2/3x^3 – 3x² + 1
f'(x) = 4
f'(x) = 2/3 x^-3 – 3x^2 + 1
f'(x) = 2/3 (3) x^-3-1 – 3(2)x^2-1 + 0
f'(x) = 2x^-4 – 6x^1
f'(x) = 2/x^4 – 6x
f'(x) = 4
f'(x) = 2/ 4^4 – 6.4
f'(x) = 2 / 256 – 24
= 1/128 -24
=