mokleng
1 d misalkan y = f(x) = x³ - 6x+ 9x + 1 turunkan maka y'= f'(x)=3x² - 12 x + 9 karena syarat stasioner y'= f'(x) = 0, maka => 3x² - 12x + 9 = 0 => 3 (x² - 4x + 3) = 0 => 3 (x - 1) (x - 3) = 0. x = 1,3 => titik stasioner jika x = 1 ,y = f (x) = 5 => nilai maksimum, titik balik maks. jika x = 3, maka y = f(x) = 1 => nilai minimum,titik balik min. #semoga membantu#
turunkan maka y'= f'(x)=3x² - 12 x + 9
karena syarat stasioner y'= f'(x) = 0, maka
=> 3x² - 12x + 9 = 0
=> 3 (x² - 4x + 3) = 0
=> 3 (x - 1) (x - 3) = 0. x = 1,3
=> titik stasioner jika x = 1 ,y = f (x) = 5
=> nilai maksimum, titik balik maks.
jika x = 3, maka y = f(x) = 1
=> nilai minimum,titik balik min.
#semoga membantu#