jika h(x) = (nx^2-3x+1)^2 dan h (1)=2, nilai n =

h(x) = (nx² - 3x + 1)²

h'(x) = 2(nx² - 3x + 1) × (2nx - 3)
= (2nx² - 6x + 2)(2nx - 3)
= 4n²x³ - 6nx² - 12nx² + 18x + 4nx - 6
= 4n²x³ - 18nx² + 18x + 4nx - 6

h'(1) = 4n²(1)³ - 18n(1)² + 18(1) + 4n(1) - 6
2 = 4n² - 18n + 18 + 4n - 6
2 = 4n² - 14n + 12
0 = 4n² - 14n + 10
0 = 2(2n - 5)(n - 1)
0 = (2n - 5)(n - 1)
2n - 5 = 0 atau n - 1 = 0
2n = 5 atau n = 1
n = 5/2 atau n = 1