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f(x) minimum = 1→ n > 0
f(x) minimum jika f '(x) = 0
f '(x) = 2nx - 4
0 = 2nx - 4
x = 2/ n
masukkan nilai x = 2/n ke fungsi:
f(2/n) = n(2/n)² - 4(2/n) + n - 2
1 = 4/n - 8/ n + n - 2
n - 4/n - 3 = 0 (kalikan dengan n)
n² - 3n - 4 = 0
(n - 4)(n + 1) = 0
n = 4 atau n = -1
agar f(x) maksimum haruslah n < 0
agar f(x) minimum harauslah n > 0
sehingga nilai n yang memenuhi adalah 4