Jika h(x) = (3x^2 – 4)/(6 + x), nilai h(0) + 6 h'(0) sama dengan . . . .

Nilai h(0) + 6h'(0) = 0

Pembahasan

h (x) = (3x² – 4)/(6 + x)

h (0) = – 4/6

h (0) = – 2/3

h (x) = (3x² – 4)/(6 + x)

misal

u = 3x² – 4

u' = 6x

v = 6 + x

v' = 1

h'(x) = u' v – v' u / v²

h'(x) = 6x (6 + x) – 1 (3x² – 4) / (6 + x)²

h'(x) = 36x + 6x² – 3x² + 4 / (6 + x)²

h'(x) = 3x² + 36x + 4 / (6 + x)²

h'(0) = 4/36

h'(0) = 1/9

h(0) + 6h'(0) = -2/3 + 6(1/9)

h(0) + 6h'(0) = -2/3 + 2/3

h(0) + 6h'(0) = 0

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Detil Jawaban

Kelas: 11

Mapel: Matematika

Bab: Turunan Fungsi Aljabar

Kode: 11.2.9

Kata Kunci: turunan pertama

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