Question

Soal terlampir
Jawab nomor 3.

Tidak menerima jwbn asal. Terimakasih^^​

Answer 1

Answer 2

Diketahui:
f(x) = (x²-6x)(2x-3)²
Ditanya f'(3)
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f(x) = (x²-6x)(2x-3)²
(uv)' = u'v + uv'
u = x²-6x, v = (2x-3)²
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Turunan f(x) [f'(x)] =
(x²-6x)'(2x-3)²+(x²-6x)[(2x-3)²]' =
(2x-6)(2x-3)²+(x²-6x)[2(2x-3)(2x-3)'] =
(2x-6)(2x-3)²+(x²-6x)[2(2x-3)2] =
f'(x) = (2x-6)(2x-3)²+4(x²-6x)(2x-3)
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Oleh karena itu
f'(3) = (2·3-6)(2·3-3)²+4(3²-6·3)(2·3-3)
f'(3) = (6-6)(6-3)²+4(9-18)(6-3)
f'(3) = 0 + 4(-9)(3)
f'(3) = 0 + (-108)
f'(3) = -108 ✔️