Question

fungsi f(x) =x^2+3/x-1 turun pada interval​

Answer 1

jawaban:

f(x) = (x²+3) / (x -1)

u = x²+3 --> u' = 2x

v = x - 1 --> v' = 1

f' (x) = (u'v - uv') / (v)²

f' (x) = { 2x(x-1) - (x²+3)(1)} /(x-1)²

f'(x)= (2x² -2x - x²-3)/(x-1)²

f'(x) = (x² - 2x - 3)/ (x-1)²

f '(x) = (x- 3)(x +1)/(x-1)² , x ≠ 1

fungsi turun jika f '(x) < 0

(x-3)(x+1) / (x -1)² < 0 dengan x≠ 1

krn (x-1)² selalu > 0 maka (x-3)(x+1) < 0

x < 3 atau x > -1 dan x≠1

HP -1 < x < 1 atau 1 < x < 3

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