Soal yang Akan Dibahas
Fungsi dengan persamaan $ f(x) = \frac{2x+a}{x + 2b} $ memenuhi $ f^\prime (1) = 1 $
dan $ f(b) = -\frac{2}{3} $. Nilai $ b $ yang memenuhi adalah ....
A). $ -1 \, $ B). $ -\frac{4}{5} \, $ C). $ -\frac{2}{3} \, $ D). $ -\frac{1}{4} \, $ E). $ \frac{1}{2} $
A). $ -1 \, $ B). $ -\frac{4}{5} \, $ C). $ -\frac{2}{3} \, $ D). $ -\frac{1}{4} \, $ E). $ \frac{1}{2} $
$\spadesuit $ Konsep Dasar Turunan Fungsi
*). Turunan fungsi pecahan :
$ y = \frac{U}{V} \rightarrow y^\prime = \frac{U^\prime . V - U. V^\prime}{V^2} $
*). Turunan fungsi pecahan :
$ y = \frac{U}{V} \rightarrow y^\prime = \frac{U^\prime . V - U. V^\prime}{V^2} $
$\clubsuit $ Pembahasan
*). Menyusun persamaan pertama : $ f(b) = -\frac{2}{3} $
$\begin{align} f(x) & = \frac{2x+a}{x + 2b} \\ f(b) & = -\frac{2}{3} \\ \frac{2b+a}{b + 2b} & = -\frac{2}{3} \\ \frac{2b+a}{3b} & = -\frac{2}{3} \\ \frac{2b+a}{b} & = -\frac{2}{1} \\ 2b+a & = -2b \\ a & = -4b \, \, \, \, \, \, \, \text{...pers(i)} \end{align} $
*). Menentukan turunan dan $ f^\prime (1) = 1 $ :
$\begin{align} f(x) & = \frac{2x+a}{x + 2b} = \frac{U}{V} \\ U & = 2x + a \rightarrow U^\prime = 2 \\ V & = x + 2b \rightarrow V^\prime = 1 \\ f^\prime (x) & = \frac{U^\prime . V - U. V^\prime}{V^2} \\ f^\prime (x) & = \frac{2(x+2b) - (2x+a).1}{(x+2b)^2} \\ f^\prime (x) & = \frac{4b - a}{(x+2b)^2} \\ f^\prime (1) & = 1 \\ \frac{4b - a}{(1+2b)^2} & = 1 \, \, \, \, \, \text{...dari pers(i)} \\ \frac{4b - (-4b)}{4b^2 + 4b + 1} & = 1 \\ 8b & = 1 + 4b + 4b^2 \\ 0 & = 4b^2 - 4b + 1 \\ 0 & = (2b-1)^2 \\ 0 & = 2b - 1 \\ b & = \frac{1}{2} \end{align} $
Jadi, nilai $ b = \frac{1}{2} . \, \heartsuit $
*). Menyusun persamaan pertama : $ f(b) = -\frac{2}{3} $
$\begin{align} f(x) & = \frac{2x+a}{x + 2b} \\ f(b) & = -\frac{2}{3} \\ \frac{2b+a}{b + 2b} & = -\frac{2}{3} \\ \frac{2b+a}{3b} & = -\frac{2}{3} \\ \frac{2b+a}{b} & = -\frac{2}{1} \\ 2b+a & = -2b \\ a & = -4b \, \, \, \, \, \, \, \text{...pers(i)} \end{align} $
*). Menentukan turunan dan $ f^\prime (1) = 1 $ :
$\begin{align} f(x) & = \frac{2x+a}{x + 2b} = \frac{U}{V} \\ U & = 2x + a \rightarrow U^\prime = 2 \\ V & = x + 2b \rightarrow V^\prime = 1 \\ f^\prime (x) & = \frac{U^\prime . V - U. V^\prime}{V^2} \\ f^\prime (x) & = \frac{2(x+2b) - (2x+a).1}{(x+2b)^2} \\ f^\prime (x) & = \frac{4b - a}{(x+2b)^2} \\ f^\prime (1) & = 1 \\ \frac{4b - a}{(1+2b)^2} & = 1 \, \, \, \, \, \text{...dari pers(i)} \\ \frac{4b - (-4b)}{4b^2 + 4b + 1} & = 1 \\ 8b & = 1 + 4b + 4b^2 \\ 0 & = 4b^2 - 4b + 1 \\ 0 & = (2b-1)^2 \\ 0 & = 2b - 1 \\ b & = \frac{1}{2} \end{align} $
Jadi, nilai $ b = \frac{1}{2} . \, \heartsuit $
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