Nomor 16
$ \displaystyle \lim_{x \to 3} \frac{\sqrt{x+4} - \sqrt{2x+1}}{x-3} $ adalah ....
$\spadesuit \, $ Kalikan sekawannya
$\begin{align} & \displaystyle \lim_{x \to 3} \frac{\sqrt{x+4} - \sqrt{2x+1}}{x-3} \\ & = \displaystyle \lim_{x \to 3} \frac{\sqrt{x+4} - \sqrt{2x+1}}{x-3} . \frac{\sqrt{x+4} + \sqrt{2x+1}}{\sqrt{x+4} + \sqrt{2x+1}} \\ & = \displaystyle \lim_{x \to 3} \frac{\sqrt{x+4} - \sqrt{2x+1}}{x-3} . \frac{\sqrt{x+4} + \sqrt{2x+1}}{\sqrt{x+4} + \sqrt{2x+1}} \\ & = \displaystyle \lim_{x \to 3} \frac{(x+4)-(2x+1)}{(x-3)(\sqrt{x+4} + \sqrt{2x+1}) } \\ & = \displaystyle \lim_{x \to 3} \frac{-(x-3)}{(x-3)(\sqrt{x+4} + \sqrt{2x+1}) } \\ & = \displaystyle \lim_{x \to 3} \frac{-1}{\sqrt{x+4} + \sqrt{2x+1}} \\ & = \frac{-1}{\sqrt{3+4} + \sqrt{2.3+1}} \\ & = \frac{-1}{\sqrt{7} + \sqrt{7}} \\ & = \frac{-1}{2\sqrt{7}} = -\frac{1}{14} \sqrt{7} \end{align}$
Jadi, nilai limitnya adalah $ -\frac{1}{14} \sqrt{7} . \heartsuit $
$\begin{align} & \displaystyle \lim_{x \to 3} \frac{\sqrt{x+4} - \sqrt{2x+1}}{x-3} \\ & = \displaystyle \lim_{x \to 3} \frac{\sqrt{x+4} - \sqrt{2x+1}}{x-3} . \frac{\sqrt{x+4} + \sqrt{2x+1}}{\sqrt{x+4} + \sqrt{2x+1}} \\ & = \displaystyle \lim_{x \to 3} \frac{\sqrt{x+4} - \sqrt{2x+1}}{x-3} . \frac{\sqrt{x+4} + \sqrt{2x+1}}{\sqrt{x+4} + \sqrt{2x+1}} \\ & = \displaystyle \lim_{x \to 3} \frac{(x+4)-(2x+1)}{(x-3)(\sqrt{x+4} + \sqrt{2x+1}) } \\ & = \displaystyle \lim_{x \to 3} \frac{-(x-3)}{(x-3)(\sqrt{x+4} + \sqrt{2x+1}) } \\ & = \displaystyle \lim_{x \to 3} \frac{-1}{\sqrt{x+4} + \sqrt{2x+1}} \\ & = \frac{-1}{\sqrt{3+4} + \sqrt{2.3+1}} \\ & = \frac{-1}{\sqrt{7} + \sqrt{7}} \\ & = \frac{-1}{2\sqrt{7}} = -\frac{1}{14} \sqrt{7} \end{align}$
Jadi, nilai limitnya adalah $ -\frac{1}{14} \sqrt{7} . \heartsuit $
Nomor 17
Jika $x_1 $ dan $ x_2 $ memenuhi persamaan :
$(2\log x - 1 ) . \frac{1}{{}^x \log 10 } = \log 10 , \, x_1x_2 = .... $
$(2\log x - 1 ) . \frac{1}{{}^x \log 10 } = \log 10 , \, x_1x_2 = .... $
$\clubsuit \, $ Konsep dasar logaritma :
Definisi : $ {}^a \log b = c \Leftrightarrow b = a^c $
Sifat : $ {}^a \log b = \frac{1}{ {}^b \log a } $
$\clubsuit \, $ Misalkan $ p = \log x $
$\begin{align} (2\log x - 1 ) . \frac{1}{{}^x \log 10 } & = \log 10 \\ (2\log x - 1 ) . {}^{10} \log x & = \log 10 \\ (2\log x - 1 ) . \log x & = \log 10 \\ (2p-1).p & = 1 \\ 2p^2 - p - 1 & = 0 \\ (2p+1)(p-1) & = 0 \\ p = -\frac{1}{2} \rightarrow & \log x = -\frac{1}{2} \rightarrow x_1 = 10^{-\frac{1}{2}} \\ p = 1 \rightarrow & \log x = 1 \rightarrow x_2 = 10^1 \end{align}$
Sehingga, $ x_1.x_2 = 10^{-\frac{1}{2}}. 10^1 = 10^{-\frac{1}{2} + 1} = 10^{\frac{1}{2}} = \sqrt{10} $
Jadi, nilai $ x_1.x_2 = \sqrt{10} . \heartsuit $
Definisi : $ {}^a \log b = c \Leftrightarrow b = a^c $
Sifat : $ {}^a \log b = \frac{1}{ {}^b \log a } $
$\clubsuit \, $ Misalkan $ p = \log x $
$\begin{align} (2\log x - 1 ) . \frac{1}{{}^x \log 10 } & = \log 10 \\ (2\log x - 1 ) . {}^{10} \log x & = \log 10 \\ (2\log x - 1 ) . \log x & = \log 10 \\ (2p-1).p & = 1 \\ 2p^2 - p - 1 & = 0 \\ (2p+1)(p-1) & = 0 \\ p = -\frac{1}{2} \rightarrow & \log x = -\frac{1}{2} \rightarrow x_1 = 10^{-\frac{1}{2}} \\ p = 1 \rightarrow & \log x = 1 \rightarrow x_2 = 10^1 \end{align}$
Sehingga, $ x_1.x_2 = 10^{-\frac{1}{2}}. 10^1 = 10^{-\frac{1}{2} + 1} = 10^{\frac{1}{2}} = \sqrt{10} $
Jadi, nilai $ x_1.x_2 = \sqrt{10} . \heartsuit $
Nomor 18
Nilai $ x $ yang memenuhi :
$\log x = 4 \log (a+b) + 2\log (a-b) - 3\log (a^2-b^2) - \log \frac{a+b}{a-b} $ adalah ....
$\log x = 4 \log (a+b) + 2\log (a-b) - 3\log (a^2-b^2) - \log \frac{a+b}{a-b} $ adalah ....
$\spadesuit \, $ Konsep dasar logaritma
persamaan : $ {}^a \log f(x) = {}^a \log g(x) \rightarrow f(x) = g(x) $
Sifat : $ {}^a \log b^n = n. {}^a \log b , \, {}^a \log b + {}^a \log c = {}^a \log bc $
$ {}^a \log b - {}^a \log c = {}^a \log \frac{b}{c} $
$\spadesuit \, $ Menyederhanakan soal
$\begin{align} \log x & = 4 \log (a+b) + 2\log (a-b) - 3\log (a^2-b^2) - \log \frac{a+b}{a-b} \\ \log x & = \log (a+b)^4 + \log (a-b)^2 - \log (a^2-b^2)^3 - \log \frac{a+b}{a-b} \\ \log x & = \log \frac{(a+b)^4.(a-b)^2}{(a^2-b^2)^3.\frac{a+b}{a-b}} \\ \log x & = \log \frac{(a+b)^4.(a-b)^2.(a-b)}{[(a-b)(a+b)]^3.(a+b)} \\ \log x & = \log \frac{(a+b)^4.(a-b)^3}{(a-b)^3(a+b)^3.(a+b)} \\ \log x & = \log \frac{(a+b)^4.(a-b)^3}{(a-b)^3(a+b)^4} \\ \log x & = \log 1 \\ x & = 1 \end{align}$
Jadi, nilai $ x = 1 . \heartsuit $
persamaan : $ {}^a \log f(x) = {}^a \log g(x) \rightarrow f(x) = g(x) $
Sifat : $ {}^a \log b^n = n. {}^a \log b , \, {}^a \log b + {}^a \log c = {}^a \log bc $
$ {}^a \log b - {}^a \log c = {}^a \log \frac{b}{c} $
$\spadesuit \, $ Menyederhanakan soal
$\begin{align} \log x & = 4 \log (a+b) + 2\log (a-b) - 3\log (a^2-b^2) - \log \frac{a+b}{a-b} \\ \log x & = \log (a+b)^4 + \log (a-b)^2 - \log (a^2-b^2)^3 - \log \frac{a+b}{a-b} \\ \log x & = \log \frac{(a+b)^4.(a-b)^2}{(a^2-b^2)^3.\frac{a+b}{a-b}} \\ \log x & = \log \frac{(a+b)^4.(a-b)^2.(a-b)}{[(a-b)(a+b)]^3.(a+b)} \\ \log x & = \log \frac{(a+b)^4.(a-b)^3}{(a-b)^3(a+b)^3.(a+b)} \\ \log x & = \log \frac{(a+b)^4.(a-b)^3}{(a-b)^3(a+b)^4} \\ \log x & = \log 1 \\ x & = 1 \end{align}$
Jadi, nilai $ x = 1 . \heartsuit $
Nomor 19
Jika nilai maksimum fungsi $ y = x + \sqrt{p-2x} $ adalah 4, maka $ p = .... $
$\clubsuit \,$ Konsep dasar
Turunan : $ y = \sqrt{f(x)} \rightarrow y^\prime = \frac{f^\prime (x)}{2\sqrt{f(x)}} $
Syarat nilai maks/min : $ f^\prime (x) = 0 $
$\clubsuit \,$ Menentukan turunannya
$\begin{align} y & = x + \sqrt{p-2x} \\ y^\prime & = 1 + \frac{-2}{2\sqrt{p-2x}} \\ y^\prime & = 1 + \frac{-1}{\sqrt{p-2x}} = 0 \\ \sqrt{p-2x} & = 1 \, \, \, \text{pers(i)} \\ p-2x & = 1 \rightarrow x = \frac{p-1}{2} \end{align}$
Artinya $ f(x) $ maksimum saat $ x = \frac{p-1}{2} $
$\clubsuit \,$ Substitusi $ x = \frac{p-1}{2} $ ke fungsi diperoleh nilai maksimum yaitu $ y_{maks} = 4 $
$\begin{align} y & = x + \sqrt{p-2x} \\ y_{maks} & = f\left( \frac{p-1}{2} \right) \\ 4 & = \frac{p-1}{2} + \sqrt{p-2.\frac{p-1}{2}} \\ 4 & = \frac{p-1}{2} + \sqrt{1} \\ 4 & = \frac{p-1}{2} + 1 \\ p-1 & = 6 \rightarrow p = 7 \end{align}$
Jadi, nilai $ p = 7 . \heartsuit $
Turunan : $ y = \sqrt{f(x)} \rightarrow y^\prime = \frac{f^\prime (x)}{2\sqrt{f(x)}} $
Syarat nilai maks/min : $ f^\prime (x) = 0 $
$\clubsuit \,$ Menentukan turunannya
$\begin{align} y & = x + \sqrt{p-2x} \\ y^\prime & = 1 + \frac{-2}{2\sqrt{p-2x}} \\ y^\prime & = 1 + \frac{-1}{\sqrt{p-2x}} = 0 \\ \sqrt{p-2x} & = 1 \, \, \, \text{pers(i)} \\ p-2x & = 1 \rightarrow x = \frac{p-1}{2} \end{align}$
Artinya $ f(x) $ maksimum saat $ x = \frac{p-1}{2} $
$\clubsuit \,$ Substitusi $ x = \frac{p-1}{2} $ ke fungsi diperoleh nilai maksimum yaitu $ y_{maks} = 4 $
$\begin{align} y & = x + \sqrt{p-2x} \\ y_{maks} & = f\left( \frac{p-1}{2} \right) \\ 4 & = \frac{p-1}{2} + \sqrt{p-2.\frac{p-1}{2}} \\ 4 & = \frac{p-1}{2} + \sqrt{1} \\ 4 & = \frac{p-1}{2} + 1 \\ p-1 & = 6 \rightarrow p = 7 \end{align}$
Jadi, nilai $ p = 7 . \heartsuit $
Nomor 20
Fungsi $ f $ dengan $ f(x) = \frac{x^3}{3} - 4x $ akan naik pada interval ....
$\spadesuit \, $ Menentukan turunan
$ f(x) = \frac{x^3}{3} - 4x \rightarrow f^\prime (x) = x^2 - 4 $
$\spadesuit \, $ Syarat fungsi naik : $ f^\prime (x) > 0 $
$\begin{align} f^\prime (x) & > 0 \\ x^2 - 4 & > 0 \\ x^2 & = 4 \rightarrow x = \pm 2 \end{align}$
Jadi, interval naiknya adalah $ HP = \{ x < -2 \vee x > 2 \} . \heartsuit $
$ f(x) = \frac{x^3}{3} - 4x \rightarrow f^\prime (x) = x^2 - 4 $
$\spadesuit \, $ Syarat fungsi naik : $ f^\prime (x) > 0 $
$\begin{align} f^\prime (x) & > 0 \\ x^2 - 4 & > 0 \\ x^2 & = 4 \rightarrow x = \pm 2 \end{align}$
Jadi, interval naiknya adalah $ HP = \{ x < -2 \vee x > 2 \} . \heartsuit $
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