Math QuizzzzDiberikan fungsi dengan x ≠ 1. Jika , tentukanlah nilai dari !Jawab jangan ngasal, klu.... Question from @Adjie564

Jawab:

Penjelasan dengan langkah-langkah:

$\displaystyle f(x)=\frac{1+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}}{x-1}\\misal\:\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}=A\\Perhatikan\:bahwa :\\\sqrt{x+\underbrace{\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}_\text{A}}=A\\\sqrt{x+A}=A\\x+A={A}^{2}\\{A}^{2}-A-x=0\\\\Dengan\: rumus \: abc : \\A=\frac{1\pm\sqrt{1-4\times(-x)}}{2(1)} \\\\A=\frac{1\pm\sqrt{1+4x}}{2} \\,\:karena\:A\:pasti\:selalu\:positif,\:maka\\A=\frac{1+\sqrt{1+4x}}{2}$

Sehingga f(x) menjadi :

$\displaystyle f(x)=\frac{1+\frac{1+\sqrt{1+4x}}{2}}{x-1},x\geq \frac{-1}{4}\\mencari\:invers\:dari\:g(f(x)), misal\:g(f(x))=y\\y= ^2\log{(x^2+4)},syarat\:numerus\rightarrow\:x^2+4>0\rightarrow x\in R\\{2}^{y}=x^2+4\\x=\pm\sqrt{2^y-4}\\(g o f)^{-1} =\pm\sqrt{2^x-4}\\karena\:x\geq \frac{-1}{4},maka\:(g o f)^{-1} =\sqrt{2^x-4}$ $\displaystyle f(x)=\frac{1+\frac{1+\sqrt{1+4x}}{2}}{x-1},x\geq \frac{-1}{4},x\neq 1\\mencari\:invers\:dari\:g(f(x)), misal\:g(f(x))=y\\y= ^2\log{(x^2+4)},syarat\:numerus\rightarrow\:x^2+4>0\rightarrow x\in R\\{2}^{y}=x^2+4\\x=\pm\sqrt{2^y-4}\\(g\:o\:f)^{-1}(x)=\pm\sqrt{2^x-4}\\karena\:x\geq \frac{-1}{4},maka\:(g\:o\:f)^{-1}(x) =\sqrt{2^x-4}\\(g\:o\:f)^{-1}(x)=f^{-1}(g^{-1}(x))\\f^{-1}(g^{-1}(x))=\sqrt{2^x-4}\\f^{-1}(g^{-1}(3))=\sqrt{2^3-4}\\f^{-1}(g^{-1}(3))=2$

$\displaystyle karena\:f^{-1}(g^{-1}(3))=2,maka\\\\f(2)=g^{-1}(3)\\g^{-1}(3)=f(2)=\frac{1+\frac{1+\sqrt{1+4\times2}}{2}}{2-1}\\g^{-1}(3)=f(2)=\frac{1+\frac{1+3}{2}}{1}\\g^{-1}(3)=f(2)=3\\g^{-1}(3)=3$