$\begin{aligned} (fog)(x) &=f\left(g\left(x\right)\right)\\
&=f\left(2x-1\right)\\
&=\left(2x-1\right)^{2}+1\\
&=4x^{2}-4x+1+1\\
&=4x^{2}-4x+2\\
\end{aligned}
$

$\begin{aligned}
(gof)(x)=g\left(f\left(x\right)\right)&=\left(f\left(x\right)\right)^{2}-1=4x^{2}+4x\\
&=\left(f\left(x\right)\right)^{2}=4x^{2}+4x+1\\
&=\left(2x+1\right)^{2}\leftrightarrow f(x)=2x+1\\
\end{aligned}$
Jadi $f(x-2)=2(x-2)+1=2x-3$

$(fog)(x)=x^{2}+3x+1$

$f(x+1)=x^{2}+3x+1$

$f(x+1)=\left(x+1\right)^{2}+\left(x+1\right)-1$

$f(x)=x^{2}+x-1$

$f(x+1)=x^{2}-1$

$f(x+1)=\left(x+1\right)^{2}-2(x+1)$

$f(x)=x^{2}-2x$

$(gof)(x)=g\left(f\left(x\right)\right)=g\left(x^{2}-2x\right)=2\left(x^{2}-2x\right)=2x^{2}-4x$

$(fog)(a)=f(g(a))=a^{2}-3$

$(gof)(a)=g(f(a))=\left(a-3\right)^{2}$

$(fog)(a)=(gof)(a)$

$a^{2}-3$ = $\left(a-3\right)^{2}$

$a^{2}-3=a^{2}-6a+9$

$6a=12\rightarrow a=2$

$g^{-1}(x)=2x\rightarrow g(x)=\frac{x}{2}$

$(f^{-1}og)(x+1)=x-1$

$(f^{-1}og)(x+1)=(x+1)-2$

$(f^{-1}og)(x)=x-2$

$f^{-1}\left(\frac{x}{2}\right)=x-2$

$f^{-1}\left(\frac{x}{2}\right)=2\left(\frac{x}{2}\right)-2$

$f^{-1}(x)=2x-2\rightarrow f(x)=\frac{x+2}{2}=\frac{1}{2}x+1$

$\begin{aligned} (f^{-1}og^{-1}oh^{-1})(x) &=\left(hogof\right)^{-1}(x)\\
&=2x-4\rightarrow(hogof)(x)=\frac{x+4}{2}\\
(hogof)(x) &=\frac{x+4}{2}\\
(hog)(f(x) )&=\frac{f(x)-3}{2f(x)+1}\\
&=\frac{x+4}{2}\\
\end{aligned}
$
Substitusikan $x=8$, maka :

$\frac{f(8)-3}{2f(8)+1}=\frac{8+4}{2}=6$

$f(8)-3=12f(8)+6\rightarrow11f(8)=-9\rightarrow f(8)=-\frac{9}{11}$

$f\left(g\left(x\right)\right)=2x+7$

$3\left(g\left(x\right)\right)^{2}-9\left(g\left(x\right)\right)+14=2x+7$

$3\left(g\left(x\right)\right)^{2}-9\left(g\left(x\right)\right)=2x-7$

$3\left\{ g\left(x\right)-\frac{3}{2}\right\} ^{2}-\frac{27}{4}=2x-7$

$3\left\{ g\left(x\right)-\frac{3}{2}\right\} ^{2}=\frac{8x-1}{4}$

$\left\{ g\left(x\right)-\frac{3}{2}\right\} ^{2}=\frac{8x-1}{12}$

$g\left(x\right)-\frac{3}{2}=\sqrt{\frac{8x-1}{12}}$

$g(x)=\sqrt{\frac{8x-1}{12}}+\frac{3}{2}$

$\left(gof\right)(x)=g(f(x))=g\left(x^{2}\right)=\frac{8}{x^{2}+2}$

Misalkan (gof)(x) = y, maka

$y\left(x^{2}+2\right)=8$

$x^{2}y+2y=8$

$x^{2}y=8-2y$

$x^{2}=\frac{8-2y}{y}\rightarrow x=\sqrt{\frac{8-2y}{y}}$

$\left(gof\right)^{-1}(x)=\sqrt{\frac{8-2x}{x}}$

Domain dari $\left(gof\right)^{-1}(x)$ adalah $\frac{8-2x}{x}\geq0$

Jadi domainnya adalah $0 < x\leq4$

$\left(gof\right)(x)=x^{2}-\frac{3}{2}x+3$

$g\left(2x^{2}-3x+1\right)=x^{2}-\frac{3}{2}x+3$

$g\left(2x^{2}-3x+1\right)=\frac{1}{2}\left(2x^{2}-3x+1\right)+\frac{5}{2}$

$g(x)=\frac{1}{2}x+\frac{5}{2}=\frac{x+5}{2}$

$g(-1)=\frac{-1+5}{2}=2$

$\left(fog\right)(-1)=f\left(g(-1)\right)=f(2)=2\left(2\right)^{2}-3\left(2\right)+1=3$

$\begin{aligned} f\left(\frac{3}{x}\right) &=\frac{^{3}log\left(\frac{3}{x}\right)}{1-2.^{3}log\left(\frac{3}{x}\right)}\\
&=\frac{^{3}log3-^{3}logx}{1-2\left(^{3}log3-^{3}logx\right)}\\
&=\frac{1-^{3}logx}{1-2+2.^{3}logx}\\
&=\frac{1-^{3}logx}{-1+2.^{3}logx}\\
&=\frac{^{3}logx-1}{1-2.^{3}logx}\\
\end{aligned}
$

$\begin{aligned} \mbox{Jadi}\ f(x)+f\left(\frac{3}{x}\right) &=\frac{^{3}logx}{1-2.^{3}logx}+\frac{^{3}logx-1}{1-2.^{3}logx}\\
&=\frac{2.{}^{3}logx-1}{1-2.^{3}logx}\\
&=\frac{-\left(1-2.{}^{3}logx\right)}{1-2.^{3}logx}=-1\\
\end{aligned}
$

$(fog)(x)=f\left(g\left(x\right)\right)=\frac{1}{x-2}\sqrt{x^{2}-4x+5}$

$\sqrt{\left(g\left(x\right)\right)^{2}+1}=\sqrt{\frac{x^{2}-4x+5}{\left(x-2\right)^{2}}}$ (kedua ruas dikuadratkan)

$\left(g\left(x\right)\right)^{2}+1=\frac{x^{2}-4x+5}{\left(x-2\right)^{2}}=\frac{x^{2}-4x+5}{x^{2}-4x+4}$

$\begin{aligned} \left(g(x)\right)^{2} &=\frac{x^{2}-4x+5}{x^{2}-4x+4}-1\\
&=\frac{x^{2}-4x+5}{x^{2}-4x+4}-\left(\frac{x^{2}-4x+4}{x^{2}-4x+4}\right)\\
&=\frac{1}{x^{2}-4x+4}\\
&=\left(\frac{1}{x-2}\right)^{2}\\
\end{aligned}
$

$g(x)=\frac{1}{x-2}$

Jadi $g(x-3)=\frac{1}{(x-3)-2}=\frac{1}{x-5}$

$f(x+3)=x+5$

$f(x+3)=(x+3)+2$

$f(x)=x+2$

$(hogof)\left(x-1\right)=\frac{5x+1}{2x-5}$

$\left(hog\right)\left(f\left(x-1\right)\right)=\frac{5x+1}{2x-5}$

Misal $u=x-1\rightarrow x=u+1$

Ganti variabel x dengan variabel u

$\left(hog\right)\left(f\left(u\right)\right)=\frac{5\left(u+1\right)+1}{2\left(u+1\right)-5}=\frac{5u+6}{2u-3}$

$\left(hogof\right)\left(x\right)=\frac{5x+6}{2x-3}$

* $\left(gof\right)\left(x\right)=g\left(f(x)\right)=g\left(x+2\right)=\frac{(x+2)+1}{2(x+2)-1}=\frac{x+3}{2x+3}$

$\left(hogof\right)\left(x\right)=h\left\{ \left(gof\right)(x)\right\} =h\left(\frac{x+3}{2x+3}\right)=\frac{5x+6}{2x-3}$

Misal $p=\frac{x+3}{2x+3}$

$2px+3p=x+3$

$2px-x=3-3p$

$x\left(2p-1\right)=3-3p$

$x=\frac{3-3p}{2p-1}$

Ganti variabel x dengan variabel p :

$h\left(p\right)=\frac{5\left(\frac{3-3p}{2p-1}\right)+6}{2\left(\frac{3-3p}{2p-1}\right)-3}=\frac{-3p+9}{-12p+9}=\frac{-p+3}{-4p+3}$

$h\left(2a-1\right)=\frac{-\left(2a-1\right)+3}{-4\left(2a-1\right)+3}=\frac{-2a+4}{-8a+7}=10$

$-2a+4=-80a+70$

$78a=66\rightarrow a=\frac{11}{13}$

$f(x)=\frac{1+x}{1-x}$

$(fof)(x)=f\left(\frac{1+x}{1-x}\right)=\frac{1+\left(\frac{1+x}{1-x}\right)}{1-\left(\frac{1+x}{1-x}\right)}=-\frac{1}{x}$

$(fofof)(x)=f\left(fof(x)\right)=f\left(-\frac{1}{x}\right)=\frac{1+\left(-\frac{1}{x}\right)}{1-\left(-\frac{1}{x}\right)}=\frac{x-1}{x+1}$

$(fofofof)(x)=f(fofof)(x)=\frac{1+\left(\frac{x-1}{x+1}\right)}{1-\left(\frac{x-1}{x+1}\right)}=x$

Untuk $x=1$, maka :

$f(1)+1.f(0)=1+1\rightarrow f(1)+f(0)=2$……(1)

Untuk x = 0, maka :

$f\left(0\right)+0.f\left(1\right)=1\rightarrow f(0)=1$ …..(2)

Substitusikan $f(0)=1$ ke pers (1) sehingga diperoleh $f(1)=1$

Karena $f$ terdapat $99$ kali. Selanjutnya fokuskan pada komposisi $f$ yang jumlahnya ganjil.

Lihatlah pada $f(x)$, $(fofof)(x)$, $(fofofof)(x)$, ……Memiliki pola kebalikan negatif. Fokuskan pada komposisi $f$ yang anggotanya negatif. Selanjutnya $f(x)$ kita katakan suku pertama, $(fof)(x)$ suku kedua dan seterusnya.

Adapun rumus untuk komposisi dari $f$ yang anggotanya ganjil yaitu:

$\left(-1\right)^{n+1}\left(\frac{1-x}{1+x}\right)^{\left(-1\right)^{n}}$

Adapaun $(fofofofof………of)$ \{$f$ terdapat $99$ kali) adalah komposisi $f(x)$, $(fofof)(x)$, $(fofofof)(x)$ yang berada pada urutan ke $45$.

Jadi komposisi dari $(fofofofof………of)$ \{$f$ terdapat $99$ kali) = $\left(-1\right)^{n+1}\left(\frac{1-x}{1+x}\right)^{\left(-1\right)^{n}}=\left(-1\right)^{46}\left(\frac{1-x}{1+x}\right)^{\left(-1\right)^{45}}=\frac{1+x}{1-x}$