$y=5x+4$

$y-4=5x$

$x=\frac{y-4}{5}$

Jadi $f^{-1}(x)=\frac{x-4}{5}$

$y=\sqrt{x}+1$

$\sqrt{x}=y-1$ (kuadratan kedua ruas)

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

Jadi $f^{-1}(x)=x^{2}-2x+1$ untuk $\geq1$

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

$y-2=\left(1-x^{3}\right)^{\frac{1}{5}}$ (kedua ruas dipangkatkan 5)

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

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

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

Jadi invers $f^{-1}(x)=\left\{ 1-\left(x-2\right)^{5}\right\} ^{\frac{1}{3}}$

$f(x)=2x^{\frac{2}{3}}$ dan $g(x)=x^{\frac{3}{2}}$

$\begin{aligned} (gof)(x) &=g\left(f\left(x\right)\right)\\
&=g\left(2x^{\frac{2}{3}}\right)\\
&=\left(2x^{\frac{2}{3}}\right)^{\frac{3}{2}}\\
&=2^{\frac{3}{2}}\cdot x\\
&=2\sqrt{2}x\\
\end{aligned}
$

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

$(gof)^{-1}(\sqrt{2})=\frac{\sqrt{2}}{2\sqrt{2}}=\frac{1}{2}$

$y=3^{x}-10$

$y+10=3^{x}$

$x=^{3}log\left(y+10\right)$

Jadi $f^{-1}\left(x\right)=^{3}log\left(x+10\right)$

$f(x)=x^{3}+2x^{2}+4x+8$

$y=\left(x+2\right)^{3}$

$x+2=\sqrt[3]{y}$

$x=\sqrt[3]{y}-2$

Jadi $f^{-1}(x)=\sqrt[3]{x}-2$

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

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

Ingat bentuk $\left(gof\right)^{-1}(x)=(f^{-1}og^{-1})(x)$

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

$f(x)=10^{x}\rightarrow f^{-1}(x)=^{10}logx$

$f^{-1}(g(x))=f^{-1}\left(^{10}logx^{2}\right)=^{10}log\left(^{10}logx^{2}\right)$

$y=\frac{9}{\sqrt{2x+1}}$ (kuadratkan kedua ruas)

$y^{2}=\frac{81}{2x+1}$

$2xy^{2}+y^{2}=81$

$2xy^{2}=81-y^{2}$

$x=\frac{81-y^{2}}{2y^{2}}$

$f^{-1}(x)=g(x)=\frac{81-x^{2}}{2x^{2}}$

Jadi $g(3)=\frac{81-3^{2}}{2\left(3\right)^{2}}=\frac{72}{18}=4$

Misal $y=\frac{1}{2}x+1\rightarrow x=2y-2$

$f(\frac{1}{2}x+1)=3x+2\rightarrow f(y)=3\left(2y-2\right)+2=6y-4$

$f(x)=6x-4$

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

$g(3x+2)=x-1\rightarrow f(p)=\frac{p-2}{3}-1=\frac{p-5}{3}$

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

$\left(hogof\right)^{-1}(x)=x-1\rightarrow\left(hogof\right)(x)=x+1$

$\left(gof\right)\left(x\right)=g\left(6x-4\right)=\frac{(6x-4)-5}{3}=\frac{6x-9}{3}=2x-3$

$\left(hogof\right)(x)=h\left(\left(gof\right)\left(x\right)\right)=h\left(2x-3\right)=x+1$

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

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

$\left\{ fo\left(f^{-1}og^{-1}\right)^{-1}o\left(gof\right)^{-1}oh\right\} $

$=\left\{ fogofof^{-1}og^{-1}oh\right\} (x-3)$

$=x^{2}-4x+10$

Ingat sifat invers dan sifat identitas $fof^{-1}=gog^{-1}=I$

$\left\{ fogo(fof^{-1})og^{-1}oh\right\} (x-3)$

$=\left\{ fogo(I)og^{-1}oh\right\} (x-3)$

$=\left\{ fo(gog^{-1})oh\right\} (x-3)$

$=\left(foh\right)\left(x-3\right)$

$=x^{2}-4x+10$

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

$\left(foh\right)(x)=x^{2}+2x+7$

$2\left(h\left(x\right)\right)+3=x^{2}+2x+7$

$2h\left(x\right)=x^{2}+2x+4$

$h\left(x\right)=\frac{1}{2}x^{2}+x+2$

$\left(f^{-1}og^{-1}oh^{-1}\right)^{-1}(a)=\left(hogof\right)(a)$

Ingat bentuk khusus $f(x)=\frac{ax+b}{cx+d}$, maka $f^{-1}(x)=\frac{-dx+b}{cx-a}$

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

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

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

$\begin{aligned} \left(gof\right)(x) &=g\left(f(x)\right)\\
&=g\left(\frac{x+1}{2x-1}\right)\\
&=\frac{2\left(\frac{x+1}{2x-1}\right)+2}{\left(\frac{x+1}{2x-1}\right)-1}\\
&=\frac{\frac{2x+2+4x-2}{2x-1}}{\frac{x+1-2x+1}{2x-1}}\\
&=\frac{6x}{2-x}\\
\end{aligned}$

$\begin{aligned} \left(hogof\right)(a) &=h\left(gof(a)\right)\\
&=h\left(\frac{6a}{2-a}\right)\\
&=\frac{\left(\frac{6a}{2-a}\right)+1}{\left(\frac{6a}{2-a}\right)+2}\\
&=\frac{\frac{6a+2-a}{2-a}}{\frac{6a+4-2a}{2-a}}\\
&=\frac{5a+2}{4a+4}=\frac{7}{8}\\
\end{aligned}$

$\frac{5a+2}{4a+4}=\frac{7}{8}$

$\frac{5a+2}{a+1}=\frac{7}{2}\rightarrow10a+4=7a+7\rightarrow3a=3\rightarrow a=1$

$f(x)=\sqrt{3}^{\sqrt{3}log\left(3^{3x-3}+3\right)}$

$f(x)=3^{3x-3}+3$

$y-3=\frac{3^{3x}}{3}$

$3y-9=3^{3x}$

$3x=^{3}log\left(3y-9\right)$

$x=\frac{^{3}log\left(3y-9\right)}{3}$

$f^{-1}(x)=\frac{^{3}log\left(3x-9\right)}{3}$

$f^{-1}(6)=\frac{^{3}log\left(18-9\right)}{3}=\frac{^{3}log9}{3}=\frac{2}{3}$

$y=3+\sqrt{2+^{5}log\left(x^{2}-4x+6\right)}$

$y-3=\sqrt{2+^{5}log\left(x^{2}-4x+6\right)}$

$\left(y-3\right)^{2}=2+^{5}log\left(x^{2}-4x+6\right)$

$\left(y-3\right)^{2}-2={}^{5}log\left(x^{2}-4x+6\right)$

$5^{\left(y-3\right)^{2}-2}=x^{2}-4x+6$

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

$\left(x-2\right)^{2}=5^{\left(y-3\right)^{2}-2}-2$

$x-2=\pm\sqrt{5^{\left(y-3\right)^{2}-2}-2}$

$x=2\pm\sqrt{5^{\left(y-3\right)^{2}-2}-2}$

$f^{-1}(x)=2\pm\sqrt{5^{\left(x-3\right)^{2}-2}-2}=2\pm\sqrt{5^{x^{2}-6x+7}-2}$

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

$2y=a^{x}+\frac{1}{a^{x}}=\frac{a^{2x}+1}{a^{x}}$

$2a^{x}y=a^{2x}+1$

$2a^{x}y-a^{2x}-1=0\rightarrow a^{2x}-2ya^{x}+1=0$

Gunakan rumus abc untuk mencari $a^{x}$

$a_{1,2}^{x}=\frac{2y\pm\sqrt{4y^{2}-4}}{2}=\frac{2y\pm2\sqrt{y^{2}-1}}{2}=y\pm\sqrt{y^{2}-1}$

$a^{x}=y\pm\sqrt{y^{2}-1}$

$x=^{a}log\left(y\pm\sqrt{y^{2}-1}\right)$

$f^{-1}(x)=^{a}log\left(x\pm\sqrt{x^{2}-1}\right)$