1,2,3 caranya gimana ya?

1.
[tex]\displaystyle h(x)=\frac{g(x)}{f(x)}=\frac{2x+\sqrt{x}-6}{\sqrt{x}+2}=\frac{(\sqrt{x}+2)(2\sqrt{x}-3)}{\sqrt{x}+2} \\ h(x)=2\sqrt{x}-3 \\ 2\sqrt{x}=h(x)+3 \\ \sqrt{x}=\frac{h(x)+3}{2} \\ x=\left({\frac{h(x)+3}{2}}\right)^2 \\ x=\frac{h^2(x)+6h(x)+9}{4} \\ $Maka,$ \\ h^{-1}(x)=\frac{x^2+6x+9}{4}[/tex]

2.
[tex]h(x)=2x+1 \\ (f\circ h)(x)=8x^2+2x+10 \\ a. \\ f(2x+1)=8x^2+2x+10 \\ f(2x)=8(x-1)^2+2(x-1)+10 \\ f(x)=8(\frac{x-1}{2})^2+2(\frac{x-1}{2})+10 \\ f(x)=8\times\frac{x^2-2x+1}{4}+x-1+10 \\ f(x)=2(x^2-2x+1)+x-1+10 \\ f(x)=2x^2-4x+2+x-1+10 \\ f(x)=2x^2-3x+11 [/tex]
[tex]b. \\ y=2x^2-3x+11 \\ \frac{y}{2}=x^2-\frac{3}{2}x+\frac{11}{2} \\ \frac{y}{2}+\frac{9}{16}=x^2-2\times\frac{3}{4}x+\frac{9}{16}+\frac{11}{2} \\ \frac{y}{2}+\frac{9}{16}=(x-\frac{3}{4})^2+\frac{11}{2} \\ (x-\frac{3}{4})^2=\frac{y}{2}+\frac{9}{16}-\frac{11}{2} \\ (x-\frac{3}{4})^2=\frac{y}{2}-\frac{79}{16} \\ x-\frac{3}{4}=\pm\sqrt{\frac{y}{2}-\frac{79}{16}} \\ x=\frac{3}{4}\pm\sqrt{\frac{y}{2}-\frac{79}{16}}[/tex]

3.
[tex]a. \\ g^{-1}(x+4)=x+6 \\ g^{-1}(x)=x-4+6 \\ g^{-1}(x)=x+2 \\ x=g^{-1}(x)-2 \\ g(x)=x-2 \\ \\ (f\circ g)(x)=x^2-4x+8 \\ f(x-2)=x^2-4x+8 \\ f(x)=(x+2)^2-4(x+2)+8 \\ f(x)=x^2+4x+4-4x-8+8 \\ f(x)=x^2+4 \\ x^2=f(x)-4 \\ x=\pm\sqrt{f(x)-4} \\ f^{-1}(x)=\pm\sqrt{x-4}[/tex]
[tex]b. \\ (g\circ f)(x)=g(f(x)) \\ (g\circ f)(x)=f(x)-2 \\ (g\circ f)(x)=(x^2+4)-2 \\ (g\circ f)(x)=x^2+2 \\ \\ x^2=(g\circ f)(x)-2 \\ x=\pm\sqrt{(g\circ f)(x)-2} \\ (g\circ f)^{-1}(x)=\pm\sqrt{x-2}[/tex]