Kak, Bantu aq ngerjain tugas ini dong, please banget kak..!!! :'( 1.f(x) = x2 - 2 dan g(x) = 2x + 1, maka komposisi f(g(x)) = ... 2. Jika f(g(x))= -x+3 maka f(x

1.
[tex]f(x)=x^2-2 \\ g(x)=2x+1 \\ f(g(x)) = f(2x+1) \\ f(g(x))=(2x+1)^2-2 \\ f(g(x))=4x^2+4x+1-2 \\ f(g(x))=4x^2+4x-1[/tex]

2.
Deskripsi soal kurang lengkap

3.
[tex]f(g(x))=-x \\ g(x)=\frac{x+1}{x-1} \\ f(\frac{x+1}{x-1})=-x \\ f(x)=-(\frac{x+1}{x-1}) \\ f(x)=\frac{x+1}{1-x}[/tex]

4.
[tex](3f-2g)(x)=3f(x)-2g(x) \\ (3f-2g)(x)=3(2x-1)-2(x^2) \\ (3f-2g)(x)=-2x^2+6x-3[/tex]

5.
[tex]f(x)=2x+4 \\ (f\circ g)(x)=x^2-4x-7 \\ 2g(x)+4=x^2-4x-7 \\ g(x)=x^2-4x-11 \\ y=x^2-4x-11 \\ y=(x-2)^2-15 \\ (x-2)^2=y+15 \\ x-2=\pm\sqrt{y+15} \\ x=2\pm\sqrt{y+15} \\ g^{-1}(x)=2\pm\sqrt{x+15}[/tex]

6.
[tex]\displaystyle f(x)=\frac{3x+4}{2x-1} \\ \\ y=\frac{3x+4}{2x-1} \\ \\ 2xy-y=3x+4 \\ 2xy-3x=y+4 \\ x(2y-3)=y+4 \\ x=\frac{y+4}{2y-3} \\ \\ f^{-1}(x)=\frac{x+4}{2x-3},x\neq \frac{3}{2}[/tex]

7.
[tex]f(x+1)=6x+5 \\ f(x)=6(x-1)+5 \\ f(x)=6x-6+5 \\ f(x)=6x-1 \\ y=6x-1 \\ 6x=y+1 \\ x=\frac{1}{6}(y+1) \\ f^{-1}(x)=\frac{1}{6}(x+1)[/tex]