Jika f(1/x+1)=x+3/x+1, maka nilai a-3 agar f^-1(a+1)=2=

[tex]f( \frac{1}{x} + 1) = \frac{x + 3}{x + 1} \\ y = \frac{x + 3}{x + 1} \\ y \times (x + 1) = x + 3 [/tex]
[tex]xy + y = x + 3 \\ y - 3 = x - xy \\ x(1 - y) = y - 3[/tex]
[tex] {f}^{ - 1} (x) = \frac{ \frac{1}{x} + 1 - 3}{1 - ( \frac{1}{x} + 1) } \\ = \frac{ \frac{1}{x} - 2}{ \frac{1}{x} }[/tex]
[tex] = ( \frac{1}{x} - 2)x = 1 - 2x[/tex]
[tex] {f}^{ - 1} (a + 1) = 2 \\ 1 - 2(a + 1) = 2 \\ - 2a - 1 = 2 \\ - 2a = 3 \\ a = \frac - {3}{6} [/tex]
[tex]a = - \frac{3}{2} [/tex]
[tex]a - 3 = - \frac{3}{2} - 3 = - 4 \frac{1}{2} [/tex]