QUESTION 1
We have [tex]f(x)=\frac{2x-1}{x+2}[/tex]
Let [tex]y=\frac{2x-1}{x+2}[/tex]
Interchange x and y.
[tex]x=\frac{2y-1}{y+2}[/tex]
Solve for y.
First, cross multiply;
[tex]x(y+2)=2y-1[/tex]
Expand now:
[tex]xy+2x=2y-1[/tex]
Group the y-terms on the LHS
[tex]xy-2y=-2x-1[/tex]
Factor y on the left hand side;
[tex](x-2)y=-2x-1[/tex]
Divide both sides by (x-2).
[tex]y=\frac{-2x-1}{x-2}[/tex]
[tex]f^{-1}(x)=\frac{-2x-1}{x-2}[/tex]
[tex]\boxed{f(x)=\frac{2x-1}{x+2}\to f^{-1}(x)=\frac{-2x-1}{x-2}}[/tex]
QUESTION 2
Given: [tex]f(x)=\frac{x-1}{2x+1}[/tex]
Let [tex]y=\frac{x-1}{2x+1}[/tex]
Interchange x and y.
[tex]x=\frac{y-1}{2y+1}[/tex]
Solve for y
[tex]x(2y+1)=y-1[/tex]
[tex]2xy+x=y-1[/tex]
[tex]2xy-y=-x-1[/tex]
[tex](2x-1)y=-x-1[/tex]
[tex]y=\frac{-x-1}{2x-1}[/tex]
[tex]f^{-1}(x)=\frac{-x-1}{2x-1}[/tex]
[tex]f^{-1}(x)=\frac{-x-1}{2x-1}[/tex]
[tex]\boxed{f(x)=\frac{x-1}{2x+1}\to f^{-1}(x)=\frac{-x-1}{2x-1}}[/tex]
QUESTION 3
Given : [tex]f(x)=\frac{2x+1}{2x-1}[/tex]
We let [tex]y=\frac{2x+1}{2x-1}[/tex]
Interchange x and y.
[tex]x=\frac{2y+1}{2y-1}[/tex]
Solve for y;
[tex]x(2y-1)=2y+1[/tex]
[tex]2xy-x=2y+1[/tex]
[tex]2xy-2y=x+1[/tex]
[tex](2x-2)y=x+1[/tex]
[tex]y=\frac{x+1}{2x-2}[/tex]
[tex]f^{-1}(x)=\frac{x+1}{2(x-1)}[/tex]
[tex]\boxed{f(x)=\frac{2x+1}{2x-1}\to f^{-1}(x)=\frac{x+1}{2(x-1)}}[/tex]
