Answer:
Part 1) [tex]f(x)=\frac{2x-1}{x+2}[/tex] -------> [tex]f^{-1}(x)=\frac{-2x-1}{x-2}[/tex]
Part 2) [tex]f(x)=\frac{x-1}{2x+1}[/tex] -------> [tex]f^{-1}(x)=\frac{-x-1}{2x-1}[/tex]
Part 3) [tex]f(x)=\frac{2x+1}{2x-1}[/tex] -----> [tex]f^{-1}(x)=\frac{x+1}{2(x-1)}[/tex]
Part 4) [tex]f(x)=\frac{x+2}{-2x+1}[/tex] ----> [tex]f^{-1}(x)=\frac{x-2}{2x+1}[/tex]
Part 5) [tex]f(x)=\frac{x+2}{x-1}[/tex] -------> [tex]f^{-1}(x)=\frac{x+2}{x-1}[/tex]
Step-by-step explanation:
Part 1) we have
[tex]f(x)=\frac{2x-1}{x+2}[/tex]
Find the inverse
Let
y=f(x)
[tex]y=\frac{2x-1}{x+2}[/tex]
Exchange the variables x for y and t for x
[tex]x=\frac{2y-1}{y+2}[/tex]
Isolate the variable y
[tex]x=\frac{2y-1}{y+2}\\ \\ xy+2x=2y-1\\ \\xy-2y=-2x-1\\ \\y[x-2]=-2x-1\\ \\y=\frac{-2x-1}{x-2}[/tex]
Let
[tex]f^{-1}(x)=y[/tex]
[tex]f^{-1}(x)=\frac{-2x-1}{x-2}[/tex]
Part 2) we have
[tex]f(x)=\frac{x-1}{2x+1}[/tex]
Find the inverse
Let
y=f(x)
[tex]y=\frac{x-1}{2x+1}[/tex]
Exchange the variables x for y and t for x
[tex]x=\frac{y-1}{2y+1}[/tex]
Isolate the variable y
[tex]x=\frac{y-1}{2y+1}\\ \\2xy+x=y-1\\ \\2xy-y=-x-1\\ \\y[2x-1]=-x-1\\ \\y=\frac{-x-1}{2x-1}[/tex]
Let
[tex]f^{-1}(x)=y[/tex]
[tex]f^{-1}(x)=\frac{-x-1}{2x-1}[/tex]
Part 3) we have
[tex]f(x)=\frac{2x+1}{2x-1}[/tex]
Find the inverse
Let
y=f(x)
[tex]y=\frac{2x+1}{2x-1}[/tex]
Exchange the variables x for y and t for x
[tex]x=\frac{2y+1}{2y-1}[/tex]
Isolate the variable y
[tex]x=\frac{2y+1}{2y-1}\\ \\2xy-x=2y+1\\ \\2xy-2y=x+1\\ \\y[2x-2]=x+1\\ \\y=\frac{x+1}{2(x-1)}[/tex]
Let
[tex]f^{-1}(x)=y[/tex]
[tex]f^{-1}(x)=\frac{x+1}{2(x-1)}[/tex]
Part 4) we have
[tex]f(x)=\frac{x+2}{-2x+1}[/tex]
Find the inverse
Let
y=f(x)
[tex]y=\frac{x+2}{-2x+1}[/tex]
Exchange the variables x for y and t for x
[tex]x=\frac{y+2}{-2y+1}[/tex]
Isolate the variable y
[tex]x=\frac{y+2}{-2y+1}\\ \\-2xy+x=y+2\\ \\-2xy-y=-x+2\\ \\y[-2x-1]=-x+2\\ \\y=\frac{-x+2}{-2x-1} \\ \\y=\frac{x-2}{2x+1}[/tex]
Let
[tex]f^{-1}(x)=y[/tex]
[tex]f^{-1}(x)=\frac{x-2}{2x+1}[/tex]
Part 5) we have
[tex]f(x)=\frac{x+2}{x-1}[/tex]
Find the inverse
Let
y=f(x)
[tex]y=\frac{x+2}{x-1}[/tex]
Exchange the variables x for y and t for x
[tex]x=\frac{y+2}{y-1}[/tex]
Isolate the variable y
[tex]x=\frac{y+2}{y-1}\\ \\xy-x=y+2\\ \\xy-y=x+2\\ \\y[x-1]=x+2\\ \\y=\frac{x+2}{x-1}[/tex]
Let
[tex]f^{-1}(x)=y[/tex]
[tex]f^{-1}(x)=\frac{x+2}{x-1}[/tex]
