Question

Drag the tiles to the boxes to form correct pairs. Not all tiles will be used.

Answer 1

Answer:

Part 1) $f(x)=\frac{2x-1}{x+2}$ -------> $f^{-1}(x)=\frac{-2x-1}{x-2}$

Part 2) $f(x)=\frac{x-1}{2x+1}$ -------> $f^{-1}(x)=\frac{-x-1}{2x-1}$

Part 3) $f(x)=\frac{2x+1}{2x-1}$ -----> $f^{-1}(x)=\frac{x+1}{2(x-1)}$

Part 4) $f(x)=\frac{x+2}{-2x+1}$ ----> $f^{-1}(x)=\frac{x-2}{2x+1}$

Part 5) $f(x)=\frac{x+2}{x-1}$ -------> $f^{-1}(x)=\frac{x+2}{x-1}$

Step-by-step explanation:

Part 1) we have

$f(x)=\frac{2x-1}{x+2}$

Find the inverse  

Let

y=f(x)

$y=\frac{2x-1}{x+2}$

Exchange the variables x for y and t for x

$x=\frac{2y-1}{y+2}$

Isolate the variable y

$x=\frac{2y-1}{y+2}\\ \\ xy+2x=2y-1\\ \\xy-2y=-2x-1\\ \\y[x-2]=-2x-1\\ \\y=\frac{-2x-1}{x-2}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{-2x-1}{x-2}$

Part 2) we have

$f(x)=\frac{x-1}{2x+1}$

Find the inverse  

Let

y=f(x)

$y=\frac{x-1}{2x+1}$

Exchange the variables x for y and t for x

$x=\frac{y-1}{2y+1}$

Isolate the variable y

$x=\frac{y-1}{2y+1}\\ \\2xy+x=y-1\\ \\2xy-y=-x-1\\ \\y[2x-1]=-x-1\\ \\y=\frac{-x-1}{2x-1}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{-x-1}{2x-1}$

Part 3) we have

$f(x)=\frac{2x+1}{2x-1}$

Find the inverse  

Let

y=f(x)

$y=\frac{2x+1}{2x-1}$

Exchange the variables x for y and t for x

$x=\frac{2y+1}{2y-1}$

Isolate the variable y

$x=\frac{2y+1}{2y-1}\\ \\2xy-x=2y+1\\ \\2xy-2y=x+1\\ \\y[2x-2]=x+1\\ \\y=\frac{x+1}{2(x-1)}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{x+1}{2(x-1)}$

Part 4) we have

$f(x)=\frac{x+2}{-2x+1}$

Find the inverse  

Let

y=f(x)

$y=\frac{x+2}{-2x+1}$

Exchange the variables x for y and t for x

$x=\frac{y+2}{-2y+1}$

Isolate the variable y

$x=\frac{y+2}{-2y+1}\\ \\-2xy+x=y+2\\ \\-2xy-y=-x+2\\ \\y[-2x-1]=-x+2\\ \\y=\frac{-x+2}{-2x-1} \\ \\y=\frac{x-2}{2x+1}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{x-2}{2x+1}$

Part 5) we have

$f(x)=\frac{x+2}{x-1}$

Find the inverse  

Let

y=f(x)

$y=\frac{x+2}{x-1}$

Exchange the variables x for y and t for x

$x=\frac{y+2}{y-1}$

Isolate the variable y

$x=\frac{y+2}{y-1}\\ \\xy-x=y+2\\ \\xy-y=x+2\\ \\y[x-1]=x+2\\ \\y=\frac{x+2}{x-1}$

Let

$f^{-1}(x)=y$

$f^{-1}(x)=\frac{x+2}{x-1}$