Question

Sorry if this is too much but I'm desperate right now.

Answer 1

Answer:

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

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

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

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

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

Step-by-step explanation:

1.

$f(x)=y=\dfrac{2x-1}{x+2}\\ \\y(x+2)=2x-1\\ \\yx+2y=2x-1\\ \\yx-2x=-1-2y\\ \\x(y-2)=-1-2y\\ \\x=\dfrac{-1-2y}{y-2}\\ \\y=f^{-1}(x)=\dfrac{-2x-1}{x-2}$

2.

$f(x)=y=\dfrac{x-1}{2x+1}\\ \\y(2x+1)=x-1\\ \\2xy+y=x-1\\ \\2xy-x=-1-y\\ \\x(2y-1)=-1-y\\ \\x=\dfrac{-1-y}{2y-1}\\ \\y=f^{-1}(x)=\dfrac{-x-1}{2x-1}$

3.

$f(x)=y=\dfrac{x+2}{-2x+1}\\ \\y(-2x+1)=x+2\\ \\-2xy+y=x+2\\ \\-2xy-x=2-y\\ \\x(-2y-1)=2-y\\ \\x=\dfrac{2-y}{-2y-1}\\ \\y=f^{-1}(x)=\dfrac{2-x}{-2x-1}=\dfrac{x-2}{2x+1}$

4.

$f(x)=y=\dfrac{2x+1}{2x-1}\\ \\y(2x-1)=2x+1\\ \\2xy-y=2x+1\\ \\2xy-2x=1+y\\ \\x(2y-2)=1+y\\ \\x=\dfrac{1+y}{2y-2}\\ \\y=f^{-1}(x)=\dfrac{1+x}{2(x-1)}$

5.

$f(x)=y=\dfrac{x+2}{x-1}\\ \\y(x-1)=x+2\\ \\xy-y=x+2\\ \\xy-x=2+y\\ \\x(y-1)=2+y\\ \\x=\dfrac{2+y}{y-1}\\ \\y=f^{-1}(x)=\dfrac{x+2}{x-1}$