Q:

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = quantity x minus nine divided by quantity x plus five. and g(x) = quantity negative five x minus nine divided by quantity x minus one.

Accepted Solution

A:
We are given the functions f and g with the rules:

[tex]\displaystyle{ f(x)= \frac{x-9}{x+5} [/tex]

and 

[tex]\displaystyle{ g(x)= \frac{-5x-9}{x-1} [/tex].


First, we prove that f(g(x))=x:

[tex]\displaystyle{ f(g(x))= \frac{g(x)-9}{g(x)+5}=\frac{\frac{-5x-9}{x-1}-9}{\frac{-5x-9}{x-1}+5}=\frac{\frac{-5x-9}{x-1}-9\cdot\frac{x-1}{x-1}}{\frac{-5x-9}{x-1}+5\cdot\frac{x-1}{x-1}}\\\\ [/tex]

[tex]=\displaystyle { \frac{ \frac{-5x-9-9x+9}{x-1} }{\frac{-5x-9+5x-5}{x-1}}= \frac{-5x-9-9x+9}{-5x-9+5x-5} = \frac{-14x}{-14}=x. [/tex]


Similarly, we prove that g(f(x))=x:

[tex]\displaystyle{ g(f(x))= \frac{-5f(x)-9}{f(x)-1}= \frac{-5\cdot \frac{x-9}{x+5}-9\cdot \frac{x+5}{x+5}}{\frac{x-9}{x+5}-\frac{x+5}{x+5}}= \frac{-5(x-9)-9(x+5)}{x-9-(x+5)} [/tex]

[tex]=\displaystyle{ \frac{-5x+45-9x-45}{x-9-x-5}= \frac{-14x}{-14}=x. [/tex]