Question

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = (x-8)/(x+7) g(x) = (-7x-8)/(x-1) I'm running out of time. I'd appreciate if someone could just help me with this.

Answer 1

$f(x)=\frac{x-8}{x+7} \\ g(x)=\frac{-7x-8}{x-1} \\ \\ f(g(x))=\frac{\frac{-7x-8}{x-1}-8}{\frac{-7x-8}{x-1}+7}=\frac{\frac{-7x-8}{x-1}-\frac{8(x-1)}{x-1}}{\frac{-7x-8}{x-1}+\frac{7(x-1)}{x-1}}=\frac{\frac{-7x-8}{x-1}-\frac{8x-8}{x-1}}{\frac{-7x-8}{x-1}+\frac{7x-7}{x-1}}=\frac{\frac{-7x-8-8x+8}{x-1}}{\frac{-7x-8+7x-7}{x-1}}= \\ =\frac{\frac{-13x}{x-1}}{\frac{-13}{x-1}}=\frac{-13x}{(x-1)} \times \frac{(x-1)}{-13}=x \\ \Downarrow \\ f(g(x))=x \\ \checkmark$

$g(f(x))=\frac{-7(\frac{x-8}{x+7})-8}{\frac{x-8}{x+7}-1}=\frac{\frac{-7(x-8)}{x+7}-\frac{8(x+7)}{x+7}}{\frac{x-8}{x+7}-\frac{1(x+7)}{x+7}}=\frac{\frac{-7x+56}{x+7}-\frac{8x+56}{x+7}}{\frac{x-8}{x+7}-\frac{x+7}{x+7}}= \\ =\frac{\frac{-7x+56-8x-56}{x+7}}{\frac{x-8-x-7}{x+7}}=\frac{\frac{-15x}{x+7}}{\frac{-15}{x+7}}=\frac{-15x}{(x+7)} \times \frac{(x+7)}{-15}=x \\ \Downarrow \\ g(f(x))=x \\ \checkmark$