Question

dfsd

Answer 1

I've been stuck on this one for hours! :(

Answer 2

@mathmale

Answer 3

You're suppose to check that when you plug g(x) into f(x) it equals x and vice versa. f(g(x)) would be: f(-7x-8/x-1)=(-7x-8)/(x-1)-8/(-7x-8)(x-1)+7\[f \left(\begin{matrix}-7x-8 \\ x-1\end{matrix}\right)=\frac{\left(\begin{matrix}-7x-8 \\ x-1\end{matrix}\right)-8}{\left(\begin{matrix}-7x-8 \\ x-1\end{matrix}\right)+7}\]
Which if you simplify it completely it equals x. You can check g(f(x) in exactly the same way.
\[g \left(\begin{matrix}x-8 \\ x+7\end{matrix}\right)=\frac{ -7\left(\begin{matrix}x-8 \\ x+7\end{matrix}\right)-8 }{ \left(\begin{matrix}x-8 \\ x+7\end{matrix}\right) -1}\]
If you need the steps I can do those too.

Answer 4

Simplify f(x) and g(x) first:

\[f(x) = \frac{x - 8}{x + 7}\] \(= \dfrac{x + 7 - 15}{x + 7}\\= \dfrac{x + 7}{x + 7} - \dfrac{15}{x + 7} \\= 1 - \dfrac{15}{x + 7}\)

Answer 5

\[g(x) = \frac{-7x - 8}{x - 1}\]

\(= \dfrac{-7x + 7}{x - 1} -\dfrac{15}{x - 1} \\ = -7 - \dfrac{15}{x - 1}\)

Answer 6

Then input the respective functions

\[f(g(x)) = 1 - \dfrac{15}{g(x) + 7}\]


\[g(f(x)) = -7 - \frac{15}{f(x) - 1}\]