Question

Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

f(x) = (x-7)/(x+2)
g(x)= (-2x-7)/(x-1)

Answer 1

for f(g(x)), plug in "g(x)" wherever you see "x" in the function f(x):
\[\Large f(\color{red}{x})=\frac{\color{red}{x}-7}{\color{red}{x}+2} \]
\[\Large f(\color{red}{g(x)})=\frac{\color{red}{g(x)}-7}{\color{red}{g(x)}+2} =\frac{\color{red}{\frac{-2x-7}{x-1}}-7}{\color{red}{\frac{-2x-7}{x-1}}+2}\]

Answer 2

And do the same type of procedure for g(f(x)):
plug in "f(x)" wherever you see "x" in g(x)

Answer 3

So then g(f(x)) would be (f(x)-7)/(f(x)-1)

Answer 4

But wouldn't that mean thatI'm saying -7/2 is the inverse of -9/1?

Answer 5

g(f(x)) would be (-2f(x)-7)/(f(x)-1)

Answer 6

I'm not sure what you mean by your second question

Answer 7

I mean, how does that prove that f and g are inverses?

Answer 8

well when you reduce your expression, it should give you just "x"

Answer 9

I tried that, I can't seem to figure out how to get just x.

Answer 10

If g is an inverse of f, you could write this as \(f^{-1}\), so, \(f(g(x))=f(f^{-1}(x))=x\)

Answer 11

But I need to prove that it's an inverse, I can't just assume that. How do I reduce the f(g(x)) to get x?

Answer 12

\[\Large f(\color{red}{g(x)})=\frac{\color{red}{\frac{-2x-7}{x-1}}-7}{\color{red}{\frac{-2x-7}{x-1}}+2}=\frac{\frac{-2x-7}{x-1}-\frac{7(x-1)}{x-1}}{\frac{-2x-7}{x-1}+\frac{2(x-1)}{x-1}}=\\ ~ \\ =\Large\frac{\frac{-2x-7-7(x-1)}{x-1}}{\frac{-2x-7+2(x-1)}{x-1}}=\frac{-2x-7-7x+7}{-2x-7+2x-2}=\frac{-9x}{-9}=x \]

Answer 13

And for g(f(x))?

Answer 14

It will be the same idea as above.. just try and follow what I did in those steps and you should be able to do the same ones for g(f(x))

Answer 15

Thanks :)

Answer 16

:)