Question

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

f(x) = quantity x minus seven divided by quantity x plus two. and g(x) = quantity negative two x minus seven divided by quantity x minus one.

Answer 1

Please help guys!!

Answer 2

-2x-7

Answer 3

Ah, just noticed that.
\[f(x)=\frac{x-7}{x+2}~~~~~g(x)=\frac{-2x-7}{x-1}\]

Answer 4

ya thats right ^

Answer 5

Alright, so we need to take the whole function of g(x) and plug it into every single x within f(x). Bit tricky, so it just means we need to be careful. Alright, so this is the first set-up

\[\frac{ \frac{ -2x-7 }{ x-1 }-7 }{ \frac{ -2x-7 }{ x-1 }+2 }\]So what I would do is make the top portion and the bottom portion all into one fraction. This means I need a common denominator. So to do this, I'll take x-1 as a common denominator and multiply it up into the numerator and the imaginary 1 denominator of -7 and 2. Doing this I have:
\[\frac{ \frac{ -2x-7-7(x-1) }{ x-1 } }{ \frac{ -2x-7+2(x-1) }{ x-1 } }\]
From here if I divide the top fraction and the bottom fraction, (x-1) will cancel out dueto me flipping the bottom fraction and multiplying. So that leaves me with:
\[\frac{ -2x-7-7(x-1) }{ -2x-7+2(x-1) }\]
Now if I multiply everything out and combine like terms I will have:
\[\frac{ -2x-7-7x+7 }{ -2x-7+2x-2 } = \frac{ -9x }{ -9 }= x\]

So that is the first part. We have to check both ways to confirm they are actually inverses of each other, though. Take a look at this part and see if you can make sense of what I did.

Answer 6

okay I think I get what you did

Answer 7

Alrighty, cool. Now the second part. I plug f(x) into every single x within g(x) and see if I can simplify it down to x. So here is the set-up:
\[\frac{ -2(\frac{ x-7 }{ x+2 })-7 }{ (\frac{ x-7 }{ x+2 })-1 }\]Now for this I'm going to do the same thing, multiply the denominator of x+2 into the numerators and the imaginary 1 denominators of -7 and -1

Answer 8

\[\frac{ -2\frac{ (x-7)-7(x+2) }{ x+2 } }{ \frac{ x-7-1(x+2) }{ x+2 } }\]Once again, if I flip and multiply, the x+2s on bottom will cancel out, giving me:
\[\frac{ -2(x-7)-7(x+2) }{ x-7-1(x+2) }\]

Answer 9

Now I just multiply things out and simplify:
\[\frac{ -2x+14-7x-14 }{ x-7-x-2 }=\frac{ -9x }{ -9 }=x\]

Answer 10

Thank you so much! I finally understand haha thanks

Answer 11

Yeah, tricky algebra, easy to mess up xD Glad you get it ^_^