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. and g(x) = -7x-8/x-1
Please help only question on the test i dont know how to do :(

Answer 1

Well for f(g(x)) and g(f(x)) do you understand what this means?

Answer 2

Yea i understand the process, im just not sure with this particular problem, the x's on both denominators and numorators confused me

Answer 3

Okay, I'm not sure if you are missing parenthesis in the two functions. Are these the correct equations?\[f(x) = \frac {x-8} {x+7}\]\[g(x) = \frac{-7x-8}{x-1}\]

Answer 4

Yes sir

Answer 5

Okay so for f(g(x)) what are you getting?

Answer 6

Something like (
-7(x-8/x+7) -8)/((x-8/x+7)-1)

Answer 7

Right, so without pulling anything out or rearranging you basically get\[\frac{\frac{7x + 8}{x - 1}+ 8}{\frac{7x + 8}{x - 1} - 7}\]What you have to do is get ride of the x-1 on the top and bottom by multiplying x-1 so it will cancel. So like this:\[\frac{\frac{7x + 8}{x - 1}+ 8}{\frac{7x + 8}{x - 1} - 7}*\frac{x-1}{x-1}\]Multiply this out in the numerator and denominator and you will find it is indeed \[\frac{-15x}{-15} = x\]

Answer 8

I did actually pull out a negative in the first step, sorry about that but it is still correct.

Answer 9

So now do the same thing, but with g(f(x))?

Answer 10

Essentially, your multiplications will be slightly different. Instead of \[\frac{x-1}{x-1}\]You will have \[\frac{x+7}{x+7}\]multiplied to cancel the secondary fractions out.

Answer 11

Does this make sense?

Answer 12

Yea it makes sence, i just didnt realize how you divided 1 fraction with x over another, like you did above, but now it makes sence

Answer 13

Okay, as long as you understand it. Did you get it to work out correctly?