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Mathematics 18 Online
OpenStudy (anonymous):

Please help? :C Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

OpenStudy (anonymous):

\[\frac{ \frac{ -3x -7 }{ x-1 } -7}{x+3 }\]

OpenStudy (anonymous):

Sorry i do not know give it some goodies so that people want it like if you answer it i will fan you and give you a medal

OpenStudy (anonymous):

That's poopy. :p

OpenStudy (anonymous):

@whpalmer4

OpenStudy (anonymous):

@jdoe0001 Please?

OpenStudy (anonymous):

It might help if I put the question correctly... \[f(x)=\frac{ x-7 }{ x+3 } g(x)=\frac{ -3x-7 }{ x-1 }\]

OpenStudy (anonymous):

@sourwing

OpenStudy (jdoe0001):

lemme do the first one, that is f( g(x) ) = x

OpenStudy (anonymous):

well just plug g in for x of f, then simply and do the other way around

OpenStudy (anonymous):

Thank you. :) I can't seem to figure out how to get past (-10x)/(x-1)/(x+3).

OpenStudy (anonymous):

I did, but I can't get it to the point where it equals x...

OpenStudy (jdoe0001):

one sec

OpenStudy (jdoe0001):

\(\bf f(x)=\cfrac{ x-7 }{ x+3 } \qquad {\color{red}{ g(x)}}=\cfrac{ -3x-7 }{ x-1 } \\ \quad \\ \quad \\ f(\quad g(x)\quad )=\cfrac{ ({\color{red}{ \frac{ -3x-7 }{ x-1 }}})-7 }{ ({\color{red}{ \frac{ -3x-7 }{ x-1 }}})+3 }\implies \cfrac{ \frac{ -3x-7 }{ x-1 }-7 }{ \frac{ -3x-7 }{ x-1 }+3 } \\ \quad \\ f(\quad g(x)\quad )=\cfrac{\frac{-3x-7-7(x-1)}{x-1}}{\frac{-3x-7+3(x-1)}{x-1}}\implies \cfrac{\frac{-3x-7-7x+7}{x-1}}{\frac{-3x-7+3x-3}{x-1}} \\ \quad \\ f(\quad g(x)\quad )=\cfrac{-3x\cancel{-7}-7x\cancel{+7}}{\cancel{x-1}}\cdot \cfrac{\cancel{x-1}}{\cancel{-3x}-7\cancel{+3x}-3}\) see what that gives you

OpenStudy (anonymous):

Oh.. You put it where both x's are. Hang on.

OpenStudy (jdoe0001):

yes

OpenStudy (anonymous):

Okay, so you wind up with -10x/-10, which simplifies down to just x. Right?

OpenStudy (whpalmer4):

yes, \[\frac{-10x}{-10} = x\]

OpenStudy (jdoe0001):

yeap

OpenStudy (jdoe0001):

so \(\bf {\color{red}{ f(x)}}=\cfrac{ x-7 }{ x+3 } \qquad g(x)=\cfrac{ -3x-7 }{ x-1 } \\ \quad \\ \quad \\ g(\quad f(x)\quad )=\cfrac{ -3({\color{red}{ \frac{ x-7 }{ x+3 }}})-7 }{ ({\color{red}{ \frac{ x-7 }{ x+3 }}})-1 }\implies \cfrac{ \frac{ -3x+21 }{ x+3 }-7 }{ \frac{ x-7 }{ x+3 }-1 }\)

OpenStudy (jdoe0001):

the previous compound function, that is f( g(x) ) gave us "x" thus that means that they're indeed inverse of each other thus this one should too

OpenStudy (anonymous):

Oh, gawsh. Thank you so much! God bless you!

OpenStudy (jdoe0001):

yw

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