Question

Please help me with this one question! I need to get this done today.

1. 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 and g(x) = -2x - 7/x - 1?

Answer 1

\(\bf f(x)= \cfrac{x-7}{x}+2\qquad g(x)=\cfrac{-2x-7}{x}-1\quad ?\)

Answer 2

That's right..

Answer 3

or -> \(\bf f(x)= \cfrac{x-7}{x+2}\qquad g(x)=\cfrac{-2x-7}{x-1}\quad ?\)

Answer 4

What do I do from there?

Answer 5

Oh no. Sorry, the second one is right.

Answer 6

you pretty much just INSERT one function into the other, thus f( g(x) )

Answer 7

Um. I don't know how to do that..

Answer 8

@jdoe0001 Could you help me with the steps please?

Answer 9

\(\bf f(x)= \cfrac{x-7}{x+2}\qquad {\color{red}{ g(x)}}=\cfrac{-2x-7}{x-1}\quad
\\ \quad \\ \quad \\
f(\quad {\color{red}{ g(x)}}\quad )=\cfrac{\left({\color{red}{ \frac{-2x-7}{x-1}}}\right)-7}{\left({\color{red}{ \frac{-2x-7}{x-1}}}\right)+2}\)

Answer 10

and if g(x) is really the inverse function of f(x)
simplifying that , should give you "x"

Answer 11

So f(g(x))=(−2x−7x−1)−7/(−2x−7x−1)+2 is the answer?

Answer 12

no... you'd need to simplify it
lemme do the first one, so you can do the 2nd one that is g( fx) )

Answer 13

Okay :) Thank you

Answer 14

\(\bf f(\quad {\color{red}{ g(x)}}\quad )=\cfrac{\left({\color{red}{ \frac{-2x-7}{x-1}}}\right)-7}{\left({\color{red}{ \frac{-2x-7}{x-1}}}\right)+2}
\\ \quad \\
\cfrac{\frac{-2x-7}{x-1}-7}{\frac{-2x-7}{x-1}+2}\implies \cfrac{\frac{-2x-7-7(x-1)}{x-1}}{\frac{-2x-7+2(x-1)}{x-1}}
\\ \quad \\

\cfrac{-2x-7-7(x-1)}{x-1}\cdot \cfrac{x-1}{-2x-7+2(x-1)}\implies \cfrac{-2x-7-7(x-1)}{-2x-7+2(x-1)}
\\ \quad \\
\implies \cfrac{-2x-7-7x+7}{-2x-7+2x-2}\implies \cfrac{-9x}{-9}\implies x\)

Answer 15

Woah. Thanks. I'm gonna try to do the other half now...

Answer 16

Do I do it the same as that^^?

Answer 17

more or less, yes keep in mind that \(\bf {\color{red}{ f(x)}}= \cfrac{x-7}{x+2}\qquad g(x)=\cfrac{-2x-7}{x-1}\quad
\\ \quad \\ \quad \\
g(\quad {\color{red}{ f(x)}}\quad )=\cfrac{-2\left({\color{red}{ \frac{x-7}{x+2}}}\right)-7}{\left({\color{red}{ \frac{x-7}{x+2}}}\right)-1}\)

Answer 18

Okay I can't get the second half to equal x

Answer 19

Can I get some help on this please? The second part?