Question

Show that g(x) = -2x - 7/x - 1 = x

Please help a girl out!

Answer 1

What do you think? o.0

Answer 2

impossible

Answer 3

I'm serious..

Answer 4

\[\frac{-2x-7}{x-1}=x\iff x(x-1)=-2x-7\]\]

Answer 5

x*x is x^2 ?

Answer 6

Oh I see what you did! xD

Answer 7

if it is what i wrote, the solutions are complex (not real numbers)

Answer 8

I'm slightly confused.

Answer 9

Aren't there supposed to be a lot of steps?

Answer 10

this is what you wrote:
show \(g(x)=\frac{-2x-7}{x-1}=x\)
maybe it is a different question
they are not the same thing at all

Answer 11

can you post a screen shot of the question as written?

Answer 12

Earlier I was solving for f(g(x)) = x

and my steps were:
(-2x - 7/x - 1) - 7/(-2x - 7/x - 1) + 2

(-2x - 7/ x - 1) - 7/(-2x - 7/x - 1) = 2

(-2x - 7 -7(x-1)/x - 1)/(-2x - 7 + 2(x -1)/x - 1)

-2x - 7 - 7(x-1)/ x - 1 * x - 1/-2x - 7 + 2 (x - 1)

-2x - 7 - 7(x-1)/-2x - 7 + 2(x-1)

-2x - 7 - 7x + 7/-2x - 7 + 2x - 2

-9x/-9

x

Answer 13

AHHHH
perhaps the question is :
let \(g(x)=\frac{-2x-7}{x-1}\) find \(g^{-1}(x)\)?

Answer 14

\[g(x) = \frac{ -2x - 7 }{ x - 1 } = x\]

Answer 15

I was also thing that, switching the y for x and x for y ._.

Answer 16

thinking*

Answer 17

perhaps it says
\[g(x)=\frac{-2x-7}{x-1},f(x)=\frac{x-7}{x+2}\] show
\[f(g(x))=x\]???

Answer 18

It wants me to prove that the equation does indeed equal x

Answer 19

what equation?

Answer 20

The g(x) = -2x - 7/ x - 1

Answer 21

it does not equal \(x\) so you cannot show it

Answer 22

Hm. Okay. Hang on one sec...

Answer 23

do you have another function
\[f(x)=\frac{x-7}{x+2}\]?

Answer 24

It'd be better if you screen shot it, g(x) is a function.. >.>

Answer 25

Yes! I do!

Answer 26

I should probably just have posted the full question, but I got half done myself so I thought it would be less confusing if I did it this way....here's the 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 and g(x) = -2x - 7/x - 1?

Answer 27

that is what you did earlier
you showed
\[f(g(x))=x\] now i think you have to show that
\[g(f(x))=x\] does that sound right?

Answer 28

Yes.

Answer 29

how'd i guess?

Answer 30

Lol sorry...

Answer 31

\[g(f(x))\] replace \(f(x)\) by \(\frac{x-7}{x+2}\) and get
\[g\left(\frac{x-7}{x+2}\right)\] is the first step

Answer 32

So I'm writing g(x - 7/x + 2) as the first step...

Answer 33

then since
\[g(\heartsuit)=\frac{-2\heartsuit-7}{\heartsuit-1}\] you have
\[g(f(x))=\frac{-2\left(\frac{x-7}{x+2}\right)-7}{\left(\frac{x-7}{x+2}\right)-1}\] as the second step

Answer 34

What are the hearts supposed to be?

Answer 35

x's?

Answer 36

whatever the input is
if you use \(x\)'s all the time it gets confusing

Answer 37

try to think of the function not in terms of \(x\)
\[g(x)=\frac{-2x-7}{x-1}\] is the same as
\[g(\spadesuit)=\frac{-2\spadesuit-7}{\spadesuit-1}\]

Answer 38

in any case , now that we have \[g(f(x))=\frac{-2\left(\frac{x-7}{x+2}\right)-7}{\left(\frac{x-7}{x+2}\right)-1}\] it is algebra from here on in

Answer 39

Okay so so far I have:

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

-2 (x - 7/x + 2) - 7/(x - 7/x + 2) - 1

And for now I'm just using x's because it's easier to type....what's next?

Answer 40

a raft of algebra, then a raft of cancellation

Answer 41

Could you elaborate?

Answer 42

\[\frac{-2\left(\frac{x-7}{x+2}\right)-7}{\left(\frac{x-7}{x+2}\right)-1}\] clear the compound fraction by multiplying top and bottom by \(x+2\)

Answer 43

you get
\[\frac{-2(x-7)-7(x+2)}{x-7-1(x+2)}\]

Answer 44

Okay so I have this so far:

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

-2(x - 7/x + 2) - 7/(x - 7/x + 2) - 1

-2(x - 7) - 7(x + 2)/x - 7 - 1(x + 2)

Answer 45

Then what?

Answer 46

multiply out in the top and bottom

Answer 47

\[\frac{-2(x-7)-7(x+2)}{x-7-1(x+2)}\]
\[=\frac{-2x+14-7x-14}{x-7-x-2}\]

Answer 48

then combine like terms top and bottom
\[\frac{-9x}{-9}\]

Answer 49

and cancel, you are done

Answer 50

So it does equal x? I thought in the beginning you said it doesn't?

Answer 51

moral of the story
post the entire question! it didn't the way you wrote the question at the beginning, without saying to compose the functions and giving the inverse function as \(f(x)=\frac{x-7}{x+2}\)

Answer 52

Ahhhhh okay. Thank you so much and sorry for the confusion!

Answer 53

it is not that \[f(x)=\frac{-2x-7}{x-1}=x\] but rather that \[f(g(x))=g(f(x))=x\]

Answer 54

yw

Answer 55

Poor satellite lol, I felt your pain as you were trying to figure out the question.

Answer 56

Can I ask where you are from?