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I have the function

f(x)=2x/x-1

I had to find the inverse and for that I got

f(x)=2x/x-1

y(x-1)=2x

yx-y=2x

x(y-2)=y

=y/y-2

=x/x-2

Is this correct?

I also have to state which value cannot be in any range of f, would that be any number which is a negative?

f(x)=2x/x-1

I had to find the inverse and for that I got

f(x)=2x/x-1

y(x-1)=2x

yx-y=2x

x(y-2)=y

=y/y-2

=x/x-2

Is this correct?

I also have to state which value cannot be in any range of f, would that be any number which is a negative?

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#3

i clicked on here cuz i thought it was a gcse question lmaao

idk how to find the invsere of a function soz mate

lemme think gimme a min

idk how to find the invsere of a function soz mate

lemme think gimme a min

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(Original post by

i clicked on here cuz i thought it was a gcse question lmaao

idk how to find the invsere of a function soz mate

lemme think gimme a min

**vix.xvi**)i clicked on here cuz i thought it was a gcse question lmaao

idk how to find the invsere of a function soz mate

lemme think gimme a min

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#6

(Original post by

Do u mean this is a A level question? I am doing GCSE o_0

**noodlestopshop**)Do u mean this is a A level question? I am doing GCSE o_0

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#7

(Original post by

I have the function

f(x)=2x/x-1

I had to find the inverse and for that I got

f(x)=2x/x-1

y(x-1)=2x

yx-y=2x

x(y-2)=y

=y/y-2

=x/x-2

Is this correct?

I also have to state which value cannot be in any range of f, would that be any number which is a negative?

**noodlestopshop**)I have the function

f(x)=2x/x-1

I had to find the inverse and for that I got

f(x)=2x/x-1

y(x-1)=2x

yx-y=2x

x(y-2)=y

=y/y-2

=x/x-2

Is this correct?

I also have to state which value cannot be in any range of f, would that be any number which is a negative?

I made a mistake in my working out, sorry you are right

Last edited by Raven Firefly; 2 months ago

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(Original post by

I got x/2x-2

**Raven Firefly**)I got x/2x-2

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#9

**noodlestopshop**)

I have the function

f(x)=2x/x-1

I had to find the inverse and for that I got

f(x)=2x/x-1

y(x-1)=2x

yx-y=2x

x(y-2)=y

=y/y-2

=x/x-2

Is this correct?

I also have to state which value cannot be in any range of f, would that be any number which is a negative?

y=2x/x-1

x=2y/y-1

x(y-1)=2y

xy-x=2y

xy-2y=x

y(x-2)=x

y=x/x-2

f'(x)=x/x-2

So yes, you are correct.

The range of an inverse is the domain of the normal function; the domain of an inverse is the range of the normal function. Therefore, the range of the inverse is f'(x) does not equal to 1

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(Original post by

Here's my working:

y=2x/x-1

x=2y/y-1

x(y-1)=2y

xy-x=2y

xy-2y=x

y(x-2)=x

y=x/x-2

f'(x)=x/x-2

So yes, you are correct.

The range of an inverse is the domain of the normal function; the domain of an inverse is the range of the normal function. Therefore, the range of the inverse is f'(x) does not equal to 1

**GCM42**)Here's my working:

y=2x/x-1

x=2y/y-1

x(y-1)=2y

xy-x=2y

xy-2y=x

y(x-2)=x

y=x/x-2

f'(x)=x/x-2

So yes, you are correct.

The range of an inverse is the domain of the normal function; the domain of an inverse is the range of the normal function. Therefore, the range of the inverse is f'(x) does not equal to 1

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#11

**GCM42**)

Here's my working:

y=2x/x-1

x=2y/y-1

x(y-1)=2y

xy-x=2y

xy-2y=x

y(x-2)=x

y=x/x-2

f'(x)=x/x-2

So yes, you are correct.

The range of an inverse is the domain of the normal function; the domain of an inverse is the range of the normal function. Therefore, the range of the inverse is f'(x) does not equal to 1

The question asks for the range of f, rather than its inverse.. However the inverse (domain) may give a clue about problem point(s).

Think of f as a modified recriprocal. Which point(s) (asymptote ...) won't be in the range of f?

Last edited by mqb2766; 2 months ago

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#12

y = 2x / (x-1)

Swap x and y

x = 2y / (y-1)

x(y-1) = 2y

xy - x = 2y

xy - 2y = x

y(x - 2) = x

y = x / (x - 2)

So yep you're right!

Swap x and y

x = 2y / (y-1)

x(y-1) = 2y

xy - x = 2y

xy - 2y = x

y(x - 2) = x

y = x / (x - 2)

So yep you're right!

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#13

(Original post by

Do u mean this is a A level question? I am doing GCSE o_0

**noodlestopshop**)Do u mean this is a A level question? I am doing GCSE o_0

Last edited by shivsaransh1; 2 months ago

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#14

(Original post by

This is a GCSE question. To test whether the inverse function you have is right test F(x) of F-1(x) or the other way around, and you should get x if your inverse is right because they cancel each other out.

To get the range, draw the function out. It is a reciprocal function, so the only value that cant isn't part of the function is 0. It's transformed, but nothing has happened to the Y(axis) So 0 still can't be part of the

**shivsaransh1**)This is a GCSE question. To test whether the inverse function you have is right test F(x) of F-1(x) or the other way around, and you should get x if your inverse is right because they cancel each other out.

To get the range, draw the function out. It is a reciprocal function, so the only value that cant isn't part of the function is 0. It's transformed, but nothing has happened to the Y(axis) So 0 still can't be part of the

https://www.desmos.com/calculator/t3ixij1wt8

The previous hint about thinking about the domain of the inverse function is all you need. As you say, it's good to sketch it.

Last edited by mqb2766; 2 months ago

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#15

(Original post by

0 is part of the range

https://www.desmos.com/calculator/t3ixij1wt8

The previous hint about thinking about the domain of the inverse function is all you need. As you say, it's good to sketch it.

**mqb2766**)0 is part of the range

https://www.desmos.com/calculator/t3ixij1wt8

The previous hint about thinking about the domain of the inverse function is all you need. As you say, it's good to sketch it.

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#16

(Original post by

The only values that aren't part of the range are y=2 and x=1 because these are asymptotes, but finding these would be an A level question surely?

**RosieMc11**)The only values that aren't part of the range are y=2 and x=1 because these are asymptotes, but finding these would be an A level question surely?

More than required, you could rewrite the original function as

y = (2x -2 + 2)/(x-1) = 2 + 2/(x-1)

So the asymptotes / relationship to a reciprocal are explicit and the range and domain should be easy.

Last edited by mqb2766; 2 months ago

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