noodlestopshop
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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?
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noodlestopshop
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Please someone
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vix.xvi
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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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vix.xvi
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test it out wiht some values to check
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noodlestopshop
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(Original post by 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
Do u mean this is a A level question? I am doing GCSE o_0
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vix.xvi
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(Original post by noodlestopshop)
Do u mean this is a A level question? I am doing GCSE o_0
oh lol
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Raven Firefly
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(Original post by 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 got x/x-2
I made a mistake in my working out, sorry you are right
Last edited by Raven Firefly; 2 months ago
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noodlestopshop
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(Original post by Raven Firefly)
I got x/2x-2
Do you have some workings out you could post so I can see
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GCM42
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(Original post by 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?
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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noodlestopshop
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(Original post by 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
Ohhh! Thankyou so much! This just made so much sense
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mqb2766
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(Original post by 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
Agree that the OP is correct.
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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RosieMc11
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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!
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shivsaransh1
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#13
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(Original post by noodlestopshop)
Do u mean this is a A level question? I am doing GCSE o_0
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.
Last edited by shivsaransh1; 2 months ago
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mqb2766
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(Original post by 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
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.
Last edited by mqb2766; 2 months ago
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RosieMc11
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#15
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(Original post by 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.
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?
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mqb2766
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(Original post by 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?
It's sometimes hard to comment on the level of a question, but once the inverse is found, 2 is clearly a "problem point". The posted question says state, and that's probably all that's required. I'd guess they're testing to see whether you understand the relationship between the range of f and the domain of inverse f.

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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