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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 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
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 noodlestopshop)
Do u mean this is a A level question? I am doing GCSE o_0
Do u mean this is a A level question? I am doing GCSE o_0
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#7
(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 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 Raven Firefly)
I got x/2x-2
I got x/2x-2
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#9
(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 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 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
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
(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
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 noodlestopshop)
Do u mean this is a A level question? I am doing GCSE o_0
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 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
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 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.
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 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?
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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