Farmerjj
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The function f is defined by
F(X) = (x-2)/4x where X cannot = 0

Find an expression for the inverse and state it's domain.


I always get confused with these when you have a denominator to get rid of and two separate x values so any help would be greatly appreciated.
Thanks
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Muttley79
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(Original post by Farmerjj)
The function f is defined by
F(X) = (x-2)/4x where X cannot = 0

Find an expression for the inverse and state it's domain.


I always get confused with these when you have a denominator to get rid of and two separate x values so any help would be greatly appreciated.
Thanks
I usually suggest putting y = (x-2)/4x and then rearrange to make x the subject.

Post what you get and I'll help further if you get stuck.
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The Sikh Doctor
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What I do is swap the x's and y's and try making y the subject again. You end up with a nasty looking equation but you can factorise the y out and then divide through by the rest. I got:
-2/(4x-1)
To work out the domain, you can work out the range of the original equation and because this is the inverse, the range of that equation will be the domain of this equation.
Hope that helps
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Farmerjj
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(Original post by Muttley79)
I usually suggest putting y = (x-2)/4x and then rearrange to make x the subject.

Post what you get and I'll help further if you get stuck.
Got this but like 99.9% sure it's wrong. Does the y-2 count as one function?
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Muttley79
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(Original post by Farmerjj)
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Got this but like 99.9% sure it's wrong. Does the y-2 count as one function?
The first three lines are correct.
Collect the y terms on the LHS and then factorise.
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K-Man_PhysCheM
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(Original post by Farmerjj)
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Got this but like 99.9% sure it's wrong. Does the y-2 count as one function?
You made a algebraic mistake going from line 3 to line 4.

Generally, it's better to start by getting rid of fractions, by multiplying both sides by the denominator of the fraction. Then get all terms with x on one side, then factorising x out, then finally divide by the factor bit. You have used u in place of x, which is fine.

example of what I described above:

y = \dfrac{5x+1}{x-3}

xy - 3y = 5x+1 (multipled both sides by denominator and expanded).

xy - 5x = 3y + 1

x(y-5) = 3y+1

x = \dfrac{3y+1}{y-5}

Then replace y with x to write your inverse function. Now try with your function.
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Farmerjj
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(Original post by Muttley79)
The first three lines are correct.
Collect the y terms on the LHS and then factorise.
So do I take y-2 across instead of just y?
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Muttley79
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(Original post by Farmerjj)
So do I take y-2 across instead of just y?
Yes but don't divide yet.
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junita
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Size:  2.4 KB the domain : x<-1/4 or x>1/4
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Muttley79
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(Original post by junita)
....
In the MATHS forum we only post hints - not full solutions please remember this another time
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junita
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(Original post by Muttley79)
In the MATHS forum we only post hints - not full solutions please remember this another time
Oh okay, I'm sorry, I'm a new member
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Muttley79
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(Original post by junita)
Oh okay, I'm sorry, I'm a new member
OK - I could see you were new
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