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C3 function question

So the range of the function f(x)=2x^2-1/x^2+1 for x=0 to x=2 is y=-1 to y=1.4 but the domain of inv f(x) is -1<x<1.4
Why isn't it equal to or greater than -1 to equal to or less than 1.4
isnt the domain of the inverse function just the range of the original function?


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Well to start off its not a one to one function so im not sure about finding the inverse.
Secondly as f(x) has been posted it never crosses the y-axis and so im not sure if its been misposted
Original post by mattyd375
Well to start off its not a one to one function so im not sure about finding the inverse.
Secondly as f(x) has been posted it never crosses the y-axis and so im not sure if its been misposted

Hi. I think it's meant to be understood as f(x)=(2x2-1)/(x2+1).

Personally, I can't see any reason why the domain of f-1(x) shouldn't be -1 x 1.4. I've worked out the inverse function and sketched the graphs of both the original and inverse functions (for the domains) and can't see anything.

Was the question from a book, or past paper? Just wondering; if the former, is it possible the answer in the back of the book is wrong?
Even if it is as you have put it is still impossible to get an inverse function as it is not a one to one function and therefore you end up with f^-1(x)=+-((y+1)/(y-2))^1/2
Original post by mattyd375
Even if it is as you have put it is still impossible to get an inverse function as it is not a one to one function and therefore you end up with f^-1(x)=+-((y+1)/(y-2))^1/2

I think that in answering the question, you would need to specify that

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

and then you can have an inverse.
Original post by M.C. Math
I think that in answering the question, you would need to specify that

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

and then you can have an inverse.


Still dont think so as you can never have an inverse of a one to many function. Would really help if a picture of the question could be posted
Original post by mattyd375
Still dont think so as you can never have an inverse of a one to many function. Would really help if a picture of the question could be posted

Hi. The function is not a one to many. Here's a diagram.
Original post by M.C. Math
Hi. The function is not a one to many. Here's a diagram.


ah yeah your right hadnt tried drawing it out like that
Hi. Can I also add (especially for those doing Edexcel C3 tomorrow) that a one to many function can be changed to a one to one function either by changing the domain, e.g.
y=1/x : Not a function
y=1/x, {x > 0} Is a function

or by manipulating the formula, e.g.
y=√x , {x > 0} : Not a function
y=+√x, {x > 0} : Is a function (by specifying the positive square root, so now a one to one mapping)
(edited 8 years ago)

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