# Maths domain / range problem

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Thread starter 2 months ago
#1

For f-1(x) to exist / be a function, its range must have only one distinct value whenever x is put in, eg. it must only have one y value for the values of x. Since it is the inverse function, its range is equal to the domain of f(x), or, its y values are equal to the x values of f(x), right? So, if y on the inverse must have one distinct value, then for f(x) we need to find the value of y where there is only one distinct x value, or where the domain can only be one value. This occurs on f(x) at the bottom of the quadratic curve, because here, there is only one x value that can give the desired y value (its one to one). So we need to find the x value of this point, which is -3. However, it is only at THIS point that f(x) is one to one, because any point after or before x = -3 and all of a sudden it becomes many to one (as it is a quadratic of course) so why is the mark scheme saying 'hence f is one-to-one when x > -3' because that's not entirely true... it is ONLY when x = -3 right?

sorry for the wall of text to anyone looking at this, I spent a great deal of time trying to understand this question and appreciate any help I can get considerably
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2 months ago
#2
(Original post by pondering-soul)

For f-1(x) to exist / be a function, its range must have only one distinct value whenever x is put in, eg. it must only have one y value for the values of x. Since it is the inverse function, its range is equal to the domain of f(x), or, its y values are equal to the x values of f(x), right? So, if y on the inverse must have one distinct value, then for f(x) we need to find the value of y where there is only one distinct x value, or where the domain can only be one value. This occurs on f(x) at the bottom of the quadratic curve, because here, there is only one x value that can give the desired y value (its one to one). So we need to find the x value of this point, which is -3. However, it is only at THIS point that f(x) is one to one, because any point after or before x = -3 and all of a sudden it becomes many to one (as it is a quadratic of course) so why is the mark scheme saying 'hence f is one-to-one when x > -3' because that's not entirely true... it is ONLY when x = -3 right?

sorry for the wall of text to anyone looking at this, I spent a great deal of time trying to understand this question and appreciate any help I can get considerably
You're tight - that is a wall of text

It's a bit late in the evening to be thinking about this , but I think what they're saying is that because x = -3 is the minimum point (vertex) of the curve, if you restrict the domain of f(x) to x > - 3 then you just have one branch of the parabola to "invert" so you can define the inverse of f(x) properly. If you allowed (for example) x > -2 then there would be 2 values of x either side of the vertex giving the same y-value, and hence the function would not be invertible.
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Thread starter 2 months ago
#3
(Original post by davros)
You're tight - that is a wall of text

It's a bit late in the evening to be thinking about this , but I think what they're saying is that because x = -3 is the minimum point (vertex) of the curve, if you restrict the domain of f(x) to x > - 3 then you just have one branch of the parabola to "invert" so you can define the inverse of f(x) properly. If you allowed (for example) x > -2 then there would be 2 values of x either side of the vertex giving the same y-value, and hence the function would not be invertible.
oh that makes sense thank you!
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