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# C3 - Maths Functions Help watch

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1. I heard a different way to find the range of a function is to interchange the x and y and then find the domain instead. Then substitute the domain back into the original function and find what values work. I just wrote a really rough note a few months ago. I have no idea if it works or not. If anyone knows and can explain it better, any help would be appreciated.
Thanks
I heard a different way to find the range of a function is to interchange the x and y and then find the domain instead. Then substitute the domain back into the original function and find what values work. I just wrote a really rough note a few months ago. I have no idea if it works or not. If anyone knows and can explain it better, any help would be appreciated.
Thanks
3. (Original post by rsgloryandgold)
f(x) = 1 - x^2
y = 1 - x^2
interchange y and x
x = 1 - y^2
rearrange
y = sqrt 1- x
therefore the domain is x greater than or equal to zero, since you cannot sqrt a negative.
Now rearrange and substitute back into original.
y = 1 - x^2
so if the x values have to be equal or greater than zero....the largest y value is 1.

IS THIS A VALID METHOD !
What if its a new method :O ? I doubt it but lmao that would be awesome

Oh and will it work with all functions?
f(x) = 1 - x^2
y = 1 - x^2
interchange y and x
x = 1 - y^2
rearrange
y = sqrt 1- x
therefore the domain is x greater than or equal to zero, since you cannot sqrt a negative.
Now rearrange and substitute back into original.
y = 1 - x^2
so if the x values have to be equal or greater than zero....the largest y value is 1.

IS THIS A VALID METHOD !
What if its a new method :O ? I doubt it but lmao that would be awesome

Oh and will it work with all functions?
for f(x) = 1-x^2

To get the domain, you can draw the graph.

Easy, As you know what x^2 looks like, so you can reflect in the y direction to get -x^2

then shift up one.

therefore the domain should be: X E r (x equals all real numbers)

As for the range of the function:

From the graph it's always below 1 (and equal to)

So range = f(x) <= 1 (f(x) less than or equal to one)

I think what you described is to get the inverse function, thats where you rearrange to get x the subject.

F^-1(x) = sqrt (1 - x)

Domain of that (you can draw it): X >= 1 (x is greater than or equal to 1)

Range (from graph) : f^-1(x) >= 0 (f^-1(x) is greater than or equal to 0)

hope that sort of helps
I heard a different way to find the range of a function is to interchange the x and y and then find the domain instead. Then substitute the domain back into the original function and find what values work. I just wrote a really rough note a few months ago. I have no idea if it works or not. If anyone knows and can explain it better, any help would be appreciated.
Thanks
It's the same as saying that the domain of a function is the same as the range of its inverse function (what you've calculated) and the range of the function is the same as the domain of the inverse function.
6. (Original post by SeanFM)
It's the same as saying that the domain of a function is the same as the range of its inverse function (what you've calculated) and the range of the function is the same as the domain of the inverse function.
Right, and is that true for any function? That its domain is equivalent to the range of the inverse function and the range is equivalent to the domain of the inverse function?
Right, and is that true for any function? That its domain is equivalent to the range of the inverse function and the range is equivalent to the domain of the inverse function?
I suppose it's true for any function that has an inverse.

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Updated: January 16, 2016
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