Quick question about ranges/domains of functions Watch

Muppety_Kid
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Hi,

In an exam question I was asked to find the cartesian equation of a curve in the form y=f(x). I did this and got the right answer, which is:

y=\frac{8}{4+x^2}

It then asked me to state the domain of the function, and I said that x is an element of the group of real numbers (i.e. it can have any value, including negatives, because when it is squared it will be positive).

However, in the mark scheme, it says x ...0

Does this mean x is "subject to" the conditions (there are none)? If so, is my answer right?

A brief explanation of the differences between the range and domain too would be much appreciated!

Thanks
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figureeight
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I would have said xER too, maybe it's got something to do with the parametrics you took it from?
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generalebriety
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"x ...0" is almost certainly mark scheme slang, which I don't speak. It's not a mathematical statement. Maybe it means "award a mark for either x < 0, x <= 0, x > 0 or x >= 0". Anyway, the domain of that function could be more or less anything - there will have been something given in the question. For example, if you translated that from parametric to Cartesian, it will have been given in the question "x = x(t), y = y(t), t > 0" or something, and you'll have been expected to change that to Cartesians too.
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Muppety_Kid
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(Original post by generalebriety)
"x ...0" is almost certainly mark scheme slang, which I don't speak. It's not a mathematical statement. Maybe it means "award a mark for either x < 0, x <= 0, x > 0 or x >= 0". Anyway, the domain of that function could be more or less anything - there will have been something given in the question. For example, if you translated that from parametric to Cartesian, it will have been given in the question "x = x(t), y = y(t), t > 0" or something, and you'll have been expected to change that to Cartesians too.
OK...

Looking at the question, it says 0<t<=(pi/2). How does that have an effect on my domain though? :confused:

I think I'll go away and mull this over for a few minutes...
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figureeight
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Because it can give you your largest and smallest values of x?
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generalebriety
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(Original post by Muppety_Kid)
OK...

Looking at the question, it says 0<t<=(pi/2). How does that have an effect on my domain though? :confused:

I think I'll go away and mull this over for a few minutes...
Well, what were the parametric equations?
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Muppety_Kid
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(Original post by generalebriety)
Well, what were the parametric equations?
Sorry, they were:

x = 2 cot t, y = 2sin^2t
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generalebriety
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(Original post by Muppety_Kid)
Sorry, they were:

x = 2 cot t, y = 2sin^2t
Ok, so what's the minimum and maximum x can be if 0 < t <= pi/2?
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Muppety_Kid
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(Original post by generalebriety)
Ok, so what's the minimum and maximum x can be if 0 < t <= pi/2?
The minimum is 0. The maximum would be \frac{2}{tan\frac{\pi}{2}}, but I get that to be undefined. So does that mean x must be greater than or equal to 0?
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generalebriety
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(Original post by Muppety_Kid)
The minimum is 0. The maximum would be \frac{2}{tan\frac{\pi}{2}}, but I get that to be undefined. So does that mean x must be greater than or equal to 0?
No, cot pi/2 is well defined (and is 0). It's cot 0 that's not defined. Luckily, 0 isn't in the set of values that t can take. Yes, basically, you're right: x >= 0, because at t = pi/2 we have x = 0, and as t goes to 0, x goes to +infinity.
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Muppety_Kid
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(Original post by generalebriety)
No, cot pi/2 is well defined (and is 0). It's cot 0 that's not defined. Luckily, 0 isn't in the set of values that t can take. Yes, basically, you're right: x >= 0, because at t = pi/2 we have x = 0, and as t goes to 0, x goes to +infinity.
Ah! I see it now - I've never had one that requires me to apply the conditions for t to x before.

I'll add that to my notes now, thanks a lot
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