Beaconite
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domain and range (1 Viewer)
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Beaconite
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Beaconite
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how'd you work out the domain cuz? and ur range is incorrect
Beaconite
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Yeah, how'd you work out domain?
Leffife
A lover is a best friend
-_- it is literally the same thing...
π + 2y ≥ 0 and 2y ≤ π
From here,
2y ≥ - π => y ≥ - π/2 or y ≤ π/2
From here we can combine it, - π/2 ≤ y ≤ π/2 ...
I was actually going to explain how I did but now I have some doubts because you said I was wrong...
π + 2y ≥ 0 and 2y ≤ π
From here,
2y ≥ - π => y ≥ - π/2 or y ≤ π/2
From here we can combine it, - π/2 ≤ y ≤ π/2 ...
I was actually going to explain how I did but now I have some doubts because you said I was wrong...
Beaconite
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domain??? How"d u work that out?!
Sy123
This too shall pass
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Domain is the x values that can be x, for instance in =\sqrt{x} )
Our domain is
The reason why this is, because for any real numbered x, it cannot be negative otherwise undefined, in a similar sense
=x \sin^{-1} {x^2} )
Well what cant x be?
We know that x^2, no matter what you have it must be less than 1, and greater than -1 **(due to the fact that the domain of an inverse sine function states that it is defined for that part, the reason for this is, because sine is opposite over hypotenuse, taking the sin inverse of a value that states that opposite is greater than hypotenuse means that there is no such triangle because hypotenuse is always greater than opposite (since hypotensue is opposite the greatest angle (right angle)) therefore the domain of any sine inverse function must be from -1 to 1)**
What we are doing when we square the x is, we are making sure x always greater than or equal to zero, but x can be negative. Therefore our domain just stays the same as normal, since we are not restricting x in anyway other than restriction by the sine inverse function.
For example=\sin^{-1} \sqrt{x} )
Normally any sine inverse functions have the domain for -1 to 1, moreover because we have a square root on the x, the x must be positive, 'combining' these domains, gives us:
Just an illustration of what Im trying to explain.
**Do not read this astriks paragraph if you already know why inverse sine has such and such domain
Our domain is
The reason why this is, because for any real numbered x, it cannot be negative otherwise undefined, in a similar sense
Well what cant x be?
We know that x^2, no matter what you have it must be less than 1, and greater than -1 **(due to the fact that the domain of an inverse sine function states that it is defined for that part, the reason for this is, because sine is opposite over hypotenuse, taking the sin inverse of a value that states that opposite is greater than hypotenuse means that there is no such triangle because hypotenuse is always greater than opposite (since hypotensue is opposite the greatest angle (right angle)) therefore the domain of any sine inverse function must be from -1 to 1)**
What we are doing when we square the x is, we are making sure x always greater than or equal to zero, but x can be negative. Therefore our domain just stays the same as normal, since we are not restricting x in anyway other than restriction by the sine inverse function.
For example
Normally any sine inverse functions have the domain for -1 to 1, moreover because we have a square root on the x, the x must be positive, 'combining' these domains, gives us:
Just an illustration of what Im trying to explain.
**Do not read this astriks paragraph if you already know why inverse sine has such and such domain
Leffife
A lover is a best friend
I'll just say arcsin instead of sin^-1
Well, you already know that f(x) = arcsin(x) has the domain [-1,1] i.e. - 1 ≤ x ≤ 1
However, since there is an x² we know it can be like from - 1 all the way to 1.
Thus, 'x' can be from - 1 ≤ x ≤ 1. I.e. f(x) = arcsin(x²) is from - 1 ≤ x ≤ 1. And we
do all the stuff we already know that f(x) = x.arcsin(x²) will then be from - 1 to 1. And that's how I got
the domain.
Well, you already know that f(x) = arcsin(x) has the domain [-1,1] i.e. - 1 ≤ x ≤ 1
However, since there is an x² we know it can be like from - 1 all the way to 1.
Thus, 'x' can be from - 1 ≤ x ≤ 1. I.e. f(x) = arcsin(x²) is from - 1 ≤ x ≤ 1. And we
do all the stuff we already know that f(x) = x.arcsin(x²) will then be from - 1 to 1. And that's how I got
the domain.