Please help me to find the domain (1 Viewer)
- Thread starter happy10
- Start date
sida1049
Well-Known Member
- Joined
- Jun 18, 2013
- Messages
- 926
- Gender
- Male
- HSC
- 2015
The domain is all reals; x can take any real value.
How do you work this out? Well, you start by asking yourself "are there any points or regions at which the function is undefined?" Since the function is a rational function (i.e. a fraction), an obvious place to start is by asking yourself for what values of x will the denominator be zero (and cause a problem)?
But a quick glance at the denominator should tell you that the denominator can never be zero for any value of x; specifically, the denominator is always 2 or higher.
Now, for any other possible points at which the function becomes undefined, note that there is a square root. Since we're considering only real numbers, recall that we cannot find the square root of a negative number. So you should now ask yourself for what values of x is x^2 + 4 < 0 ? The answer is none. So the square root doesn't cause any problems in the function.
So there are no values of x such that the function will become undefined, and therefore the domain of the function is all reals.
Hope this helps.
How do you work this out? Well, you start by asking yourself "are there any points or regions at which the function is undefined?" Since the function is a rational function (i.e. a fraction), an obvious place to start is by asking yourself for what values of x will the denominator be zero (and cause a problem)?
But a quick glance at the denominator should tell you that the denominator can never be zero for any value of x; specifically, the denominator is always 2 or higher.
Now, for any other possible points at which the function becomes undefined, note that there is a square root. Since we're considering only real numbers, recall that we cannot find the square root of a negative number. So you should now ask yourself for what values of x is x^2 + 4 < 0 ? The answer is none. So the square root doesn't cause any problems in the function.
So there are no values of x such that the function will become undefined, and therefore the domain of the function is all reals.
Hope this helps.
integral95
Well-Known Member
- Joined
- Dec 16, 2012
- Messages
- 779
- Gender
- Male
- HSC
- 2013
The Range would be
You should think about what the lowest value you could have in your denominator, in that case it would be
That lowest value would give you the highest value in the function (the maximum)
Now think about the largest possible value in the denominator, that would be infinity since x could always increase without bound.
This means the lowest value of the function is 0. (Now it actually doesn't get to the value 0, this way the range doesn't include that value)
The reason it can't be negative is that when you take a square root, you can only take a square root of a positive number, which only gives a positive value.
You should think about what the lowest value you could have in your denominator, in that case it would be
That lowest value would give you the highest value in the function (the maximum)
Now think about the largest possible value in the denominator, that would be infinity since x could always increase without bound.
This means the lowest value of the function is 0. (Now it actually doesn't get to the value 0, this way the range doesn't include that value)
The reason it can't be negative is that when you take a square root, you can only take a square root of a positive number, which only gives a positive value.
integral95
Well-Known Member
- Joined
- Dec 16, 2012
- Messages
- 779
- Gender
- Male
- HSC
- 2013