<p>How do i find the domain of the function? For example:</p>
<p>f(x) = 2x</p>
<p>The domain is simply what x can equal.</p>
<p>In your example, f(x)=2x, x can be anything. It can be a negative number; it can be zero; it can be a million. Expressed as an interval it would be (-infinite, infinite), which means negative infinite to positive infinite (or pretty much everything).</p>
<p>Sometimes there are restrictions on what x can equal. Like if it was f(x)=1/x, then x couldn't equal zero because you can't divide by zero. Therefore the domain would be (-infinite, 0) U <a href="0,%20infinite">the U sign means united with</a>. The parentheses indicate that 0 is not included in possible values for x (if however, I had used brackets, this would have meant 0 is included). If it was f(x)=1/(x-1)(x+1), then x couldn't equal 1 or -1. So you would say the domain was (-infinite, -1) U (-1, 1) U (1, infinite). </p>
<p>And FYI, range is what f(x) or y can equal.</p>
<p>graphing it out should make it easier</p>
<p>Normally, the domain would be described as being "all real numbers."</p>
<p>I feel retarded, i still dont get it.</p>
<p>AFABound,</p>
<p>Carefully read the following definition:</p>
<p>The domain of a function is all values for which the function is defined.</p>
<p>Now, recall that in math that some values are considered "undefined." The most prominent "undefined" value in math is anything divided by zero, for example, 1/0 is considered undefined.</p>
<p>Why is division by zero undefined? I found an excellent explanation on mathforum.org and have pasted it below for you to read. It is very intuitive and easy to understand:</p>
<p>
[quote]
</p>
<p>If you know a little bit about multiplication, you could look at it this way:</p>
<p>10/2 = 5 This means that 5 x 2 = 10 </p>
<p>9/3 = 3 This means that 3 x 3 = 9 </p>
<p>5/1 = 5 This means that 5 x 1 = 5</p>
<p>5/0 = ? This would mean that the answer x 0 = 5, but
anything times 0 is always zero.
So there isn't an answer.
[/quote]
</p>
<p>Hence, division by zero is undefined. Now, when a math test asks you to find the domain of a function, let's say f(x), what they are asking you to do is to find the input value of the function (the value of x) for which the function is NOT undefined (not divided by zero)</p>
<p>So let's take the your original function
f(x) = 2x.</p>
<p>Is there any value of x for which this function would be undefined? (something divided by zero?) Never. 2 times any real number will be a real number. So the domain of this function is all real numbers. In bracket notation, you would say (-oo, +oo) where the oo represents the infinity sign. This means the domain of the function is all the negative numbers through all the positive, extending infinitely both ways. </p>
<p>Now let's say your function was f(x) = 1/x. What is the domain of the function now?</p>
<p>The answer is: all real numbers EXCEPT 0. (Because 1/0 is undefined).</p>
<p>Hope that clarifies.</p>
<p>Tip: if you want to figure out how to express the domain of 1/x in bracket notation, use jlauer95's post as a guide. I verbally stated it for you but you may be asked on a test to show that you know how to state it in bracket notation.</p>
<p>So all i would write is (-oo, oo)</p>
<p>For the function f(x)=2x, your answer (-oo,oo) would be correct.</p>
<p>Yes, I know it's more common to express that as all real numbers, but my Calculus teacher requires us to write the domain and range in interval/bracket form, although technically, both are correct.</p>
<p>how do I put the answer in brackets, this is really frustrating me.</p>
<p>You put the answer in parentheses when you don't count the numbers inside it, but you use brackets when the numbers inside are included.</p>
<p>You already have an example of one with parentheses, so I'll give you one with brackets.</p>
<p>If you had: f(x)= square root of (4-x^2), then x could not be greater than 2 or less than -2, because if it was, you would have a negative number under the radical, which you can't have. Therefore, x is between -2 and 2, but notice that x CAN be 2 and -2, but just not greater than 2 or less than 2.</p>
<p>Therefore, the domain, expressed as an interval, would be:
[-2,2]</p>
<p>As said above, if -2 and 2 didn't work, then you'd put parentheses around them, but since -2 and 2 are perfectly valid values for x, you use brackets.</p>
<p>(x+1)^2</p>
<hr>
<p>radical(2x-1)</p>
<p>the (x+1)^2 is over the radical(2x-1)</p>
<p>how do i do that?</p>
<p>the domain of that function is all real numbers greater than or equal to 1/2</p>
<p>take precalculus</p>
<p>AFA: There are two aspects to that function that could restrict its domain. First, you can't take the square root of (2x-1) if (2x-1) is negative. So x has to be greater than or equal to 1/2. Second, you also can't divide by zero, so sqrt(2x-1) can't be zero. This only happens if x=1/2.</p>
<p>So your final answer is that the domain is all real numbers x that are greater than (NOT equal to) 1/2.</p>
<p>AFAbound, always go to the basics. The domain is where the function is defined. A function is simply for every x value there is only a y value. So your objective is to find where your equation is not a function or values where the function won't work. Where does that occur? When the denominator equals zero since you are dividing by zero. So now you want to find the x value where the denominator equals zero. </p>
<p>So your first task is to set the sqrt(2x-1) to zero. And solve for x. You will get x=1/2. Now plug back x=1/2 to the denominator and you'll see you are dividing by zero, which you can't do in mathematics. This means that if you graph the equation you posted, to keep things simply, you'll see some goofy ***** occuring at x=1/2. </p>
<p>Let me summarize a little. When finding the domain you basically look for two things:
1. Where you divide by zero.
2. Where you take square root of negatives. </p>
<p>In the function you posted, #2 is also something you have to consider because you have a square root. You do not want to take the square root of a negative number because it will give you a nonreal answer, something you are not concerned with in domain. </p>
<p>So your objective is to find for what values where the sqroot(2x-1) produce real answers. In other words, you want to square root positive numbers. In your problem, think of "2x-1" as a number you want to keep positive. In order to do this, you want 2x-1 to be greater than 0 (i.e, positive). Mathematically, this is the inequality 2x-1 > 0. Solve for the inequality and you get x>1/2. </p>
<p>Now you go back and try to answer the question as a whole. For what set of values does your function occur (i.e, domain)? Well, you found two things: </p>
<ol>
<li>That the function blows up at x=1/2. </li>
<li>In order to maintain square rooting postive numbers, you can not pick values less than 1/2. If you don't believe me, test it out for your self. Plug in 1/4, 0, -1, -100, .002, -10000 and you will get a negative number. </li>
</ol>
<p>Therefore the domain is all numbers greater than 1/2. The previous poster is wrong, it can not equal 1/2 since you will be dividing by zero and the universe explodes. Mathematically you write x>1/2. </p>
<p>Now go back and start plugging in values into your function that are greater than 1/2. You will always get some number. </p>
<p>Also, note that we didn't care about the numerator (x+1)^2 because any number for x will become positive. But in many problems, you will have radicals in the numerator you have to be concerned with.</p>
<p>Anyways, good luck and I hope I helped!</p>
<p>w/e i was close enough</p>