Algebra (all content)
- Domain of a radical function
- Worked example: domain of algebraic functions
- Determine the domain of functions
- Worked example: determining domain word problem (real numbers)
- Worked example: determining domain word problem (positive integers)
- Worked example: determining domain word problem (all integers)
- Function domain word problems
Many examples of determining the domains of functions according to mathematical limitations.
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- why is |X| < 6 = -6<X<6 ? I couldn't really understand this part(34 votes)
- We know, absolute value is always positive. Just an example :
| 3 | = 3
| -3 | = 3
so, it also works for variables
| x | = x
| -x | = x
now we apply this condition/term to this problem
| x | < 6 or | -x | < 6
x < 6 or -x < 6
x < 6 or x > -6
-6 < x < 6
I guess, like that
hope it answers your question(20 votes)
- Where does the -6 come from and how, in around3:30?(11 votes)
- Absolute value x is smaller than 6.
Based on the root,the denominator can not be zero or less than 0.
Because of the absolute value, all of the negative numbers are positive.
if x= -7 ,the denominator less than zero in the root. Undefined.(6 votes)
- What is the definition of absolute value?(5 votes)
- The absolute value is the value's distance from 0. In practical terms, it basically means you remove the negative sign if it is present.
You might find these videos useful:
- (For top part of def) isn't obvious? when x=5 the function is NOT defined.. then why still "x NOT EQUALS TO 5 " is not a domain?(5 votes)
- Both definitions of the function must be followed. The first definition is if x IS NOT equal to 5, and the second definition is for if x IS equal to five.
The second definition states that if x IS equal to five, therefore you can input 5 into the function and get pi as your value for the funciton. So you CAN have x be equal to five, you just don't use the top definition, you use the bottom definition.(7 votes)
- At around6:05, Sal says that we can't cancel the terms in the numerator and the denominator ( ( x-10 ) ) because it would "change the function" ... What did he mean by that? And why would it change the function? Please answer quick .(7 votes)
- I think probably that if we change the function by canceling out (X+10) the domain also would change so if you talking about (x+10)/(x+10)(x-9)(x-5) the domain will be x#-10 and x#9
but if we talking about 1/(x-9)(x-5) the domain will be x#9 ...
hope this be helpful ^^(0 votes)
- At @2:47, why X is greater than -6 ( -6 < X ) ?(3 votes)
- This originates from the absolute value inequality: |x| < 6
You should review the videos on the topic to better understand. Here is the link: https://www.khanacademy.org/math/algebra-home/alg-absolute-value/alg-absolute-value-inequalities/v/absolute-value-inequalities(4 votes)
- What is the absolute value of zero?(2 votes)
- The absolute value of zero is zero. This is because absolute value measures a numerical value's distance from zero on the number line, and zero is zero units away from zero.
Another way to think about absolute value is that if a number is negative, finding its absolute value is just taking the negative sign away. Since zero is not a negative value, we just leave it as zero when finding the absolute value.(4 votes)
- At4:43divide zero by what??(1 vote)
- Sal says "divide by zero" and not "divide zero by" as you seem to have understood. That means he is looking for any numbers that make the denominator equal to zero.(5 votes)
Let's do some more examples finding do mains of functions. So let's say we have a function g of x. So this is our function definition here tells us, look, if we have an input x, the output g of x is going to be equal to 1 over the square root of 6 minus -- we write this little bit neater, 1 over the square root of 6 minus the absolute value of x So like always, pause this video and see if you can figure out what what is the domain of this function. Based on this function definition what is the domain of g? What is the set of all inputs for which this function is defined? Alright. So, to think about all of the inputs that would allow this function to be defined, it may be easier to state when is this function not get defined. Well if we divide by 0 then we're not going to be defined. Or if we have a negative under the radical. So if you think about it -- if if what we have under the radical is 0 or negative -- if it is 0 or negative. If it's 0, you can take the, you can take the principal root of 0. It's going to be 0. But then you are going to divide by 0. That's going to be undefined. And if what you have another radical is negative, the principal root isn't defined for native number, at least the classic principal root is a defined for a negative number. So if 6 minus the absolute value of x is zero or negative, this thing is not going to be defined. Or another way to think about it is it's going to be defined -- so g is is defined -- is defined if -- g is defined if 6 minus, maybe I can write if and only if. Sometimes people write if and only if with two f's right there, iff. g is defined if and only if -- this is kind of mathy way of saying if and only if -- 6 minus the absolute value of x is greater than 0 it has to be, it has to be positive. If it's zero, we're going take the square root of 0 is 0. Then you divide by 0. That's undefined. And if it's less than zero, then you're taking, you're trying to find the principal root of a negative number, that's not defined. So let's see, it's g is defined if and only if this is true. And let's see. We could add the absolute value of x to both sides. We could add the absolute value of x to both sides, then that would give us 6 is greater than the absolute value of x, or that the absolute value of x is less than 6. Or we could say that, you know, let me write that way. The absolute value of x is less than 6. Another way of saying that is x would have to be less than 6 and greater than negative 6. or x is between negative 6 and 6. These two things -- these two things are equivalent. If the magnitude of x is less than 6 then x is greater than negative 6 and less than positive 6. So if we wanted to write the domain in kinda fancy domain-set notation, we could write the domain of g is going to be x, all the x's that are a member of real numbers such that negative 6 is less than x, which is less than 6, and we're done. Let's do another one. And this was gonna get even a little bit a little bit hairier, just for, just for kicks. Alright. So let's say that I have -- let's say that I have -- h of x is equal to -- and I have a kind of composite definition here. So let's say it's x plus 10 over x plus 10 times x minus 9 times x minus 5 times x minus 5 and it's this if it -- h of x is this if x does not equal 5 and it's equal to -- it's equal to pi if x is equal if x is equal to 5. So once again, pause this video. Think about what is the domain of h, or another way to think about is what made h not defined? So let's think about it. So what would make h not defined? So if some -- if x is anything other than 5, we go this clause. If it's 5, we go this clause. So in this clause up here, what would make this thing undefined? Well the most obvious thing is if we divide by 0. So what's going to cause us to divide by 0? So if -- if x is equal to if -- let me write it here -- so we're going to divide by 0 -- divide by 0. That would happen if x is equal to 9. That would happen if x is equal to negative 10. x is equal to negative 10. Now we have to be careful, would that happen if this was the only definition here, it would happen when x equals to 5. But remember when x is equal 5, we don't look at this part of the compound definition. We look at this part. So it's true that up here you would be dividing by 0 if x is equaling 5 but x equaling 5, you wouldn't even look at -- look there. For input is 5, you use this part of the definition. So you'd divide by 0. Maybe I should write it this way, divide by 0 on, I guess you could say the top -- the top clause or the top part a definition. Part of the definition. If x equals 9, X equals negative 10 or -- and that's it because x equal 5 doesn't apply to this top part. If this clause wasn't here then yes, you would write x equals 5. Now we're almost done, but some of you might say, wait, wait, wait, but look, can't I simplify this? I have x plus 10 in the numerator and x plus 10 in the denominator. Can I just simplify this and that will disappear? And you could except, if you did that, you are now creating a different function definition. Because if you just simplify this, you just said 1 over x minus 9 -- 1 over x minus 9 times x minus 5. This is now a different function. That one actually would be defined at x equals negative 10. But the one that we have started with, this one is not. This is -- you're gonna end up with 0 over 0. You gonna end up with that indeterminate form. So for this function, exactly the way it's written, it's not going to be defined with x is equal to 9 or x is equal to negative 10. So once again if you want a fan -- write in our fancy domain set notation. The domain is going to be x all the x's that are a member of the real such that x does not equal 9 and x does not equal negative 10. Any other real number x -- it's going to work including 5. If x equals 5, h of -- h of 5 is going to be equal to pi, because you default to this one over here. h of 5 so OK x is equal 5, we do that one right over there. Now if you gave x equals 9 you gonna divide by 0. If x equals negative 10, you gonna divide by 0, but that's gonna work for anything else. So that right over there is the domain.