anonymous
  • anonymous
Finding the domain of X write in interval notation f(x)= 4x+2 sq.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
since "x" is only apart of 4x, there are no restrictions to it, the domain of f(x) is all real numbers: (-inf,inf)
amistre64
  • amistre64
I noticed from another post, that you might mean that all this is under the radical sign... sqrt(4x+2) ??
anonymous
  • anonymous
yes

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amistre64
  • amistre64
it is good to use the paranthesis to indicate inclusion :) since "even" roots care about the sign of the answers, this square root doesnt want anything that turns it negative. We restrict our "domain" to ll real xs that keep this a zero or positive number.
anonymous
  • anonymous
how do i solve it?
amistre64
  • amistre64
You would set the 4x+2 = 0 and solve for x 4x+2 = 0 4x = -2 x = -2/4 x = -1/2 since this number makes everything under the radical equal to zero, any number less than this will produce a negative result and that is bad. Every number ablve this value will stay positive, to the domain is: x such that x is an element of the set [-1/2, inf)
anonymous
  • anonymous
so if I had f(x)=\[\sqrt{6-3x}\] i would add 6 to both sides
amistre64
  • amistre64
you would notice first of all that this is a square root problem, and that square roots are forbidden to have any negative value under their umbrella. So we equate the problem that is under the radical sign to zero. 6 - 3x = 0 we need this equation here to be either zero or positive numbers. to get the x variable all by itself, we need to "move" its extra baggage to the other side of the (=) sign. We first start off by subtracting 6 from both side. 6-6 -3x = 0-6 0 - 3x = -6 -3x = -6 does this make sense so far?
anonymous
  • anonymous
I get so confused when it is a negative I am writing it down step by step
anonymous
  • anonymous
I got it no negative number
amistre64
  • amistre64
good :) now we are left with -3x = -6 and in order to get rid of the "-3" that is hugging up on that x, we need to remove it by "division" like this. (-3/-3)x = -6/-3 because any number divided by itself equals 1 we get: (1)x = -6/-3 x = -6/-3 we should also know by now that whenever we have 2 negatives they become a positive. to give us: x = 6/3 = 2 So we now that when x = 2 we get a value of zero under the radical: sqrt(6 -3(2)) = sqrt(6-6) = sqrt(0) anything to the left of this value is a negative and bad for our problem. Anything that is to the right of this value is a positive and is good with our problem. <..............2..............> ------- 0 ++++++ So our domain is all the values for x such that x is equal to or greater than 2. or in notation: [2,inf)
anonymous
  • anonymous
I got it that is what i got to if i have a problem \[\sqrt{x-1} divided by 1\]
amistre64
  • amistre64
this here is a fraction with a "1" in its bottom part, a 1 in the denominator correct?
anonymous
  • anonymous
1 is the numerator
amistre64
  • amistre64
sqrt(x-1) -------- is what you have written; is this wrong? 1
anonymous
  • anonymous
yes the one is on top
anonymous
  • anonymous
I got [1,inf)
amistre64
  • amistre64
is there a divided by sign between the square root and that "1" ? sqrt(x-1) [divide sign] 1 ???
amistre64
  • amistre64
[1,inf) is correct :)
anonymous
  • anonymous
\[1 \over \sqrt{x-1?}\]
amistre64
  • amistre64
ahhhh, then in this case, we have another concern to worry about, the value of the bottom cannot equal zero.
anonymous
  • anonymous
okay why
amistre64
  • amistre64
so we have to restrict our domain of x to all values other than zero, and becasue we have a square root there, we also have to limit it to all values that are positive.
amistre64
  • amistre64
the expression 1/0 has no value in math. it is meaningless. and cannot be used to produce anything with meaning. 5 + 1/0 = ???? cant be solved since 1/0 has no value.
anonymous
  • anonymous
so it is (-inf,inf)
amistre64
  • amistre64
we still have the radical sign to worry about also, which means that we cannot have a value that is less than 0. All in all, the value of x in: sqrt(x-1) cannot make the inside go negative; and because it is all in the denominator, that value of: sqrt(x-1) cannot be equal to zero. So we get this: sqrt(x-1) has the domain [1,inf) but since x=1 makes this problem = zero: we need to remove it from the domain as well. so we get: (1,inf)
anonymous
  • anonymous
on another problem they have 1 over\[\sqrt{2x+2}\] and came up with -1/2,inf
anonymous
  • anonymous
is that because it was 2x+1
amistre64
  • amistre64
1 ---------- correct? sqrt(2x + 2)
anonymous
  • anonymous
never mind I needed to read on
anonymous
  • anonymous
Thank you you have been a lot of help very smart person
amistre64
  • amistre64
thanx :) and youre welcome
anonymous
  • anonymous
I think I can pass this part of my finals
amistre64
  • amistre64
yay!!!

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