Find the domain of the following functions: f(x)=(6x^2-2x+1)/3

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Find the domain of the following functions: f(x)=(6x^2-2x+1)/3

Mathematics
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(- oo, oo)
Domain is looking for numbers where the function is discontinuous.
it is continuoues everywhere so its from negative infinity to infinity

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Thank you!
Yeah. I wasn't saying you were wrong, just trying to put out the general way of solving the problem without just giving the answer.
oo yeah gotcha, i was giving a reason to my answer
Alright, no worries.
you guys are cool! very helpful i think i have a harder one
And that is?
f(x)=x+squartroot x^2+1
cant figure out how to do the squartroot symbol
Still looking for areas where the function is discontinuous. This one is a little different. Normally you check for values that make the square root bad. So that would be negative numbers in a square root.
Here you would check \[\sqrt{x^2+1}\] and well what do you find out?
since the x value will always be positive because it is squared it will be continuous everywhere aswell
hmm what if its sqrt root of x^2-1 instead. will it be same
Nope.
x cant be 1
Not quite. If x=1 it's still valid.
You don't want anything under the square root to be negative. So you would set what is under the square root to > or = to 0. And solve.
pj has it right
hmm still not understanding
If you plug zero in for x in \[\sqrt{x^2-1}\] you get \[\sqrt{-1}\] which is imaginary and not good.
oh ok that wat you meant
thanks
Make sense?

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