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anonymous
 5 years ago
Find the domain of the following functions:
f(x)=(6x^22x+1)/3
anonymous
 5 years ago
Find the domain of the following functions: f(x)=(6x^22x+1)/3

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Domain is looking for numbers where the function is discontinuous.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is continuoues everywhere so its from negative infinity to infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. I wasn't saying you were wrong, just trying to put out the general way of solving the problem without just giving the answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oo yeah gotcha, i was giving a reason to my answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you guys are cool! very helpful i think i have a harder one

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0f(x)=x+squartroot x^2+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cant figure out how to do the squartroot symbol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Still 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here you would check \[\sqrt{x^2+1}\] and well what do you find out?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since the x value will always be positive because it is squared it will be continuous everywhere aswell

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm what if its sqrt root of x^21 instead. will it be same

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Not quite. If x=1 it's still valid.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm still not understanding

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you plug zero in for x in \[\sqrt{x^21}\] you get \[\sqrt{1}\] which is imaginary and not good.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok that wat you meant
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