anonymous
  • anonymous
Find the domain of the following functions: f(x)=(6x^2-2x+1)/3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
(- oo, oo)
anonymous
  • anonymous
Domain is looking for numbers where the function is discontinuous.
anonymous
  • anonymous
it is continuoues everywhere so its from negative infinity to infinity

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anonymous
  • anonymous
Thank you!
anonymous
  • anonymous
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.
anonymous
  • anonymous
oo yeah gotcha, i was giving a reason to my answer
anonymous
  • anonymous
Alright, no worries.
anonymous
  • anonymous
you guys are cool! very helpful i think i have a harder one
anonymous
  • anonymous
And that is?
anonymous
  • anonymous
f(x)=x+squartroot x^2+1
anonymous
  • anonymous
cant figure out how to do the squartroot symbol
anonymous
  • anonymous
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.
anonymous
  • anonymous
Here you would check \[\sqrt{x^2+1}\] and well what do you find out?
anonymous
  • anonymous
since the x value will always be positive because it is squared it will be continuous everywhere aswell
anonymous
  • anonymous
hmm what if its sqrt root of x^2-1 instead. will it be same
anonymous
  • anonymous
Nope.
anonymous
  • anonymous
x cant be 1
anonymous
  • anonymous
Not quite. If x=1 it's still valid.
anonymous
  • anonymous
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.
anonymous
  • anonymous
pj has it right
anonymous
  • anonymous
hmm still not understanding
anonymous
  • anonymous
If you plug zero in for x in \[\sqrt{x^2-1}\] you get \[\sqrt{-1}\] which is imaginary and not good.
anonymous
  • anonymous
oh ok that wat you meant
anonymous
  • anonymous
thanks
anonymous
  • anonymous
Make sense?

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