Question

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

Answer 1

(- oo, oo)

Answer 2

Domain is looking for numbers where the function is discontinuous.

Answer 3

it is continuoues everywhere so its from negative infinity to infinity

Answer 4

Thank you!

Answer 5

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.

Answer 6

oo yeah gotcha, i was giving a reason to my answer

Answer 7

Alright, no worries.

Answer 8

you guys are cool! very helpful i think i have a harder one

Answer 9

And that is?

Answer 10

f(x)=x+squartroot x^2+1

Answer 11

cant figure out how to do the squartroot symbol

Answer 12

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.

Answer 13

Here you would check \[\sqrt{x^2+1}\] and well what do you find out?

Answer 14

since the x value will always be positive because it is squared it will be continuous everywhere aswell

Answer 15

hmm what if its sqrt root of x^2-1 instead. will it be same

Answer 16

Nope.

Answer 17

x cant be 1

Answer 18

Not quite. If x=1 it's still valid.

Answer 19

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.

Answer 20

pj has it right

Answer 21

hmm still not understanding

Answer 22

If you plug zero in for x in \[\sqrt{x^2-1}\] you get \[\sqrt{-1}\] which is imaginary and not good.

Answer 23

oh ok that wat you meant

Answer 24

thanks

Answer 25

Make sense?