## anonymous 5 years ago Find the domain of the following functions: f(x)=(6x^2-2x+1)/3

1. anonymous

(- oo, oo)

2. anonymous

Domain is looking for numbers where the function is discontinuous.

3. anonymous

it is continuoues everywhere so its from negative infinity to infinity

4. anonymous

Thank you!

5. 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.

6. anonymous

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

7. anonymous

Alright, no worries.

8. anonymous

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

9. anonymous

And that is?

10. anonymous

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

11. anonymous

cant figure out how to do the squartroot symbol

12. 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.

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

Nope.

17. anonymous

x cant be 1

18. anonymous

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

19. 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.

20. anonymous

pj has it right

21. anonymous

hmm still not understanding

22. anonymous

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

23. anonymous

oh ok that wat you meant

24. anonymous

thanks

25. anonymous

Make sense?