## zhengcl86 3 years ago determine the domain of $f(x)=\sqrt{9-x ^{2}}$

1. henryrodriguez713

2. zhengcl86

determine the domain $f(x)=\sqrt{9-x ^{2}}$

3. baldymcgee6

Do you know what the question is asking for? i.e. do you know what the domain of a function is?

4. zhengcl86

the question just says determine the domain of $f(x)=\sqrt{9-x ^{2}}$

5. baldymcgee6

Yes I know. But do YOU know what it means? What is the "domain"

6. zhengcl86

oh lol isnt it a set of functions with all possible inputs from neg to pos or infinite?

7. zhengcl86

wouldn't i be breaking down the square root of 9 - x squared (3-x)(3+x)

8. baldymcgee6

It is all the values of x that you could put into your function to get an output value. So a better question would be to ask yourself what values of x can I put into the function that makes it DNE

9. zhengcl86

what does DNE mean?

10. henryrodriguez713

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11. baldymcgee6

DNE is 'does not exist' i.e refers to a part on the functions graph that cannot be determined, because it simply does not exist. :) Basically you need to watch out for square roots, logarithms and denominators equalling zero. In your case, you have a square root, and as @henryrodriguez713 pointed out, you CANNOT have a negative number under the square root.

12. zhengcl86

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13. baldymcgee6

To get rid of the square root, simply square both sides of the equation.

14. zhengcl86

oh since i can't have a negative number, the answer would be false correct? its been 3 years since i've been in school xD

15. baldymcgee6

The function does not exist for x values less than -3 and more than 3

16. baldymcgee6

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17. henryrodriguez713

Haha. Don't worry if you haven't been in school for three years. All that really matters is that you're learning it. :D

18. zhengcl86

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19. zhengcl86

would this be the final product answer?