# Thread: domain help

1. ## domain help

Hello,

I am able to get all the other graphs on my paper along with their domain, range, zeros, symmetry with respect to origin, even/odd function, period and if it is a one-to-one function (for every y there is only one x). But I can not seem to figure this out. A graph along with the

Domain:
Range:
Zeros:
Symmetry with respect to origin (y-axis or origin):
even/odd function:
period (if it has one):
is it a one-to-one function:

would be great.

this is the equation:

$\displaystyle f(x)=\sqrt{a^2-x^2}$

2. $\displaystyle y = \sqrt{a^2 - x^2}$

$\displaystyle y^2 = a^2 - x^2$

$\displaystyle x^2 + y^2 = a^2$

does the last equation look familiar?

how does the original equation relate to the above graphically?

wow, I can not believe I missed that, thank you. It is a circle. So the

Domain: all real numbers
Range: all real numbers
Zeros: at $\displaystyle \sqrt{a},0$
Symmetry with respect to origin (y-axis or origin): both
even/odd function: both
period (if it has one):none
is it a one-to-one function:no

correct?

4. Originally Posted by OnMyWayToBeAMathProffesor

correct?
Nop,

Originally Posted by OnMyWayToBeAMathProffesor

Range: all real numbers
Square roots are not negative.

I was saying that range is all real numbers because it can be anywhere on the number line.

6. Short counterexample: consider $\displaystyle f(x)=\sqrt{4-x^2},$ $\displaystyle \exists\,x\in\text{Dom }f\,\Big|\,f(x)=-1$ ?

7. actually, $\displaystyle y = \sqrt{a^2 - x^2}$ is the upper semicircle.

look at its graph.

domain is -a < x < a

range is 0 < x < a

8. aaa, I see, that makes a big difference. I understand 'skeeter' but 'Krizalid', I am sorry, I did not understand your symbols. Thank you