How do you determine if # y=sqrt(1-x^2)# is an even or odd function?

1 Answer
Mar 6, 2018

Easy method: graph it
Rigorous method: prove that it is odd or even using the definitions

Explanation:

Easy method: graph{y=sqrt(1-x^2) [-10, 10, -5, 5]}
By inspection, we see that the graph is symmetric about the y-axis so it is even.

Rigorous method:
We show that the function is even by showing that #f(x)=f(-x)# for all #x# in the domain, which in this case is #[-1,1] sub RR#. Thus, we get begin with #y=sqrt(1-x^2)# and we want to show that #y=sqrt(1-(-x)^2)# as well. This is easy for the problem since #(-x)^2=x^2# for real #x#, so clearly, the two are equal and the function is even.

We may also show that the function is not odd by showing that #f(x)=-f(-x)# for all #x# in the domain. Thus, we want to show that #sqrt(1-x^2)=-sqrt(1-(-x)^2)#. Clearly this is false as equality only holds when #sqrt(1-x^2)=0# or #x=+-1#, which clearly is not all #x#. Thus, the function is not odd.