How do you determine if y=sqrt(1-x^2)y=1x2 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)f(x)=f(x) for all xx 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.