How to expand $f(x)=x^2$ in the interval of $(0,\pi)$ to an odd and even function with the interval of $(-\pi,\pi)$?
I know that in even function $f(x)=f(-x)$ and for an odd function $f(-x)=-f(x)$ which in this case, for even:
$f(x)= x^2$ for $0$ to $\pi$
$f(x)=-x^2$ for $-\pi$ to $0$
For odd function:
$f(x)=x^2$ for $0$ to $\pi$
$f(x)=-(-x^2)=x^2$ for $-\pi$ to $0$
Is this correct? And how do you extend this function to become an odd function and how to extend this to become an even function? I don't understand how to do this expansion.
Your answers are backwards. $f(x)=x^2$ is even as $f(-x)=(-x)^2=x^2=f(x)$. Intuitively, even means the graph is symmetric around the $y$ axis and $y=x^2$ is. Then to extend it to an odd function, for which we will use $g$, when $x \gt 0$ we need $g(-x)=-g(x)=-(x^2)$