Extend a function as odd/even periodic function Let $f$ be the function $f(x) = x^2 + 2 $,    where  $   0<x<1 $.
Extend the function $f(x)$
(1) As an odd periodic function with period $2$
(2) As an even periodic function with period $2$
(3) As a periodic function with period $1$  
I know what exactly are odd and even functions. But have no idea how to extend as such. Please explain with a correct method. 
 A: *

*Even case


Due to symmetry we need to define $f(x)=x^2+2$ on $(-1,0)$. It remains to discuss the $2$-periodicity. One starts with "special points", for example $x=0$.  Then
$$f(0)=0^2+2=\text{(2-periodicity)}=f(0+2)=f(2),$$
i.e. $f(2):=2$. To find the extension of $f$ on $(2,3)$ one continues with all $\delta\in (0,1)$, i.e.
$$f(\delta)=\delta^2+2=\text{(2-periodicity)}=f(\delta+2),$$
where $\delta+2\in (2,3)$. Similarly, considering all points $-\delta\in(-1,0)$, using $2$-periodicity one finds the extension of $f$ on the interval $(1,2)$ via $-\delta+2\in (1,2)$, as we did above. The extension of $f$ for all other points easily follow by drawing.


*

*Odd case


Due to symmetry we need to define $f(x)=-x^2-2$ on $(-1,0)$. One selects once again
the "special points", for example $x=0$ as in the even case. The analysis is completely similar, with due changes. 
A: Given a function $f:\ A\to{\mathbb R}$ and a superset $\tilde A\supset A$  extending $f$ to $\tilde A$ means defining a function $\tilde f:\ \tilde A\to{\mathbb R}$ such that the restriction of $\tilde f$ to $A$ coincides with the given $f$. Usually the extended function is denoted by $f$ again.
In your case $A=\ ]0,1[\ $. As the extended functions are required to be periodic I shall assume that $\tilde A={\mathbb R}$ is intended; but $\tilde A={\mathbb R}\setminus{\mathbb Z}$ would also be okay.
ad 1: The requirement $f(-x)=f(x)$ enforces $f(x)=x^2+1$ $\>(-1<x<0)$. The value $f(0)$ may be chosen arbitrarily; and the same holds for $f(1)=f(-1)$.  Now $f$ is specified on the full interval $[-1,1]$. Since $[-1,1]$ is a fundamental domain for the envisaged $2$-periodicity it follows that the extended $f$ is now determined on all of ${\mathbb R}$. The value of the extended function at an arbitrary point $x\in{\mathbb R}$ is obtained as follows: Write $x$ in the form $x=2k+x'$ with $x'\in[-1,1]$ and put
$$f(x):=f(x')\ .$$
When we have chosen $f(0)=0$ and $f(1)=f(-1)=2$ then the extended function will be continuous on all of ${\mathbb R}$.
ad 2: The requirement $f(-x)=-f(x)$ enforces $f(x)=-x^2-1$ $\>(-1<x<0)$, as well as $f(0)=0$.  Now $f$ is specified on the full interval $\ ]-1,1[\ $. Since $[-1,1]$ is a fundamental domain for the envisaged $2$-periodicity it follows that the extended $f$ is now determined on all of ${\mathbb R}$, apart from the odd integers. Now the condition $f(-1)=-f(1)$ together with $f(1)=f(-1)$ enforces $f(-1)=f(1)=0$. The value of the extended function at an arbitrary point $x\in{\mathbb R}$ is therefore obtained as follows: Write $x$ in the form $x=2k+x'$ with $x'\in[-1,1]$ and put
$$f(x):=f(x')\ .$$
The extended function, which was uniquely determined, is not continuous at the odd integers, since it has a jump discontinuity there.
I leave  case 3. to you.
