How do I figure out if a 'function is odd or even'?

Question

I am currently doing an highschool math problem and I do not know what the question is asking for when it asks 'state which functions are odd and which are even for the below'.

• $f(x)= x^2+1 $ ,
  Dom: $x : R$
• $h(x)= 1/(x^2)$ ,
  Dom: $x ≠ 0$
• $g(x)= x$ ,
  Dom: $x : R$
• $K(x)= 1/x$ ,
  Dom: $x ≠ 0$

• What can I do to find out if a function is odd or even?

• What is the definition of an odd function and an even function?

Answer 1

I found an explanation that was really clear, and it seem to complement the answer that Mahidevran gave.

Answer 2

$F$ is $\mathbf{even}$ if $F(x) = F(-x)$ for all $x$.

$F$ is $\mathbf{odd}$ if $F(-x) = - F(x) $ for all $x$

$\mathbf{Homework}$: Are there functions that are both even and odd?. [Hint: There is exactly one.]

Answer 3

substitute $-x$ on the function. That is if $f(-x) = f(x)$ then even, if $f(-x) = -f(x)$ then odd.

1) if $f(x) = x^{2} + 1$, then substituting $-x$ on the funciton we have $f(-x) =(-x)^{2} + 1 = x^2 + 1 = f(x)$ so even.

2) if $h(x) = \frac{1}{x^2}$ ,then $h(-x) = \frac{1}{(-x)^2} = \frac{1}{x^2} = h(x)$. So even.

3) if $g(x) = x$, then $g(-x) = -x = -g(x)$ so odd.

4) if $k(x) = \frac{1}{x}$, then $k(-x) = \frac{1}{(-x)} = -\frac{1}{x} = -k(x) $ so odd.

Answer 4

Odd function graphically they are symmetric about an origin and for even function they are symmetric about the y-axis so you can see it that way...