# Math Help - even/odd graph

1. ## even/odd graph

Are the graphs of

$\sqrt{x}$

$2^x$

$log_2x$

even or odd? I don't see any symmetry with respect to y axis or origin. Thanks!

2. Originally Posted by live_laugh_luv27
Are the graphs of

$\sqrt{x}$

$2^x$

$log_2x$

even or odd? I don't see any symmetry with respect to y axis or origin. Thanks!
A function $f(x)$ is even if $f(-x) = f(x)$

A function $f(x)$ is odd if $f(-x) = -f(x)$

can you continue?

3. so,

even

odd

even

?

4. I'm still not exactly sure how to set that up, but have another ?...even if a function does not have symmetry in respect to the y-axis, can it still be even? similarly, even if the function does not have symmetry in respect to the origin, can it still be odd?
thanks!

5. Originally Posted by live_laugh_luv27
so,

even

odd

even
What does any of the above have to do with the basic definitions of odd & even?
$\text{Odd functions are such that }f(-x)=-f(x)$

$\text{Even functions are such that }f(-x)=f(x)$

6. Originally Posted by live_laugh_luv27
I'm still not exactly sure how to set that up, but have another ?...even if a function does not have symmetry in respect to the y-axis, can it still be even? similarly, even if the function does not have symmetry in respect to the origin, can it still be odd?
thanks!
no, even and odd follow the exact definitions i gave you. you don't even have to look at the graphs.

example, state whether (a) $f(x) = x^2$, (b) $f(x) = \sin x$ and (c) $f(x) = \frac {x^2}{x^3 + 1}$

(a) $f(x) = x^2$ is even, since

$f(-x) = (-x)^2 = x^2 = f(x)$

(b) $f(x) = \sin x$ is odd, since

$f(-x) = \sin (-x) = - \sin x = - f(x)$

(c) $f(x) = \frac {x^2}{x^3 + 1}$ is neither even nor odd, since

$f(-x) = \frac {(-x)^2}{(-x)^3 + 1} = \frac {x^2}{1 - x^3} \ne f(x) \text{ or } -f(x)$

notice that i didn't even draw any graphs. now, try again

7. Originally Posted by Plato
What does any of the above have to do with the basic definitions of odd & even?
$\text{Odd functions are such that }f(-x)=-f(x)$

$\text{Even functions are such that }f(-x)=f(x)$
so these answers are incorrect?
- even

- odd

- even

8. Originally Posted by live_laugh_luv27
so these answers are incorrect?
- even

- odd

- even
yes, they are wrong. did you try what i said?

9. = $\sqrt{-x}$ = odd

= $2^{-x}$ = odd?

= $\frac{ln(-x )}{ln2}$ ?

10. Originally Posted by live_laugh_luv27
= $\sqrt{-x}$ = odd
?! are you telling me that, for example, $\sqrt 2 = \sqrt {-2}$? and if so, does that comply with the definition of an odd function as defined in my and Plato's posts?

= $2^{-x}$ = odd?
interesting. say $x = 1$, is it true that $2^1 = 2^{-1}$ or $-2^1 = 2^{-1}$ ?

= $\frac{ln(-x )}{ln2}$ ?
tell me, what's $\log_2 (-5)$, say

11. = = $-\sqrt{x}$ = $-f(x)$ ?

is not true

is $\frac{ln(-5)}{ln(2)}$ ?

12. Originally Posted by live_laugh_luv27
= = $-\sqrt{x}$ = $-f(x)$ ?

is not true

is $\frac{ln(-5)}{ln(2)}$ ?
ok, as far as real numbers are concerned, the square root function and the logarithm function are not defined for negative real numbers.... you should know this. please look this up and make sure you get it

and clearly $2^1 \ne 2^{-1}$ nor does $-2^1 = 2^{-1}$ so that $f(x) = 2^x$ is neither even nor odd.