Are the graphs of

$\displaystyle \sqrt{x}$

$\displaystyle 2^x$

$\displaystyle log_2x$

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

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- Aug 9th 2009, 09:27 AMlive_laugh_luv27even/odd graph
Are the graphs of

$\displaystyle \sqrt{x}$

$\displaystyle 2^x$

$\displaystyle log_2x$

even or odd? I don't see any symmetry with respect to y axis or origin. Thanks! - Aug 9th 2009, 09:40 AMJhevon
- Aug 9th 2009, 10:10 AMlive_laugh_luv27
- Aug 9th 2009, 10:15 AMlive_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! - Aug 9th 2009, 10:17 AMPlato
- Aug 9th 2009, 10:27 AMJhevon
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) $\displaystyle f(x) = x^2$, (b) $\displaystyle f(x) = \sin x$ and (c) $\displaystyle f(x) = \frac {x^2}{x^3 + 1}$

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

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

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

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

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

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

notice that i didn't even draw any graphs. now, try again - Aug 9th 2009, 10:29 AMlive_laugh_luv27
so these answers are incorrect?

http://www.mathhelpforum.com/math-he...19f85201-1.gif - even

http://www.mathhelpforum.com/math-he...84e8019b-1.gif - odd

http://www.mathhelpforum.com/math-he...069c3f11-1.gif - even - Aug 9th 2009, 10:31 AMJhevon
- Aug 9th 2009, 10:33 AMlive_laugh_luv27
http://www.mathhelpforum.com/math-he...19f85201-1.gif = $\displaystyle \sqrt{-x}$ = odd

http://www.mathhelpforum.com/math-he...84e8019b-1.gif = $\displaystyle 2^{-x}$ = odd?

http://www.mathhelpforum.com/math-he...069c3f11-1.gif = $\displaystyle \frac{ln(-x )}{ln2}$ ? - Aug 9th 2009, 10:36 AMJhevon
?! are you telling me that, for example, $\displaystyle \sqrt 2 = \sqrt {-2}$? and if so, does that comply with the definition of an odd function as defined in my and Plato's posts?

Quote:

Quote: - Aug 9th 2009, 10:52 AMlive_laugh_luv27
http://www.mathhelpforum.com/math-he...19f85201-1.gif = http://www.mathhelpforum.com/math-he...a3ae9cbf-1.gif = $\displaystyle -\sqrt{x}$ = $\displaystyle -f(x)$ ?

http://www.mathhelpforum.com/math-he...2950423e-1.gif is not true

http://www.mathhelpforum.com/math-he...f1a02f81-1.gif is $\displaystyle \frac{ln(-5)}{ln(2)}$ ? - Aug 9th 2009, 10:57 AMJhevon
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 $\displaystyle 2^1 \ne 2^{-1}$ nor does $\displaystyle -2^1 = 2^{-1}$ so that $\displaystyle f(x) = 2^x$ is neither even nor odd.