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!
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) $\displaystyle f(x) = x^2$, (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
?! 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?
interesting. say $\displaystyle x = 1$, is it true that $\displaystyle 2^1 = 2^{-1}$ or $\displaystyle -2^1 = 2^{-1}$ ?
tell me, what's $\displaystyle \log_2 (-5)$, say
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.