When is a function neither odd or even?For example why is x^3/2 neither odd or even?

When the domain is R\{0}, is the range always R+?

How do u find {x:x^3/2>x^2 } and {x:x^-3/2<x^-2}?

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- Jan 3rd 2008, 12:23 AMchanelimanFamilies of functions
When is a function neither odd or even?For example why is x^3/2 neither odd or even?

When the domain is R\{0}, is the range always R+?

How do u find {x:x^3/2>x^2 } and {x:x^-3/2<x^-2}? - Jan 3rd 2008, 12:27 AMJhevon
$\displaystyle \frac {x^3}2$ is odd. did you mean to say $\displaystyle x^{3/2}$ ?

a function f(x) is even if f(-x) = f(x)

a function is odd if f(-x) = -f(x)

otherwise, it is neither

Quote:

When the domain is R\{0}, is the range always R+?

Quote:

How do u find {x:x^3/2>x^2 } and {x:x^-3/2<x^-2}?

can you solve $\displaystyle x^{3/2} > x^2$ for x? (is the power 3/2? type clearly) - Jan 3rd 2008, 12:34 AMchaneliman
yea i typed it right

- Jan 3rd 2008, 12:46 AMJhevon
- Jan 3rd 2008, 12:50 AMchaneliman
i can't solve x^3/2>x^2 for x. Does it have something 2 do with logs

? - Jan 3rd 2008, 04:46 AMcolby2152
- Jan 3rd 2008, 08:25 AMThePerfectHacker
- Jan 3rd 2008, 08:26 AMcolby2152
- Jan 3rd 2008, 11:48 AMJhevon
- Jan 3rd 2008, 12:01 PMcolby2152
- Jan 3rd 2008, 02:30 PMchaneliman
i understand it till the point u got 0<x<1

- Jan 3rd 2008, 02:33 PMJhevon
we have $\displaystyle 1 > \sqrt{x}$

note that the domain of the square root function is $\displaystyle x \ge 0$, so we must have that for $\displaystyle \sqrt{x}$ to make sense. however, we drop the $\displaystyle x = 0$ part, because that does not work in the original inequality.

now, if we square both sides, we get: $\displaystyle 1 > x$

so we have $\displaystyle 0< x$ and $\displaystyle x < 1$, hence $\displaystyle 0 < x < 1$ - Jan 3rd 2008, 03:15 PMchaneliman
then shouldn't it be 0>x if the domain of the square root function is more then or equal to 0

- Jan 3rd 2008, 03:27 PMJhevon