# Families of functions

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• January 3rd 2008, 12:23 AM
chaneliman
Families 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}?
• January 3rd 2008, 12:27 AM
Jhevon
Quote:

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

$\frac {x^3}2$ is odd. did you mean to say $x^{3/2}$ ?

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+?
consider the function f(x) = 1/x

Quote:

How do u find {x:x^3/2>x^2 } and {x:x^-3/2<x^-2}?
set up your inequalities and solve for x

can you solve $x^{3/2} > x^2$ for x? (is the power 3/2? type clearly)
• January 3rd 2008, 12:34 AM
chaneliman
yea i typed it right
• January 3rd 2008, 12:46 AM
Jhevon
Quote:

Originally Posted by chaneliman
yea i typed it right

i do not know what you are refering to. you should quote the part of my message that you are responding to so i can respond
• January 3rd 2008, 12:50 AM
chaneliman
i can't solve x^3/2>x^2 for x. Does it have something 2 do with logs
?
• January 3rd 2008, 04:46 AM
colby2152
Quote:

Originally Posted by chaneliman
i can't solve x^3/2>x^2 for x. Does it have something 2 do with logs
?

$x^{3/2} > x^2$

Divide by $x^{3/2}$

$1 > \sqrt{x}$

$0 < x < 1$
• January 3rd 2008, 08:25 AM
ThePerfectHacker
Quote:

Originally Posted by colby2152
$x^{3/2} > x^2$

Divide by $x^{3/2}$

$1 > \sqrt{x}$

$0 \le x < 1$

Before you divide by $x^{3/2}$ you should say that $x^{3/2} > 0$ because otherwise the inequality is flipped.
• January 3rd 2008, 08:26 AM
colby2152
Quote:

Originally Posted by ThePerfectHacker
Before you divide by $x^{3/2}$ you should say that $x^{3/2} > 0$ because otherwise the inequality is flipped.

True, but I kept that as an unstated assumption since it holds in the end o fthe solution with the square root.
• January 3rd 2008, 11:48 AM
Jhevon
Quote:

Originally Posted by colby2152
[tex]
$0 \le x < 1$

another technical note, which i'm sure was just a typo or something on your part. we can't have x = 0, since dividing by x^{3/2} would be invalid, and also the original inequality would not hold in the first place. we need x > 0 here for it to make sense
• January 3rd 2008, 12:01 PM
colby2152
Quote:

Originally Posted by Jhevon
another technical note, which i'm sure was just a typo or something on your part. we can't have x = 0, since dividing by x^{3/2} would be invalid, and also the original inequality would not hold in the first place. we need x > 0 here for it to make sense

I didn't see that at last glance, but I originally had a regular greater than sign, but changed it for some reason.;)
• January 3rd 2008, 02:30 PM
chaneliman
i understand it till the point u got 0<x<1
• January 3rd 2008, 02:33 PM
Jhevon
Quote:

Originally Posted by chaneliman
i understand it till the point u got 0<x<1

we have $1 > \sqrt{x}$

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

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

so we have $0< x$ and $x < 1$, hence $0 < x < 1$
• January 3rd 2008, 03:15 PM
chaneliman
then shouldn't it be 0>x if the domain of the square root function is more then or equal to 0
• January 3rd 2008, 03:27 PM
Jhevon
Quote:

Originally Posted by chaneliman
then shouldn't it be 0>x if the domain of the square root function is more then or equal to 0

yes. more than or equal to zero is $x \ge 0$. we reject the = (for the afore mentioned reason) and make it $x > 0$. we cannot have x < 0, because the square root is not defined for such x's