1. ## Functions?

For the j(x) one, i need to know how to put in the negative sign....

I'm trying to find out if those functions are even or odd

Also, how do I find the domain of a function that doesn't have a denominator?

Like say, y + x/y = 3, or sq. root y = x^2 + 1

2. Originally Posted by realintegerz

For the j(x) one, i need to know how to put in the negative sign....

I'm trying to find out if those functions are even or odd

Also, how do I find the domain of a function that doesn't have a denominator?

Like say, y + x/y = 3, or sq. root y = x^2 + 1
For $\displaystyle j(x)$ what you have is perfectly acceptable. If you like you could distribute the negative by multiplying everything in the numerator by -1.

i.e. $\displaystyle j(x) = -\frac{x^3 + 2}{x^2 + 1}= \frac{-x^3 - 2}{x^2 + 1}$.

$\displaystyle m(-x)$ is fine. To find $\displaystyle -m(-x)$, just take the negative of $\displaystyle m(-x)$. So do exactly what I suggested for $\displaystyle j(x)$.

What you've done with $\displaystyle n(x)$ is fine too. Just do what I suggested with $\displaystyle j(x)$ and $\displaystyle m(x)$. It's even, because if you do a little more work, you can get $\displaystyle n(-x) = n(x)$.

To find the domain of a function, it is usually easiest to get y in terms of x.

So for the first one...

$\displaystyle y + \frac{x}{y} = 3$

Multiply everything by y

$\displaystyle y^2 + x = 3y$

Get all the y's on one side...

$\displaystyle y^2 - 3y = -x$

Complete the square...

$\displaystyle y^2 - 3y + \left(-\frac{3}{2}\right)^2 = \left(-\frac{3}{2}\right)^2 - x$

$\displaystyle \left(y - \frac{3}{2}\right)^2 = \frac{9}{4} - x$.

Now solve for y...

$\displaystyle y - \frac{3}{2} = \pm \sqrt{\frac{9}{4} - x}$

$\displaystyle y = \frac{3}{2} \pm \sqrt{\frac{9}{4} - x}$.

Now you should be able to analyse the right hand side to find the domain.

Clearly what is under the square root can't be negative, so...

$\displaystyle \frac{9}{4} - x \geq 0$

$\displaystyle x \leq \frac{9}{4}$.

So the domain is $\displaystyle \left(- \infty , \frac{9}{4} \right ]$.

For the second one, just square both sides and you have y = something. Analyse the right hand side.

3. i got y = x^4 + 1 for the 2nd one

i still dont get how i find the domain for that though.....

do i take the 4th root of it or ?

4. Originally Posted by realintegerz
i got y = x^4 + 1 for the 2nd one

i still dont get how i find the domain for that though.....

do i take the 4th root of it or ?
Actually it's $\displaystyle y = (x^2 + 1)^2 = (x^2 + 1)(x^2 + 1) = x^4 + 2x^2 + 1$

This is a polynomial function. Polynomials are continuous over $\displaystyle \mathbf{R}$. So the domain is $\displaystyle \mathbf{R}$.