Functions?

**realintegerz** · September 28th 2008, 06:52 PM

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

**Prove It** · September 29th 2008, 03:12 AM

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 $j(x)$ what you have is perfectly acceptable. If you like you could distribute the negative by multiplying everything in the numerator by -1.

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

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

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

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

So for the first one...

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

Multiply everything by y

$y^2 + x = 3y$

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

$y^2 - 3y = -x$

Complete the square...

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

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

Now solve for y...

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

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

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...

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

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

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

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

**realintegerz** · September 29th 2008, 02:41 PM

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 ?

**Prove It** · September 30th 2008, 01:11 AM

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 $y = (x^2 + 1)^2 = (x^2 + 1)(x^2 + 1) = x^4 + 2x^2 + 1$

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

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