# Thread: Function Help

1. ## Function Help

Decide whether the given relation defines y as a function of x. Give the domain...

So, would you just put a number in y like 1 and keep incrementing and cubing the value to see if any x-values match or not?

y=x^3

What about something like this...

y= Square root of 4x+2

2. I would plug numbers into y then solve for x to find the domain
$
y=x^3
$

As for,
$
y=\sqrt{4x+2}
$

I would clear the radical from said equation entirely by squaring both sides. Then, I would start plugging numbers into $y^2$ and solve for x. You have to be careful though, and then check your results in the origional equation for any extraneous solutions. Give it a shot. Start with -2 on up to 2 and see what happens. Remember to check your solutions.

3. That is
$
(y)^2=(\sqrt{4x+2})^2
$

is equivalent to
$
y^2=4x+2
$

this is because
$
(\sqrt{4x+2})^2 = 4x+2
$

and
$
(y)^2 = y^2
$

4. Ok, thanks! I thought it was something like that, but the book doesn't explain it very well.

5. Bam ba lam!!