1. Domain

Sorry I had to write it out, I don't know how to put it on my computer and I don't have a scanner.

Find the domain:
f(x) = cube root of (t + 4)
f(x) = fourth root of (x squared + 3x)
f(x) = x - 5 / (square root of (x squared - 9))

2. Hello,
Originally Posted by superpike
Sorry I had to write it out, I don't know how to put it on my computer and I don't have a scanner.
It's ok, we can easily read it. You put the parentheses correctly

Find the domain:
f(x) = cube root of (x + 4)
f(x) = fourth root of (x squared + 3x)
f(x) = x - 5 / (square root of (x squared - 9))
Remember the domain is all the values x can have and still the function is defined at this point !

What is under an even-th root should be positive. For odd-th root, it's ok. That is to say you don't care about the sign of x+4 in the cube (3) root (and hence the domain is...), but you'll care about the sign of x²+3x under the fourth root, or the sign of x²-9 under the square (2) root.

When you have a quotient, the denominator can't be 0. So find the values of x for which it is 0. Then remoove them from the possible values of x.

Now please, give it a try, or ask questions if explanations weren't clear enough.

3. 1) all x values for which x is greater than or equal to 4
2) all real numbers greater than or equal to 0 except 0 or -3
3) all real numbers except -3 and 3

Am I right?

4. Originally Posted by superpike
1) all x values for which x is greater than or equal to 4
Hmm... I told you it didn't matter for the cube root. All values of x will fit in !
And t+4>0 ---> t> -4, not t>4
2) all real numbers greater than or equal to 0 except 0 or -3
This is not a quotient.

$\sqrt[4]{x^2+3x}=\sqrt[4]{x(x+3)}$
x(x+3) has to be positive, that is to say if x is negative, then x+3 is negative. If x is positive, x+3 is positive.
This gives 2 possibilities. But I notice that it is quite difficult to give you only tips if you've never done it before... I hoped you would know

$\begin{array}{c|ccc} & -\infty ~~~~~~ & -3 ~~~~~~ & 0 ~~~~~~ + \infty \\ \hline
x & ~~~~~~ - & \vline ~~~~~~ + & \vline ~~~~~~ + \\ \hline
x+3 & ~~~~~~ - & \vline ~~~~~~ - & \vline ~~~~~~ + \\ \hline \hline
x(x+3) & ~~~~~~ + & \vline ~~~~~~ - & \vline ~~~~~~ +
\end{array}$

Try to understand this table please.

3) all real numbers except -3 and 3
Try the reasoning above.
And indeed, -3 and 3 are not possible values

Am I right?
Unfortunately no... But it will come to the Truth lol

5. So then 1 is all real numbers.

I do not understand the chart at all, sorry. Is that the infinity sign? What's its relevance?

6. The chart tells you in what intervals the quantities x, x+3, and x(x+3) will be positive or negative.

7. I understand that part of the chart (at least I think), but I don't know how Moo generated the top half or what s/he is exactly doing or even what it means.
I looked at Moo's first post closer and realized that x = radical negative three if I make f(x) = 0... does that help at all?

I'm sorry, my teacher never taught me how to do this...

8. Originally Posted by superpike
I understand that part of the chart (at least I think), but I don't know how Moo generated the top half or what s/he is exactly doing or even what it means.
I looked at Moo's first post closer and realized that x = radical negative three if I make f(x) = 0... does that help at all?

I'm sorry, my teacher never taught me how to do this...
Basically, the first part of making this chart is to find out where x and x + 3 are equal to zero. We know that x is zero at x = 0 and x + 3 is zero at x = -3. Now, for all values other than x = 0 and x = 3, the product of x(x+3) will be either positive or negative, because both of the factors will be nonzero. All the chart says is that for all values of x < -3, both x and x + 3 will be negative, which will make the product positive. For -3 < x < 0, x + 3 will be positive but x will be negative so the product will be negative. And for all values of x > 0, both x and x+3 will be positive so their product will be positive.

9. So does that make the domain -3 < x < 0, because the value when inserted in has to be positive as it is in the chart?

10. Not quite, notice that using Moo's chart in the interval you chose

$x = -$

$x + 3 = +$

So (+)(-) = -

11. Originally Posted by Moo
Hmm... I told you it didn't matter for the cube root. All values of x will fit in !
And t+4>0 ---> t> -4, not t>4

This is not a quotient.

$\sqrt[4]{x^2+3x}=\sqrt[4]{x(x+3)}$
x(x+3) has to be positive, that is to say if x is negative, then x+3 is negative. If x is positive, x+3 is positive.
This gives 2 possibilities. But I notice that it is quite difficult to give you only tips if you've never done it before... I hoped you would know

$\begin{array}{c|ccc} & -\infty ~~~~~~ & -3 ~~~~~~ & 0 ~~~~~~ + \infty \\ \hline
x & ~~~~~~ - & \vline ~~~~~~ {\color{red}-} & \vline ~~~~~~ + \\ \hline
x+3 & ~~~~~~ - & \vline ~~~~~~ {\color{red}+} & \vline ~~~~~~ + \\ \hline \hline
x(x+3) & ~~~~~~ + & \vline ~~~~~~ - & \vline ~~~~~~ +
\end{array}$

Try to understand this table please.

Try the reasoning above.
And indeed, -3 and 3 are not possible values

Unfortunately no... But it will come to the Truth lol
Made a correction for you Moo