# Finding the Domain Problems:

• Jan 9th 2007, 01:57 PM
qbkr21
Finding the Domain Problems:

1. 5throot(-4+3x)

The function is defined on the interval from _________ to __________

2. 6throot(-4+3x)

The function is defined on the interval from _________ to __________

3. The domain of the function sqrt(x(x-4)) in interval notation is _________

Thanks so much!!
• Jan 9th 2007, 02:03 PM
topsquark
Quote:

Originally Posted by qbkr21

1. 5throot(-4+3x)

The function is defined on the interval from _________ to __________

2. 6throot(-4+3x)

The function is defined on the interval from _________ to __________

3. The domain of the function sqrt(x(x-4)) in interval notation is _________

Thanks so much!!

For an "even" root we require that the argument is non-negative. For an "odd" root we don't have that restriction. So the answer to 1 is all real numbers and the answer to 2 is $[0, \infty)$.

3 is a tad more difficult, but all we need do is determine where $x(x-4) \geq 0$. This happens for $( - \infty, 0) \cup (4, \infty)$ as you can verify by several means.

-Dan
• Jan 9th 2007, 02:09 PM
qbkr21
Except for #3 you must use bracket, also I am unable to figure out how you got #2 could you please explain a bit more?
• Jan 9th 2007, 02:28 PM
topsquark
Quote:

Originally Posted by qbkr21
Except for #3 you must use bracket, also I am unable to figure out how you got #2 could you please explain a bit more?

Yes, the answer for 3 should be $( - \infty, 0] \cup [4, \infty)$ because of the $\geq$ relation. Thank you.

For 2 we require that the argument under the 6th root be non-negative for the same reason we require the same for the square root: $x^{1/6}$ is not defined for negative x. (Or if you wish to speak more loosely, a negative x will return an imaginary value.)

So we see that I goofed again. :o The argument of the 6th root is $-4 + 3x$ so we require that $-4 + 3x \geq 0$ or the solution for x is the interval: $[4/3, \infty )$.

(I was obviously doing this problem too fast. My apologies!)

-Dan