# Domain/Function Notation/Symmetry...

• November 12th 2009, 05:24 PM
anni88
Domain/Function Notation/Symmetry...
Please help! I hit a snag while attempting these, and will try to explain why, but it might just be easier to compare my answer to others. And I dont know how to do symbols so I will try to explain it. Thanks!
Find the domain of the following function...
n(x)=
The square root of (2-3x) over 3x+1
got the 2 domains but not sure if I can have a negative part? since its a square root..

For the function f(x)=square root of x^2-2x, g(x)= 2 over x^2-4, find:
g [f(2)]

For the following functions, determine whether their graphs are symmetric about y-axis, origin, or neither:

f(x)= -4 over x^3

okay last one sorry!..
so theres a graph and it is asking for zeros of f. (how do I write that? just how many times it hits the x axis? or what those points are) and intervals where f is increasing, decreasing or constant. and I just don't know how I would write that answer either. and last but not last it wants to know the intervals where f is positive, and where it is negative. is it negative only when under the x axis? or for it to be positive does it hvae to be in the 1st quadrant of the graph?
thanks in advance for any help!!!
• November 12th 2009, 08:48 PM
Bacterius
I'll explain the first question. You have (is that right ? There is confusion in your question, is it the whole fraction under square root ?) :

$f(x) = \frac{\sqrt{2 - 3x}}{3x + 1}$

Now you want to find the domain of this function. That is, for which values of $x$ is $f(x)$ defined.

Step 1 - look out for any special operations that are not defined for all real numbers (for example, $2x$ is defined for any real number, but $\sqrt{x}$ is not defined for negative numbers).

Here, you have two such operations : the square root cannot take any negative number, and the denominator of the fraction cannot be equal to $0$.

Step 2 - identify the expressions that may lead to an undefined function.

First issue, the square root : $2 - 3x$ cannot be negative, so it must be positive or equal to $0$. Thus, we have :

$2 - 3x \geq 0$

$-3x \geq -2$

$3x \leq 2$

$x \leq \frac{2}{3}$

Second issue, the denominator, which is $3x + 1$, cannot be zero. Thus, we have :

$3x + 1 \neq 0$

$3x \neq -1$

$x \neq \frac{-1}{3}$

Step 3 - put together the information and extract the domain.

Remind the information found :

$x \leq \frac{2}{3}$
$x \neq \frac{-1}{3}$

Therefore, the domain of $f(x)$ would be :

$(- \infty; \frac{-1}{3} ] U [ \frac{-1}{3} ; \frac{2}{3} ]$

Step 4 - check your answer on a graph-drawer software (or your calculator). Ensure the domain you found matches the one found by the software, and be sure you put the right interval signs ( [,],(,) ), especially on a zero-denominator restriction (be sure to exclude it on BOTH sides of the interval).

Does it help ?

PS : if I misunderstood your equation, I made sure I gave you enough tips for you to be able to work out the domain by yourself following the 4 steps.
• November 13th 2009, 03:37 AM
stapel
2) To learn about symmetry, try here.

3) To learn what it means for slope to be "increasing" or "decreasing", try here. Then note that the meaning of the terms, in your context of a curvy line, is the same: "increasing" means "going up" and "decreasing" means "going down", regardless of whether the curve is above or below the x-axis.

Yes, the zeroes are the x-intercepts, where the line crosses or touches the x-axis. (Wink)