the problems are on here

http://www.montgomeryschoolsmd.org/S.../PreCalc09.pdf

- Aug 27th 2009, 12:46 PMyitsonggI do not understand problmes 9 to 15 how do you determine domains and ranges again?
the problems are on here

http://www.montgomeryschoolsmd.org/S.../PreCalc09.pdf - Aug 27th 2009, 12:54 PMRobLikesBrunch
Let's look at 9 first:

$\displaystyle y=(x+1)^2-3$

A function's domain is all the possible x values that work within a function. A domain is often restricted by functions like $\displaystyle \frac{x}{x-1}\$where $\displaystyle x \neq 1$ because that would result in dividing by 0. I know this is a rather...imprecise definition, but I think we can work with it. The range is all the possible y values.

So looking here, is there an possible x value that won't work? No, there isn't. You could put any negative number as x or any positive number as x, and add one to it and square it without yielding an undefined answer. So the domain is simply the set of all real numbers or $\displaystyle (-\infty,\infty)$

In the range, on the other hand, we must look at all the possible resulting y-values. Notice that if we square something, the result will*always*be positive; so the lowest possible number that y can be is -3, which occurs when x = -1. However, as we know that the domain of x includes all real numbers, the (x+1)^2 part of the function can become infinitely large, mitigating the -3, such that y goes to infinity as x goes to infinity. This means that the range is simply $\displaystyle [-3, \infty)$

See if you can try the other problems from here. If you need more examples, go here: http://mathforum.org/library/drmath/...ain_range.html

Edit: I did some more for you:

10. $\displaystyle y=x^3$

The domain will be $\displaystyle (-\infty,\infty)$ and the range will also be $\displaystyle (-\infty,\infty)$. This is because you can cube any real number, so no x-value results in an undefined function, and the output of a cube results in the same sign as you started with (if you cube a negative, you get a negative), such that the entire spectrum of the real numbers can also be represented. You can see this with a graph of y=x^3: http://www.wolframalpha.com/input/?i=y%3Dx^3.

11. $\displaystyle y=\sqrt{x}$

We cannot find the square root of a negative number (as any number squared yields a positive number), so that means the domain of x is limited to the positive numbers. This means the domain is $\displaystyle [0,\infty)$, and that correspondingly, the range is also limited to $\displaystyle [0,\infty)$.