I have totally forgotten everything over the summer and this is our first hw which is due tonight in 2 hours.....

Find the Domain and Range of the function:

1) 11x^2 - 9

2) Square root of X + 16

Find the domain of the function:

3) (x+19)(x-18)

2. Do you know the "Vertical Line Test"? It's not really a technically brilliant plan, but it solve amny "function" problems. If you can see the graph, is there a vertical line that hits it twice? If not, it's a function.

Domain and Range - Well, first off, you can't have a function or a non-function without at at least two variables. This make the expression 11x^2 - 1 not really a candidate for Domain and Range, since it expresses no relationship. If we assume you mean y = 11x^2 - 1, then we can talk.

Domain is stuff that goes in.
Range is stuff that comes out.

It is often easier to hunt down what does NOT work.

If you really need help on all this for a quiz in the very near future, it is unlikely that this review will be of any value.

3. Originally Posted by redraider717
I have totally forgotten everything over the summer
Do not insult my intellegence. I know that you still remember some things.
and this is our first hw which is due tonight in 2 hours.....
Why did you wait untill the last two hours.

1) 11x^2 - 1
I start you off. The domain is all x. Because any value works. The range are the y values. Now this is a parabola and the turning point is at -b/2a. In this case the turning point occurs at -(0)/2(11)=0. This means x=0 is the turning point. And therefore the minimum point because 11, the leading coefficient, is positive and this is a parabola opening up. So the mimimum value is 11(0)^2 - 1 = -1. So -1 is the smallest value. Which means the range is $y\geq -1$.

2) Sqare root of X + 16
$\sqrt{x+16}$.
We require that the $x+16\geq 0$ otherwise we get a square root of a negative number which is bad. Thus, $x\geq -16$. Now the range is the output values. Note that the square root of a number always gives back a value which is non-negative (meaning 0 or positive). Thus, the range is $y\geq 0$.

Now you try the other ones. And I hope you try them because otherwise .

4. It says those are wrong. The answer is going to be something like [#,infinity)U(-infinity,-#] I think.

5. Originally Posted by redraider717
It says those are wrong. The answer is going to be something like [#,infinity)U(-infinity,-#] I think.
No they are not wrong. They need to be written differently.

$x\geq -16$ means $x\in [-16,\infty)$.

$y\geq 0$ means $y\in [0,\infty)$.

$1 means $x\in (1,2)$.

$y<0$ means $y\in (-\infty,0)$.

Now take the answers I explained and write them in this notation.

6. Still says they're wrong, the answer isn't supposed to have y < or anything in it, I also tried just [0,infinity), still wrong. Oh well.

7. oops first one is 11x^2 - 9 not 11x^2 - 1.

8. second one is square root of x only + 16 not square root of (x+16)

edit: sorry for the triple post

9. Originally Posted by redraider717
Still says they're wrong, the answer isn't supposed to have y < or anything in it, I also tried just [0,infinity), still wrong. Oh well.
did you not see that TPH gave you the interval interpretations?

Originally Posted by redraider717
oops first one is 11x^2 - 9 not 11x^2 - 1.
just look back at what TPH did and do the same

10. Finally figured out the first one. Now I just have to get the 2nd and 3rd.

11. Originally Posted by redraider717
second one is square root of x only + 16 not square root of (x+16)

edit: sorry for the triple post
Note that:

the domain is the set of all inputs (usually x-values) for which a function is defined

the range is the set of all outputs (usually y-values) for which a function is defined.

Now, look at $y = \sqrt {x} + 16$

what are the restrictions (if any) on the $x$'s?
what are the restrictions (if any) on the $y$'s?

12. X can't be negative and does y even have any restrictions?

edit: Ok I got the domain of it [0,infinity)

Originally Posted by redraider717

Find the domain of the function:

3) (x+19)(x-18)
note that $(x + 19)(x - 18) = x^2 + x - 342$

TPH showed you how to deal with quadratics already. try working on this one and see where you get

Originally Posted by redraider717
X can't be negative and does y even have any restrictions?

edit: Ok I got the domain of it [0,infinity)
well, if x has restrictions, would it be so strange for y to have some as well?

$y = \sqrt {x} + 16$

the smallest possible value for x is 0 as you said, which means, the minimum value for y would be:

$y = \sqrt {0} + 16 = 16$

so $y \geq 16$

how would you write that in interval notation?

14. Ah, [16, infinity)

15. Originally Posted by redraider717
Ah, [16, infinity)
yes, that's the range

now try the last problem

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