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)
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)
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.
Do not insult my intellegence. I know that you still remember some things.
Why did you wait untill the last two hours.and this is our first hw which is due tonight in 2 hours.....
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 $\displaystyle y\geq -1$.1) 11x^2 - 1
$\displaystyle \sqrt{x+16}$.2) Sqare root of X + 16
We require that the $\displaystyle x+16\geq 0$ otherwise we get a square root of a negative number which is bad. Thus, $\displaystyle x\geq -16$. 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 $\displaystyle y\geq 0$.
Now you try the other ones. And I hope you try them because otherwise .
No they are not wrong. They need to be written differently.
$\displaystyle x\geq -16$ means $\displaystyle x\in [-16,\infty)$.
$\displaystyle y\geq 0$ means $\displaystyle y\in [0,\infty)$.
$\displaystyle 1<x<2$ means $\displaystyle x\in (1,2)$.
$\displaystyle y<0$ means $\displaystyle y\in (-\infty,0)$.
Now take the answers I explained and write them in this notation.
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 $\displaystyle y = \sqrt {x} + 16$
what are the restrictions (if any) on the $\displaystyle x$'s?
what are the restrictions (if any) on the $\displaystyle y$'s?
Please do not delete your questions, even after they have been answered.
note that $\displaystyle (x + 19)(x - 18) = x^2 + x - 342$
TPH showed you how to deal with quadratics already. try working on this one and see where you get
well, if x has restrictions, would it be so strange for y to have some as well?
$\displaystyle y = \sqrt {x} + 16$
the smallest possible value for x is 0 as you said, which means, the minimum value for y would be:
$\displaystyle y = \sqrt {0} + 16 = 16$
so $\displaystyle y \geq 16$
how would you write that in interval notation?