Help me with Pre-Calc.

[Guest]

That is obviously Dave Sinkula so go back to http://cboard.cprogramming.com/index.php? and stay there.

[Sin]

Domain and Range of a Function: Domain and Range

Thanks for the link. The first persons post in the left link is pretty much what I am trying to ask. Il use this when I study tonight. My test is tomorro. So wish me luck after I study.

I think through after school lessons, this forum , the internet lessons, and my text book; I can safely say that I might be able to learn how to do this.

~Thanks everyone~

[leas5040]

Simply put: The domain of x is all reals unless there is a value(s) that make x undefined, or where there are no x values.

Take this set:

f(x)={ 2x, x<4}
{ 3x, x>4}

Notice that there is no value at x=4!! So the domain of x is:
{x|x R, x!=4} or all reals except at x=4.

Range works the same way: If there are undefined values, then the range is affected. Otherwise, it is all real numbers as well.

[Alvaro]

Range works the same way: If there are undefined values, then the range is affected. Otherwise, it is all real numbers as well.

Hummm... nope! Range is more complicated. The range of f(x)=x^2 is not all reals.

[Sin]

Range works the same way: If there are undefined values, then the range is affected. Otherwise, it is all real numbers as well.

Hummm... nope! Range is more complicated. The range of f(x)=x^2 is not all reals.

The range is easily found with the caculator for x^2.

Oh and by the way I was able to study enough to understand how to find the domain and range.

for this:
f(x)= {2x, X < 2}
{3x, X >= 2}

The domain would be all reals.
Its almost like common sense when finding the domain for that one. The calculator can also aid to find the domain.

f(x)= 1/(X+2)(X-1)
the domain would be every real number but not equal to -2 and 1.
Easy.

and for F(x)= 1/Square root of (x+1)
The domain would be X >= -1.
Thanks everyone for the help........

[Alvaro]

The range is easily found with the caculator for x^2.

How would you do that with a calculator?

f(x)= 1/(X+2)(X-1)
the domain would be every real number but not equal to -2 and 1.
Easy.

Yes, except you are missing some parenthesis in that expression. It should be 1/((X+2)(X-1)).

and for F(x)= 1/Square root of (x+1)
The domain would be X >= -1.

Here you got the `X' and `x' mixed up (Math, like C++, is usually case-sensitive), and x=-1 is not in the domain.

[leas5040]

Hummm... nope! Range is more complicated. The range of f(x)=x^2 is not all reals.

Sure, because y<0 is undefined or isn't on the interval. So the range is [0, infinity) or non-negative numbers.

[Sin]

Here you got the `X' and `x' mixed up (Math, like C++, is usually case-sensitive), and x=-1 is not in the domain.

x >= -1 is not in the domain?

[leas5040]

It should be all reals except x=-1, because that makes the denominator sqrt(0), which is anti-good.

[Alvaro]

Here you got the `X' and `x' mixed up (Math, like C++, is usually case-sensitive), and x=-1 is not in the domain.

x >= -1 is not in the domain?

That's not what I said. I said that x=1 is not in the domain. The domain is {x>1}.