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AR15.COM
3/14/2010 10:21:26 PM EDT
Hey, I am studying for the GMAT, and I am currently doing Domain/Range of function stuff....and I really forget how to do this stuff. I know the domain can't be things like a 0 in the denominator, a negative in a square root, and a negative absolute value (i think...), but I think I might be missing something.

Also, I totally forgot how to find the range.

An example from each section of my review questions:

f(x) = (4x-1)/(x-3)

f(x) = |x-6| -3

f(x) = 2 - sqroot(x-5)

f(x) = x^2 + 8x +11


Any help would be greatly appreciated. The book I have (from EZ solutions) doesn't do that good of a job covering this.
3/14/2010 11:01:44 PM EDT
[#1]
anyone?
3/14/2010 11:08:44 PM EDT
[#2]
Absolute value is always +.

The absolute value of -2 ( I -2 I )is 2.

Look at this
3/14/2010 11:33:52 PM EDT
[#3]
Quoted:
Hey, I am studying for the GMAT, and I am currently doing Domain/Range of function stuff....and I really forget how to do this stuff. I know the domain can't be things like a 0 in the denominator, a negative in a square root, and a negative absolute value (i think...), but I think I might be missing something.

Also, I totally forgot how to find the range.

An example from each section of my review questions:

f(x) = (4x-1)/(x-3)

f(x) = |x-6| -3

f(x) = 2 - sqroot(x-5)

f(x) = x^2 + 8x +11


Any help would be greatly appreciated. The book I have (from EZ solutions) doesn't do that good of a job covering this.


So wait...  Are you just trying to find just the range?  Or do you need help solving them too?

I can probably help you some.
3/14/2010 11:35:14 PM EDT
[#4]
Quoted:
Quoted:
Hey, I am studying for the GMAT, and I am currently doing Domain/Range of function stuff....and I really forget how to do this stuff. I know the domain can't be things like a 0 in the denominator, a negative in a square root, and a negative absolute value (i think...), but I think I might be missing something.

Also, I totally forgot how to find the range.

An example from each section of my review questions:

f(x) = (4x-1)/(x-3)

f(x) = |x-6| -3

f(x) = 2 - sqroot(x-5)

f(x) = x^2 + 8x +11


Any help would be greatly appreciated. The book I have (from EZ solutions) doesn't do that good of a job covering this.


So wait...  Are you just trying to find just the range?  Or do you need help solving them too?

I can probably help you some.


I think I have domains down...but not sure where to begin on the ranges
3/14/2010 11:53:44 PM EDT
[#5]
I'll see what the good folks on campus think in the morning.
3/15/2010 12:03:07 AM EDT
[#6]
Quoted:
Quoted:
Quoted:
Hey, I am studying for the GMAT, and I am currently doing Domain/Range of function stuff....and I really forget how to do this stuff. I know the domain can't be things like a 0 in the denominator, a negative in a square root, and a negative absolute value (i think...), but I think I might be missing something.

Also, I totally forgot how to find the range.

An example from each section of my review questions:

f(x) = (4x-1)/(x-3)

f(x) = |x-6| -3

f(x) = 2 - sqroot(x-5)

f(x) = x^2 + 8x +11


Any help would be greatly appreciated. The book I have (from EZ solutions) doesn't do that good of a job covering this.


So wait...  Are you just trying to find just the range?  Or do you need help solving them too?

I can probably help you some.


I think I have domains down...but not sure where to begin on the ranges


Well, we know that the range is all the points of the y-axis contained on the graph.  So for example, if you graph out your example number two, you would move it 6 units left and 3 down...  and the graph would open upwards...   So  D: (infinity, infinity) and R: [-3,infinity).

Man...  I'm raking my memory right now and can't remember too much, it's been a couple years since I took Calc. and I'm quickly realizing that I should probably start brushing up on it too cause I'll be taking my GMAT soon enough as well.


These are a lot easier to do if you have the actual graph to look at, but I don't believe that is the case...

Sorry, I thought I could help but I don't think it's going to turn out that way tonight.
3/15/2010 12:04:17 AM EDT
[#7]
I'll read through this thread after I have my "dinner".



In the mean time I suggest that you read through my sigline.

3/15/2010 12:09:41 AM EDT
[#8]
Couldn't you just solve for x and y and chart points working that way?


So for f(x)= (4x-1)/(x-3)

Let x=4 and solve for f(4), which would be 15..  so then you would have your first point of (4,15).[/span]  And then just keep charting enough until you're able to make a graph?




Edit:  I can't even make simple calculations right now...  I'm going to bed.
3/15/2010 5:48:52 AM EDT
[#9]
I seriously had dreams about these all night.
3/15/2010 8:11:11 AM EDT
[#10]
Anyone? I'm going to a math guy's office hours I know at 11.
3/15/2010 8:11:12 AM EDT
[#11]
Anyone? I'm going to a math guy's office hours I know at 11.
3/15/2010 8:21:51 AM EDT
[#12]
Lot easier if you just post the complete question from the practice exam.
3/15/2010 9:10:13 AM EDT
[#13]
a negative in a square root


Depends on the number system for the problems.

sqrt(-1) = i

i^2 = -1

i = 1angle 90 degrees (polar) = (0,1) Cartesian
3/15/2010 9:28:58 AM EDT
[#14]
Domain is anything left to right on the x axis. This can be anything from - infinity to infinity.

Range is the max and the min. of the Y axis and can run from - infinity to infinity

In your first example, the domain can be anything because there are no restrictions. So the answer would be {X|X = R} or (-infinity to infinity)

For an absolute value, you cannot have a negative absolute value

For square root you would have to set the discriminant greater than zero. and solve.

The quadratic has no restrictions.

Remember that f(x) is referring to Y. and whatever you plug in for the f(x) value, you would plug in for every value of x on the other side of the equation.
3/15/2010 9:42:44 AM EDT
[#15]
Quoted:
Lot easier if you just post the complete question from the practice exam.


Um, those were the complete questions.
3/15/2010 9:43:30 AM EDT
[#16]
Quoted:
Domain is anything left to right on the x axis. This can be anything from - infinity to infinity.

Range is the max and the min. of the Y axis and can run from - infinity to infinity

In your first example, the domain can be anything because there are no restrictions. So the answer would be {X|X = R} or (-infinity to infinity)

For an absolute value, you cannot have a negative absolute value

For square root you would have to set the discriminant greater than zero. and solve.

The quadratic has no restrictions.

Remember that f(x) is referring to Y. and whatever you plug in for the f(x) value, you would plug in for every value of x on the other side of the equation.



Thanks, man.
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