Functions

cole92

Junior Member
Joined
Mar 30, 2006
Messages
65
I was out for a lesson in my Pre Cal class due to a Dr. appt and I am a lesson behind now. I am working with functions and need a lot of help please. There are about 4 types of questions asked on my homework (each section with about 10 problems). If i could get some help on one from each I would be greatly appreciative so that I can do the rest. I hate not understanding things, but I just don't know how to teach myself math, and my book isn't very good at teaching...

First Style: Find Domain and Range algebraically:

I know how to do ones like f(x)= Sq. Root (x-1) by making it be = or > than 0...but I am having issues on ones without radicals. For example, 2x^2 + 3. By graphing I know the D/R are ( - infinity, infinity ) and [3, infinity ) respectively, but how do I do it algebraically?

Second Style: Use a utility to graph the piecewise-defined function.

f(x) = 2x + 3 when x < 0
f(x) = 3 - x when x > or = 0

For one, I am not sure what a piecewise function is because my teacher never taught me, and if I was supposed to learn them in Algebra 2 I didn't then either....I'm sorry that I have no work for that problem but frankly, I just have no idea

Third Style: Determine algebraically whether a function is even, odd, or neither:

Ex. t^2 + 2t - 3
since is is ^2 I would think it is even, but I am probably incorrect, as I don't know if you can judge it by that. Do I need to factor? (t + 3)(t - 1)? I tried that, but I don't think it gets me anywhere. Any help would be great.

One more!! Style four: Fidn the coordinates of a second point on the graph of a function f if the given point is on the graph and the function is (a) even and (b) odd:

Ex. (-3/2 , 4)
Unfortunately, this is like Style 2 for me. I have never even seen a question of this style (or maybe it is the wording that confuses me), thus I have no idea of where to even start. By "finding another point" I want to use point-slope or something, but there is no slope. I am just completely lost.


I am sorry for a lack-of-work, but I just don't know how to do any of these. I would like to figure out how so that I can get caught back up in class, any help would be greatly appreciated and I will check back soon!

THANKS IN ADVANCE TO ANYONE WHO CAN HELP A LOST STUDENT!! :]
 
cole92 said:
First Style: Find Domain and Range algebraically:

....but how do I do it algebraically?
You find the domain by finding the "problems" (such as negatives inside square roots) and setting the domain to be "everything else". For polynomials, there are no problems! This should be sufficient for "finding" the domain of a polynomial.

As for the range, you pretty much always do this from the domain and the graph, at this level of study.

cole92 said:
Second Style: Use a utility to graph the piecewise-defined function.
You'll have to check your owners manual for how to restrict the inputs on the functions. But I believe you just do Y1 as the first piece of the function (with some inequality restricting the display) and Y2 as the second piece.

cole92 said:
Third Style: Determine algebraically whether a function is even, odd, or neither:
To learn how to show "algebraically" whether a function is odd, even, or neither, try some online lessons. The process is pretty straightforward, but read through the worked examples, in order to get a feel for it.

cole92 said:
Style four: Fidn the coordinates of a second point on the graph of a function f if the given point is on the graph and the function is (a) even and (b) odd:
If a function is even (you'll learn in the lessons you study), then f(-x) = f(x). So, for instance, if the point (x, f(x)) = (2, 1) is on the graph, then the point (-x, f(-x)) = (-x, f(x)) = (-2, 1) will also be. If a function is odd, then f(-x) = -f(x), so, if (x, f(x)) = (2, 1) is on the graph, then (-x, -f(x)) = (-2, -1) will be, too.

Hope that helps! :D

Eliz.
 
Thanks that was awesome! I wasn't grasping the f(x) = f(-x) type of concepts for even odd and neither. I am doing the rest of my problems right now and so far no issues! Thanks a ton!
 
Top