| Discussion: | All Topics in PreCalculus for Properties of Rational Functions |
| Topic: | Need some quick help |
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| Subject: | RE: Need some quick help |
| Author: | ycc |
| Date: | Jul 31 2004 |
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function. To say a function, we have to tell clearly what are the domain and
co-domain. (i.e. the region for the starting value and the region for the
ending value.)
A function means that for any value x in the domain, there is one and only one
value y in the codomain such that f(x)=y.
(Please check your boook if you are still unsure the definition.)
Please see my responses inserted.
On Jul 31 2004, Tryin wrote:
> Hey Folks,
I have just went out and purchased some software to
> help educate myself with Algabra. However, I do need some quick
> help on the following problems. If anyone can help me, It would be
> deeply appreciated. I'm trying to help my kid but I am not doing a
> good job at all. Thanks
I assume the domain and co-domain are the set of real numbers.
1. Which of the following are
> functions? Explain your reasoning for a, b, and c.
a. f(x) = 2
> if x>1
f(x) = -1 otherwise
YES. Every x has one and only one corresponding real value.
b. f(x) = 5 if x>0
> f(x) = -5 if x<0
f(x) = 5 or -5 if x = 0
NO. There are two values for f(0), which is not allowed. (Every x should have
only one corresponding value f(x).)
c. f(x) =
> 10/x
NO. f(0)=10/0 is not a real number. (Every x should have a corresponding value
f(x).)
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