You are not logged in.
Please have a look at Operations with Functions and tell me anything wrong or lacking, thanks.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
Offline
Looks mostly good, but in your 'But Not Always' section you've got an (x-1)² where there should be an (x-3)², and there's an 'ex' that looks like a typo.
I'd also say that if f(x) = (x-3)²/(x-3) then f(3) is undefined, and we can't just simplify the function to make it x-3. There would be a limit at that point, but not a value.
Why did the vector cross the road?
It wanted to be normal.
Offline
Thank you.
You are right about the "But Not Always" example ... if I can't think of a correct example I will abandon that section.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
Offline
How about f(x) = x + 1/x and g(x) = x
(fg)(x) = x²+1
???
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
Offline
Hi MathsisFun;
Some functions have a removable singularity.
Def: When the function is bounded in a neighbourhood around a singularity, the function can be redefined at the point to remove it; hence it is known as a removable singularity. In contrast, whe a function tends to infinity as z approaches 0; thus, it is not bounded and the singularity is not removable,
According to this and I found other sites that agree. Just google for removable singularity.
f(x) = sin(x) / x , f(0) is defined and is equal to 1 This is the example they all give for a removable singularity.
Last edited by bobbym (2009-10-08 11:17:40)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline