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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

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

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

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

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,115

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)*

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