The domain of a function in interval notation

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The domain of the function k(y)=1/(2y+1) is determined by the condition that the denominator cannot equal zero. This means that 2y + 1 ≠ 0, leading to y ≠ -0.5. Therefore, the domain in interval notation is expressed as (-∞, -0.5) ∪ (-0.5, ∞). Understanding the domain is crucial for identifying all allowable values of y in the function. The discussion highlights the importance of recognizing restrictions in functions, particularly with fractions and radicals.
nightshade
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1. what is the domain of the function k(y)=1/(2y+1)? Express your answer in interval notation


2.



3. I find that 0.5 is a possibility in this function, but since i have not done functions at school, i do not really know much more of this question.
 
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There are generally only two restrictions you have to look for when determining domain. If there is a fraction, the denominator cannot equal 0, and if there is an even numbered radical, the inside cannot be negative.
 
nightshade said:
1. what is the domain of the function k(y)=1/(2y+1)? Express your answer in interval notation


2.



3. I find that 0.5 is a possibility in this function, but since i have not done functions at school, i do not really know much more of this question.

"0.5 is a possibility" for what? It certainly is not the domain because it is not a interval. Are you claiming that 0.5 is in the domain?
 
So what really is meant by the domain? Is it asking for all possible values of y?
 
I mean that 0.5 could be y
 
All possible values for x .
 
nightshade said:
So what really is meant by the domain? Is it asking for all possible values of y?
Good question! You can't solve a problem about the "domain" if you don't know what that word means. Did you consider looking the word up in your textbook?

The "domain" of a function, f(x), is the set of all allowable values lf x. The set of all possible values of the function, y if we have y= f(x), is the "range" of the function.
 
Nightshade,

the domain of f(x)=\frac{2}{x-1} for example is all real numbers except 1, because when x is 1 the denominator goes to 0, and the equation becomes undefined.

This can be notated as D: x≠1 or D:(-∞,1)\cup(1,∞)
 
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