another function problem

moronatmath

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Feb 14, 2006
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Can I please get some help with this homework problem?

What is the domain of this function? (in interval notation)
f(x)=(x+20)/(x^2-16)



What is the domain of this function? (in interval notation)
f(x)=sqrt(2x-2)
 
moronatmath said:
Can I please get some help with this homework problem?


The domain of a function refers to the input or x-values that will give you a real number for the value of the function. Cases that will not give a real value would include a 0 in the denominator or a negative value under the radical sign.

What is the domain of this function? (in interval notation)
f(x)=(x+20)/(x^2-16)

The denominator of this equation is x^2 - 16. Any value of x that will cause the denominator to become zero should be excluded from the domain. So set the denominator equal to zero and solve for x.

x^2 - 16 = 0
x^2 = 16
x = +/- 4

The domain of your function would be: All real numbers not equal to 4 or -4.

What is the domain of this function? (in interval notation)
f(x)=sqrt(2x-2)

The domain for this function can be determined by setting 2x - 2 </= to 0.

2x - 2 </= 0
2x </= 2
x </= 1

If x = 1, the number under the radical sign is 0. The sqrt(0) = 0, which is a real number. Any value of x less than 1 will result in a negative value under the radical sign, which will not give a real number answer.

The domain for this function: All real numbers less than or equal to 1.
 
How do I convert x = +/- 4 into interval notation?

Also, how do I convert "All real numbers less than or equal to 1" into interval notation?

Thank you very much for all your help so far!
 
You could read up a bit on Interval Notation.

\(\displaystyle (-\infty,-4)\,\bigcup\,(-4,4)\,\bigcup\,(4,\infty)\)

\(\displaystyle (-\infty,1]\)
 
You have (] as a set of brackets, TK; shame shame double shame... :roll:
 
"set of brackets"?? I thought we were writing interval notation?

Along with [genious]atmath, one of us needs to "read up a bit on Interval Notation". I'm not naming names, but...
 
moronatmath said:
What is the domain of this function? (in interval notation)
f(x)=sqrt(2x-2)
I don't understand the disagreement, but the correct answer to the above is:
\(\displaystyle x \ge 1\mbox{ in set notation } \quad [1,\infty )\).
 
Wow. To sum up:

tkhunny gets to read up on "Interval Notation", having confused it with "Set Notation".

tkhunny gets to read up on relying on other folks' work, having taken kngmn's work as correct.

kngmn gets to read up on "Domain", having solved for what was EXCLUDED, rather than what was INCLUDED. Even then, it wasn't quite right. Zero (0) is in the wrong group.

Dennis gets bonus points for knowing SOMETHING was wrong.

pka gets the Gold Medal for fixing the whole mess.

Ack!
 
tkhunny said:
You could read up a bit on Interval Notation.

\(\displaystyle (-\infty,-4)\,\bigcup\,(-4,4)\,\bigcup\,(4,\infty)\)

\(\displaystyle (-\infty,1]\)

So should I go with this anwser?

pka said:
moronatmath said:
What is the domain of this function? (in interval notation)
f(x)=sqrt(2x-2)
I don't understand the disagreement, but the correct answer to the above is:
\(\displaystyle x \ge 1\mbox{ in set notation } \quad [1,\infty )\).
In set notation? What about in interval notation?

Tk, what do you think?
 
That is both set notation and interval notation.
To the mathematician there is no difference: \(\displaystyle \L
\{ x:x \ge 1\} = [1,\infty )\)

The equals means that they are the same.
 
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