functions, domains and like that

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kelicraig

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Nov 8, 2007
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Given a domain, what is the procedure to finding the algebraic expression.

The specific domain is (-infinity, 3].

I'd prefer to know the process and seek the answer myself. Thank you for any assistance. If possible email me because don't know if I'll get back to this website again.
 
Um. . .I'm not sure. Here are the instructions that came with this particular assignment:
"Find algebraic representation for a function with the following domain: (-infinity, 3]

(Where I spelled out 'infinity' there was the actual infinity symbol but I don't want to take the time to figure out how to change the keyboard character or font and stuff)

So, I don't know. Am I trying to take an interval and put it into inequality form? I haven't heard it put that way before. I'm just trying to find out how to do what the instruction said. If you want to just give me the answer, that'll work but I'd rather know what to do in case it comes up again. So. . .any thoughts, suggestions?
Thanks
Keli
 
kelicraig said:
Um. . .I'm not sure. Here are the instructions that came with this particular assignment:
"Find algebraic representation for a function with the following domain: (-infinity, 3]

(Where I spelled out 'infinity' there was the actual infinity symbol but I don't want to take the time to figure out how to change the keyboard character or font and stuff)

So, I don't know. Am I trying to take an interval and put it into inequality form? I haven't heard it put that way before. I'm just trying to find out how to do what the instruction said. If you want to just give me the answer, that'll work but I'd rather know what to do in case it comes up again. So. . .any thoughts, suggestions?
Thanks
Keli
Click to expand...

Think about "square-root" functions.

Square root of a negative number is "imaginary".

so \(\displaystyle sqrt{x-a}\) will have a domain of 'a to infnity" -

because if 'x' is less than 'a' - the function under square root (x-a) will become negative and cease to exist in real domain.
 
I'm still not getting it. I had the bit that it was going to be a square root in there somewhere but other than that I've just been plugging in random values into a TI-83 calculator and then seeing what the table looks like. NOT the way to do math but I'm not a rocket surgeon :? So I did know enough that I'm dealing with square roots. That will get me the negative infinity part of the domain. But I don't know how to stop the domain at positive three. I understand that the 'a' in your example can't be less than x because that would be the square root of a negative number and that's not a number. Maybe I do just need the answer but then need to know how it came to be. :wink:
 
If you visualize the graph of \(\displaystyle y = \sqrt{x}\), you can see that the domain of its function is \(\displaystyle [0,\infty)\). What can we do to \(\displaystyle y = \sqrt{x}\) so that the domain goes from \(\displaystyle (-\infty, 0]\)? Imagine what the graph would have to look like. If you figure this out, you could probably figure out your answer.
 
I thought the answer would have something to do with introduction a negative in there some where so I tried multipying the square root of x by a -1 and then I divided by -1 then I divided -1 by the square root of x. and on each of these variations I've also been trying to find the highest value of the domain which is 3. I'm still flailing. My thought was this: The square root of x times -1, that's the -infinity part of my domain which is all value < 0 so the I put a +3 and there's my other value. Didn't work. Now what?
 
\(\displaystyle y = -\sqrt{x}\)

That doesn't give you the domain of \(\displaystyle (-\infty, 0]\). If you tried x = -1, then there isn't even a y value (real) since you're still getting a square root of an imaginary number. But you're on the right track in applying the negative sign. Play with it some more. There aren't too many places where you can add a negative sign.

As for getting the 3 as your end of the domain, think about your transformations:
f(x + a) = shifts a units to the left of where f(x) was
f(x - a) = shifts a units to the right f(x) was
just be careful if there's a negative sign applied to x
 
y=Square Root (-x+3) !!!! I didn't try that because I thought I couldn't put a '-' under the square root. I am sooooo confused. I had tried ever variation of the square root sign, a negative, the variable x and a 3 both positive and negative. So you can see how my mind works (or doesn't work) when it comes to math. I have some intuitive abilities but just can't put the pieces together. Thanks for all the help. [/tex][/code]
 
kelicraig said:
I thought I couldn't put a '-' under the square root.
Click to expand...
You cannot have a negative value inside a square root, but you're quite welcome to have a negative sign inside! :D

Eliz.
 
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