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  #1  
Old April 1st, 2004, 10:28 PM
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Default Math Question

Allright, one of you geniuses help me to understand this.

I am getting back into school after being out for a few years. Taking a Trig class right now. This question really isn't trig though. More of a review of algebra and geometry before jumping into the trig part of the class. It's probably something I would understand if I had taken algebra more recently. But I don't want to sound stupid and ask the teacher.

Question is find the domain and range of
f(x)=sqrt(x+4) (The square root of x+4)

The domain is [-4,infinty)
The correct answer for the range is [0,infinity)

My question is, why isn't the range all real numbers? (-infinity,infinity) In class the teacher worked the problem and he said plug in any domain value for x and see what you get for f(x) is never less then zero.

But isn't there always a positive AND negative number for any square root? If x is 0 for example you get f(0)=sqrt(4) Isn't that 2 or -2? Can someone explain why you don't count the -2?

Geoschmo
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  #2  
Old April 1st, 2004, 10:32 PM
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Default Re: Math Question

You can not have numbers less than -4 in the domain, as then you would have the square root of a negative number, which is impossible in normal math. This leads to being unable to have a negative number in the range, as there is no way to get a negative number by taking a square root. Any value in the domain leads to a number that is either 0 or positive.

When you have y ^ 2 = x, y can be positive or negative because either a positive or negative value squared can lead to x. But when you have y = x ^ (1/2), you do not have negative numbers, as this is not at all the case of y ^ 2 = x. In the square root case (ignoring complex math, which you are not doing), you are limited to taking square roots of positive numbers (or 0), and square roots are always positive.

[ April 01, 2004, 20:35: Message edited by: Imperator Fyron ]
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Old April 1st, 2004, 10:33 PM
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Default Re: Math Question

In math class, sqrt(x) is defined as "The positive square root of x."
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Old April 1st, 2004, 10:34 PM

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Default Re: Math Question

I got a headache of just reading this thread.
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Old April 1st, 2004, 10:47 PM

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Default Re: Math Question

Why do I have the picture in my head of Scotty speaking to a modern day computer (they went back in time) "Computer?"..."Computer on"...then he had to do it by hand, just think when we can just ask a computer these tedious questions and can focus on the "ideas" yea right, thats the ticket...
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Old April 1st, 2004, 10:55 PM
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Default Re: Math Question

Ok, so my problem was simple I guess. I forgot the true mathematical definition of square root. I thought that the sqrt(x²)=x, but that is incorrect. Actually the sqrt(x²)=|x|
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Old April 2nd, 2004, 06:29 AM
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Default Re: Math Question

You guys in five short Posts or so just totally ruined any desire I might have had to return to school.
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Old April 2nd, 2004, 08:33 AM
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Default Re: Math Question

Quote:
Originally posted by Atrocities:
You guys in five short Posts or so just totally ruined any desire I might have had to return to school.
Atrocities, you are killing me old boy. Wait for me to get my hands on a camera and I'll show you the most painful math you will ever see.

(I will be back with a picture in a few days; I promise you)
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Old April 2nd, 2004, 04:31 PM
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Default Re: Math Question

I might have to take another look at my math books to be able to do trig again, but from what I remember about trig and algebra they were easier when I went back to school to get my degree.
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Old April 2nd, 2004, 04:33 PM

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Default Re: Math Question

Quote:
Originally posted by Atrocities:
You guys in five short Posts or so just totally ruined any desire I might have had to return to school.
Should I post some of my old calc homework and really kill any desire?

Or maybe some statatics..

SD = sqrt((sum of y deviations - y average)^2)/n)
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